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Infima in the d.r.e. degrees
Annals of Pure and Applied North-Holland
207
Logic 62 (1993) 207-263
Infima in the d.r.e. degrees D. Kaddah* SPL,
113 Steuart St., San Francisco,
CA 94105, USA
Communicated by A. Nerode Received 3 December 1992
Abstract Kaddah, 207-263.
D.,
Infima
in the
d.r.e.
degrees,
Annals
of Pure
and
Applied
Logic
62 (1993)
This paper analyzes several properties of infima in D,, the n-r.e. degrees. We first show that, for every n > 1, there are n-r.e. degrees a, b, and c, and an (n + 1)-r.e. degree x such that a < x < b, c and, in D,, b A e = a. We also prove a related result, namely that there are two d.r.e. degrees that form a minimal pair in D,, for each n < w, but that do not form a minimal pair in D,. Next, we show that every low r.e. degree branches in the d.r.e. degrees. This result does not extend to the low, r.e. degrees. We also construct a non-low r.e. degree a such that every r.e. degree b c a branches in the d.r.e. degrees. Finally we prove that the nonbranching degrees are downward dense in the d.r.e. degrees.
1. Introduction For n 2 1, a set D c w is defined to be (n + 1)-r.e. if there is an r.e. set A and an n-r.e. set B so that D =A - B, where a 1-r.e. set is defined to be an r.e. set. A 2-r.e. set is also called a difieerence of r.e. sets, abbreviated d.r.e. Equivalently we can define an n-r.e. set D as one for which there is a recursive function f so that for every x, D(x) = limf(x, S The definition
s),
f(x,
0) = 0,
of an co-r.e. set is obtained
and
I{s:f(x,
by replacing
S) #tf(x,
s + l)}/ c 12.
the third condition
above
by I{s:f(x,
S) #f(x,
S + l)}[ Gx.
The set of all n-r.e. Turing degrees Turing degrees is denoted by D,.
is denoted
by D, and the set of all w-r.e.
Correspondence to: D. Kaddah, SPL, 113 Steuart St., San Francisco, * The author would like to thank S. Lempp for his many helpful selection of the author’s Ph. D. thesis. 016%0072/93/$06.00
0
1993 -
Elsevier
Science
Publishers
CA 94105, comments.
B.V. All rights reserved
USA. This paper
forms
a
208
D. Kaddah
The d.r.e.
sets, and more
generally
the difference
hierarchy,
were first studied
by Putnam [15] and Ershov [7]. Cooper [2] showed that R# D, by building a properly d.r.e. degree (i.e., a d.r.e. degree which contains no r.e. set). In fact, this result easily extends to show D, # D, for nz # II. Much recent work in this area has compared and contrasted the structures (R, G) and (D2, c). The elementary theories of these partial orders were shown to be undecidable by Harrington more,
and
Shelah
the .X,-elementary
[lo]
and
Slaman
equivalence
(unpublished),
of the partial
orders
respectively.
Further-
is immediate
from the
proof of the decidability of the &-theory of R. Differences between the partial orders first appeared at the &level. Arslanov [l] proved that 0’ does not have the anticupping property in D2, in contrast to the independent results of Cooper and Yates (unpublished) that 0’ does have this property in R. Cooper, Lempp, and Watson [5] extended Arslanov’s result to show that every nonrecursive d.r.e. degree cups to any high r.e. degree above it by a low d.r.e. degree. Then a Z2 difference was demonstrated by the Diamond Theorem (Downey [6]), which says that the diamond (and more generally the l-n-l) lattice can be embedded in D2 preserving 0 and 1. This contrasts with Lachlan’s Nondiamond Theorem for R. Jiang [ll] has claimed that Downey’s theorem can be extended by replacing 0’ with any high r.e. degree. Slaman (unpublished) further sharpened the comparison between the partial orders by showing that R is not a Z,-substructure of DZ. One of the most basic results in r.e. degree theory is the Sacks Density Theorem [la], and researchers naturally tried to extend this to Dz. Partial positive results included Lachlan’s unpublished proof that the d.r.e. degrees are downward dense and Arslanov’s proof [l] that there is a properly d.r.e. degree above every incomplete r.e. degree. Cooper, Lempp, and Watson [5] extended these results to show that the properly d.r.e. degrees are dense among the r.e. degrees in D2. Then Cooper [3] proved the density of the low, d.r.e. degrees as a corollary to his splitting theorem. The general density problem was finally solved by Cooper, Harrington, Lachlan, Lempp, and Soare [4], who constructed a maximal incomplete d.r.e. degree. Jiang [ll] showed that this proof works above any low r.e. degree. In this paper we extend the study of the d.r.e. degrees by examining infima in Dz. Although the r.e. degrees form a proper upper semilattice, Lachlan proved that if two r.e. degrees have a meet in the Turing degrees, then they have the same meet in R. This is a corollary of his stronger result. Theorem (Lachlan [13]). Zf b, c are r.e. degrees and d is any degree sb then there is an r.e. degree a such that d s a < b, c. In Section
2 we prove
that this result
does not extend
to the n-r.e.
and ~c,
degrees:
Theorem 2.6. For all n > 1 there are n-r.e. degrees a, b, and c, and an (n + 1)-r.e. degreexsuchthata
209
lnjima in the d. r. e. degrees
Theorem 2.7. There are d.r.e. O
use the method
In Section
degrees b and c and an mr.e.
developed
3 we look at branching
by Downey in the d.r.e.
both the branching and the nonbranching degrees Fejer [9] respectively. In contrast to Fejer’s result Theorem
degree x such that
for his Diamond are dense, we prove;
3.2. If a is a low r.e. degree then there are d.r.e.
Theorem.
In R it is known
degrees.
by Slaman
that
[19] and
degrees b and c so that
a
of Theorem
3.2 we note that:
Corollary 3.3.
The lattice S, is embeddable
in D,.
This contrasts with the proof by Lachlan and Soare [12] that S, cannot be embedded in the r.e. degrees. A related result, the existence of a properly d.r.e. branching degree, was shown by Jiang [ll]. Because the nondensity of the d.r.e. degrees precludes any sort of full density result, we instead examine downward density for two properties. Theorem 3.2 provides one method for showing that the d.r.e. branching degrees are downard dense. In Section 4 we prove that the d.r.e. nonbranching degrees are downward dense. Theorem
4.1. In D2, for every a > 0 there is a nonbranching
From the partial degrees Theorem degrees.
evidence
are branching
of Theorems
in D,. However
3.2 and 3.6 it appears this is shown
degree b 6 a. that possibly
all r.e.
to be false in:
4.8. There is a low, r.e. degree b which is nonbranching
in the d.r.e.
Recently our Theorems 3.2, 4.1, and 4.8 were independently obtained by Downey and Shore. Our notation is standard and follows Soare [21]. Capital Greek letters are used to denote partial recursive functionals, with associated uses denoted by the corresponding lowercase Greek letters. In addition the notation @(X;x, s, t) represents the value of the functional @ on argument x at substage t of stage s. A parameter defined at stage s is assumed to have the same value at stage s + 1 unless explicitly redefined.
210
D. Kaddah
2. Infima in D, We begin with a lemma which is used extensively in the rest of this paper. proof of this lemma follows from the proof of Lachlan’s result that there nonrecursive Lemma
r.e. degree
below every properly
2.1. Let a be a d.r.e.
d.r.e.
The is a
degree.
degree and d a properly
d.r.e.
degree with a < d.
Then there is a d.r.e. degree e which is r.e. in a and which satifises a < e G d. Proof.
Fix d.r.e.
sets A E a and D Ed. Let B be the r.e. set
B={(x,s):x~D,
andx$D}.
Then B s,D and D is r.e. in B. If B +A, then D is r.e. in A. Otherwise, A -+ B @A s,. D and B @A is r.e. in A. In the first case let e = deg(D) and in the second let e = deg(B @A). 0 Note that Lemma 2.1 can easily be generalized to show that there is a nonrecursive n-r.e. degree below every properly (n + 1)-r.e. degree, for n 2 1. Instead of proving Theorems 2.6 and 2.7 directly, we prove the special case n = 2 and then indicate how the proof can be extended. Theorem 2.2. There are d.r.e. a
degrees a, b, and c and a 3-r.e. degree x so that
Proof. Note that we cannot have a = 0 by the generalized version of Lemma 2.1. We will build d.r.e. sets A, B, and C, a 3-r.e. set X, and functionals 4, &, and 0 satisfying the requirements: R:
X=T,(B@A),
P,:
X # @,z(A),
N,:
@JB @A)
where
G(C@A),
= @+(C @A)
= Dj j
Dj = O(A)
e = (i,j).
Here {Dj: j E o} is some standard listing of all d.r.e. sets. The strategy for N, is a slight variant of that introduced by Downey in his Diamond Theorem [6]. Instead of trying to preserve one side of the computation as in a typical r.e. minimal pair argument, we allow both sides to be injured at once. This happens when some Pi acts by putting x into X, y,(x) into B, and yZ(x) into C. If this causes a change, say z 4 Dj before Pi acts and z E Dj afterwards, then we force a disagreement or show Dj is not d.r.e. as follows. First remove x from X and y,(x) from B, thus restoring the (B @ A)-use to what it was when z $ Dj. In order to keep r, correct we must also put the new yZ(x) into C @A. If all of N,‘s computations again agree on z, then remove y*(x) from C @ A and
Infima in the d.r.e.
degrees
211
put x into X. In order to keep 4 correct we must also put some number less than y,(x) into B. These actions restore the (C @ A)-use to what it was when z E Dj, but since Dj must be d.r.e. we have achieved our goal. Now suppose N, is below the infinite outcome of N,,. Then there may be N,,-expansionary stages between the stage s at which & acts and the stage t at which N, acts for the first time. The N,,-computations which are recognized at these intermediate stages may count on y,(x) remaining in B. Then both sides of such a computation will be injured when N, acts for the first time and, in addition, the B-side will be irrevocably injured. In order to keep 0,. correct, at stage t we will require N, to put s into A instead of putting y&x) into C. Then r, is still correct since yZJx) > s, and 0,. is correct since 0,.(z’) >s if the N,.-computation on z’ was first recognized after stage s. If N, acts again at stage u to remove s from A and put some small number into B, then the C @A-side of N,,‘s computation on z’ is restored. Finally, note that if N,, first recognizes a computation on w at some stage between stages t and u, then e,.(w) >s, so that 0 is correct on w when s is removed from A. The construction takes place on the tree 2’“, where we interpret 0 as the infinite and 1 as the finite outcome. To node (Y we assign requirement N, if la1 = 2e and P, if Ial = 2e + 1. Each node a: working on N, builds its own version of the functional 0,. At stage s + 1 of the construction we will define the approximation to the true path 6s+l, where I&+ll
m(a; s) = max{l(ct; t): t
stage if it is an a-stage and I(cu, s) > m(a(, s).
Construction Stage 0. Initialize all nodes. Stage s + 1 > 0. First update each c as follows: For each x =Zs in increasing order
such that ~&)~, redefine y,(x) large. Next define a,+, by induction until either 16s+11= s or some strategy ends the induction. Let S,,, 10 = 0. Now suppose we
D. Kaddah
212
have
defined
S,,,
r (n) = a. Define
(Y+ = S,,,
1 (n + 1) using
the
appropriate
x,,
initialize
case below and execute the actions mentioned. If Q works on P,, then pick the case below which applies. (i) (Y has no uncancelled 5 > a: and end the stage. (ii) cy has uncancelled ix_(O). (iii) a has uncancelled yl(xn) Initialize
into
B, and
witness. witness
witness
y&,)
into
Pick a new large
x,, x,,
@_(A,; @_(A,;
C. For
xa)l
witness
= 0, and x, E X.
Let
all
my+=
xol)l = 0, and x, $ X. Put x, into X,
all y ?=x,
and
i = 1, 2 declare
X(Y)?.
all 5 > (Yand end the stage.
(iv) Otherwise. Let LY+= In cases (i) and (iii) above If (Yworks on N,, where e a-expansionary stages
a-( 1). we say CYacted. = (i, j) , then let t + 1 < u + 1 be the two most recent 1 (if these exist), and pick the first case below ‘which
applies. (i) Stage s + 1 is not cu-expansionary. Let a+ = a-( 1). (ii) Some (unique) P-strategy /3 with a-(O) c /3 and witness xg executed (iii) above at stage u + 1, (Y has not been initialized since, and there is a z < l(cu, U) such that Oj,s(Z) # Dj,u(Z). Then remove xp from X and Y~,~+~(x~) from B. For all y 2x6, declare yl(y)T if y > u and otherwise reset y,(y) = ~~,~+i(y). Put u + 1 into A and declare yZ( y)t for all y 3 x6. For all N-strategies (Y’ with a’-(O) s (Y U) declare e,,(y)t. Initialize all E > (Y and end the stage. and for all y sI(a’, Note that if &(z) = 0 then LYsatisfied by this action. (iii) w executed (ii) above at stage u + 1 and has not been initialized since. Then we must have 4,U(z) = 1. Remove t + 1 from A. For all y 3xB, declare yz(y)T if y >u and otherwise reset y2(y) = ~~++~(y). Put xB into X and yl,u+l(~a) - 1 into B. Declare y,(y)? for all y 3 xp. For all N-strategies (Y’ with a’-(O) E LYand for all y L I(&, t), declare e,,(y)T if y 2 I(&, U) and otherwise reset 8,(y) = ea,,t+l ( y ). Initialize all c > LYand end the stage. (iv) Otherwise. For each y < Z(a, s) in increasing order such that define 8,(y) large. Let c~+ = cu-( 0). In cases (ii) and (iii) we say (Y acted.
6(Y,S+l(y)t,
At the end of stage s + 1 initialize
all c
with 6s+l
Injima in the d.r.e. degrees
6 10 = 0. We show by induction
Proof. By definition at a = 6 1 rz and 6(n) a-stage infinitely
is defined.
s so that (Yis never
First suppose many
213
IZ is even,
Assume
cr-expansionary
stages,
holds
this is true for all 12’< n and fix the least
again initialized say cy works
on n that the lemma
after stage s.
on N, = Nci,jj. We may assume since
otherwise
the inductive
there
are
hypotheses
are satisfied. We show that if LYacts after stage s then it has the finite outcome and satisfies N,. So assume there is such a stage and fix t + 1 > s to be the least one. Then at the previous n-expansionary stage s’ + 1< t + 1 some P-strategy /3 acted,
and at stage
t + 1 there
stage t + 1, (Yexecutes
is some
(ii) and restores
z < Z((Y,s’) so that
If Oj,s,(Z) = 0 then, at the next a-expansionary and restores the use
So if Dj,sj(z) = 0 then,
Dj,t(~) # Dj,,Y(z). At
the use
stage u + 1 > t + 1, a executes
for all u > U, Qj,,(CU @A,;
(iii)
z) = 1 and Dj,v(z) = 0, while if
Dj,s,(Z) = 1 then, for all u > t, ~i,v(B” CBA,; Z) = 1 and Oj,v(Z) = 0. In either case 6 1 (n + 1) = a_(l), N, IS satisfied, and a never acts again. Now suppose n is odd, say (Yworks on PC. At the first a-stage t B s, cx picks a new witness X, in case (i) which is never cancelled. If a: never executes case (iii) then the inductive hypotheses are satisfied for n, so assume a: does so at stage u > t. Then a: is satisfied since X, E X and @,(A; xn)j, = 0. By assumption no higher priority strategy will later act and destroy a’s diagonalization, and (Y initializes all 5 > a! at stage U, so P, is satisfied. Also, at all a-stages TV> U, a is in case (iv) and 6 r (n + 1) = cu-(0). 0 Lemma 2.4. Let cu-(0) c 6, where LXworks on N,. Then N, is satisfied. Proof. By Lemma 2.3 we may fix a least a-stage s so that cr is never again initialized, and in addition we assume a never acts after stage s, since otherwise the proof of Lemma 2.3 shows that a-( 1) c 6. We show that every p =I o! keeps 0, properly defined. First note that, if a-( 1) c /3 of if /I is a P-strategy, then /3 does not affect the correctness of O,, in the latter case because otherwise & would have acted. Fix an N-strategy p and a P-strategy /I’, with a-(O) c P-(O) E P’, so that /I’ acts at stage t + 1, causing /3 to execute case (ii) at stage u+l>t+l.Ifz
On the other hand, if z 2 /(a, t) then t + l< f3_+i(z), allowing us to correct O,(z). If /3 executes case (iii) at stage ZJ+ 1 > u + 1 then,
and p puts t + 1 into A, for z < I(a, u), p restores
214
D. Kaddah
the use
If z 2 I((Y, u) then t + 1 < Oru,v+l(z) and p removes t + 1 from A, allowing us to correct O,(z). Thus in any case O,(z) is correct at stages t + 1, u + 1, and ?J + 1. Finally we show that V, (limS e,,,(z) exists). Fix z and a smallest stage s’ >S such that 13,(z) has been defined by stage s’. Note that if e,(z) is moved by /3 after stage s’, and later /3 is initialized, where c_z~(O)G /3, then /3 will not move 0,(z) at any future stage. By Lemma 2.3, every requirement on the true path acts at most finitely often. Thus, if there is some /lo a-(O) which moves 0,(z) after stage s’, then we may fix the highest priority such and the largest stage t > s’ at which it does so. Otherwise set t = s’. Then at the next a-expansionary stage u > t, 0,(z) is (re)defined for the last time. 0 Lemma
2.5. Requirement
R is satisfied.
First we note that the functionals c for i = 1, 2 are well-defined. The uses vi(x) are defined for the first time at stage x and are only redefined when some strategy acts on a witness y Sx. By Lemma 2.3 and the procedures for initialization and witness selection, each such witness y sx will be acted on at most finitely often. Thus r, and r, are total. We now show that whenever a strategy acts it keeps q and G correctly defined. If a P-strategy executes case (iii) at stage s + 1 by putting x into X, then it also puts the markers y,(x) into the appropriate sets, allowing us to redefine them large (in particular larger than s + 1). If an iv-strategy then executes case (ii) at stage t + 1> s + 1 by removing x, then it also moves y,(x) back to Y~,~+~(x)and puts s + 1 into A, which allows us to redefine y2(x) large. If this same strategy executes case (iii) at stage u + 1 > t + 1 by putting x back into X, then it puts Y~,~+~(X)- 1 into B, removes s + 1 from A, and moves y*(x) back to its position stage t + 1. Thus the c are correctly defined at each stage. q Proof.
This concludes the proof of Theorem Using similar methods, Theorem 2.2:
2.2.
it is possible
0
to prove
the following
extensions
of
Theorem 2.6. For all n > 1 there are n-r.e. degrees a, b and c, and an (n + 1)-r.e. degreexsuchthata
Note that, as in Theorem 2.2, we cannot make a=0 in Theorem generalized version of Lemma 2.1. However we can make a = 0 in: Theorem 2.7. There are d.r.e. x
degrees b and c and a nonrecursive
2.6 by the
wr.e.
degree
Injima in the d.r.e.
The proofs
of these
theorems
only give a brief description Proof of Theorem change
we make
are similar
is to the actions
215
to that
of the necessary
2.6. We use the same
degrees
of Theorem
2.2, and so we
changes. setup
as for Theroem
of N-strategies,
where
extra
2.2. The
only
steps are needed.
Fix an Nci,jj-strategy a and assume some P-strategy 0 c a-(O) put x into X at stage s(0) + 1, forcing (Yto act at stage s(l) + 1, as in (ii) of Theorem 2.2. Assume also that Dj,,(o,(z) = 0 f Dj,,cr,(z) (th e other case is similar) and let s(0) + 1 < a-expansionary stages. Then LY executes the S(1) + 1 < * * - be consecutive following steps until there are no more cY-expansionary stages. Step 1. At stage s(1) + 1, put s(0) + 1 into A, and remove
yl,S(Oj+l(x) from B and
x from X, thus restoring the (B @ A)-use on z to what it was at stage s(0) and forcing Dj,s(*)(Z) = 0. Step 2. At stage s(2) + 1, remove s(0) + 1 from A, and put yl,S(Oj+l(x) into B and x into X, thus restoring the (C @ A)-use to what it was at stage s(1) and forcing Dj,s(3)(z) = l.
Step k > 1 odd. At stage s(k) + 1, put s(0) + restoring the (C @A)-use to what it was at 0. Step k > 2 even. At stage s(k) + 1, remove thus restoring the (C @A)-use to what
1 into A and remove x from X, thus stage s(2) and forcing Dj,s(k+r)(z) = s(0) + 1 from A and put x into X, it was at stage s(1) and forcing
Dj,s(k+l)(Z) = l.
If all IZ steps are (n + 1)-r.e. set, and earlier construction, yi in order to keep construct B to be a
executed, then (Y has forced a disagreement since Di is an thus there will be no more cu-expansionary stages. As in the a also resets (or declares t) the appropriate markers 8 and the higher priority strategies satisfied. Note that we actually 3-r.e. set and C to be a d.r.e. set, although A is n-r.e.
Proof of Theorem 2.7. Note that, by the generalized version of Lemma 2.1, we need only prove that there is no nonrecursive r.e. degree a below both b and c. Thus we will requirements:
build
d.r.e.
R:
X = T;(B),
P,:
X#{e},
N,:
pi
where
sets
B and
C and
an
w-r.e.
set X to satisfy
the
T,(C),
= @i(C) = Wj 3
y
is recursive
e = (i, j).
We use the same tree and assign nodes to here A cannot be used to reflect X-changes as still need to change the value of X(n) several each IZ. To (re)define the markers vi(n) for i =
requirements as before, although in the previous theorems. We will times so we assign use blocks to 1, 2, we search for a large unused
216
block entire
D. Kaddah
of numbers [x, x + n] in both B and C, set x(X) =x + n and reserve the block for these markers. A P-strategy (1: acts as before except that if
]cr] = 2n + 1 then LYpicks all witnesses greater than IZ. This will ensure there is enough room in the witness’s use block for a! and all higher priority N-strategies to act. Fix an N,-strategy a-(O)
(Y with ]m] = 2n, where
c p. Suppose
/3 acts at m-expansionary
into B, and y*(x) into C, and at the next
e = (i, j),
and a P-strategy
p, with
stage s(O) + 1 to put x into X, y,(x) a-expansionary
stage s(l)
+ 1 we have
x E wj,s(l) - wj,s(0). Strategy (Yremoves x from X and y,,s(0)+,(x) from B and puts yZ,so)+l(x) into C, thus restoring the B-use to what it was at stage s(O). Since M$ is an r.e. set, these will be no more a-expansionary stages. Next we examine a’s effect on higher priority N-strategies. Fix g so that g-(O) E (Yand E works on N,,, where e’ = (i’, j’). Assume s(l) + 1
at
c * * * c a,-(~)
= cy(m)
c qj=
a
s(k) + 1< s(k + 1) + 1 are consecutive most n, where stages and (yk acts at stage s(k) + 1 to create a disagreement
itself. If k is even,
cu,for
‘%kremoves x from X and yl,s(o)+,(x) - (k/2) from B and into C. If k is odd, then ak puts x into X and puts YZ,~(~)+&) - (k/2) yl,s(o)+l(x) - ((k + 1)/2) into B and removes yl,sclj+L(x) - ((k - 1)/2) from C. A use block of size at most n is needed to accomodate all of these strategies, and thus if /3 chooses a witness x > II, then the construction succeeds and X is an w-r.e.
set.
then
0
3. Branching
degrees
In this
the
and
next
section
we will
look
at branching
and
nonbranching
properties of the d.r.e. degrees. These properties have been thoroughly studied in the r.e. degrees, starting with the proof by Lachlan [13] (independently Yates also constructed a [22]) that 0 branches in R. In the same paper Lachlan nonrecursive r . e . branching degree. Next Fejer showed the density of the nonbranching r.e. degrees in [9] and the existence of an r.e. branching degree above any low r.e. degree in [8]. Finally Slaman [19] completed this analysis by proving that the r.e. branching degrees are dense. The situation in D2 is not so well-studied. We cannot hope to prove density for any property since the d.r.e. degrees themselves are not dense. Also, in contrast with R, care must be taken to specify the relevant degree structure when discussing infima. For example, by Lemma 2.2, it is possible to have two d.r.e.
Infima
in
the d.r.e.
217
degrees
degrees with a meet in Dz which does not carry over to D3. On the other hand any r.e. degree which is branching in R remains branching in D,. In [ll], Jiang proved
there
is a properly
d.r.e.
branching
degree,
and,
as his construction
follows that of the r.e. case, the meet may be taken in the Turing roundabout way we can also conclude that there is a properly d.r.e. degree
since there is a maximal
d.r.e.
degree.
In fact the maximal
[4] is maximal in D, and thus nonbranching in the o-r.e. In this section we prove two results about r.e. degrees
degrees. In a nonbranching d.r.e.
degree
degrees. which are branching
of in
the d.r.e. degrees. We begin by showing that every low r.e. degree a branches in D2. More specifically we show that a is the meet in the Turing degree of two incomparable d.r.e. degrees. This result was independently proved by Downey and Shore. Our proof draws on Fejer’s proof in [S], which in turn used a technique developed by Robinson [16] for working above a low r.e. degree. The following lemma is the basis for Robinson’s method. Lemma 3.1 (Soare such that for all j
[20]).
y fl {n: D, GA}
If A
. a low LS
r.e. set then there is a recursive function h
= Who.,fl {n: D,, GA},
M’jf~ {n: D, E A} finite 3
and
Who.,finite.
Here D,, is the finite set with canonical index n. We will briefly outline Robinson’s method below. Suppose we want to build a set B satisfying the requirement B # @=(A), where A is a given low r.e. set. The strategy selects a witness x and waits for a stage such that Qe(A; x)j, = 0. Then it tries to A-certify the use $,(A; x) by enumerating it into Wj, where D, = A r #,(A ; x) and the index i is given by the Recursion Theorem. We now enumerate Who., and A until either n enters Who., or some y < c$~(A; x) enters A. By Lemma 3.1, one of these must occur. In the first case the strategy acts by putting x into B, and in the second case it waits for a new stage such that @JA; x)l = 0 and tries again. In addition, if action is taken at stage s and later some y < $,(A; x, s) enters A, then we abandon witness x and start over. Again using Lemma 3.1 we see that the strategy acts at most finitely often and is eventually satisfied. This is the standard finite injury technique for working above a low r.e. degree. The next theorem contrasts with Fejer’s result that the nonbranching degrees are dense in R. Theorem 3.2. If a is a low r.e. degree then there are d.r.e. a
degress b and c so that
in the d.r.e. degrees using the method Theorem [6]. The same result can be
218
D. Kaddah
obtained as a corollary to Theorem 3.2 by noting that the lattice MS can be embedded in the low r.e. degrees. This embedding contrasts with the proof of Lachlan and Soare [12] that S, cannot be embedded in the r.e. degrees. Corollary 3.3. The lattice S, can be embedded Proof
of
Theorem
in the d.r.e. degrees.
3.2. Fix
an r.e. set A E a, along with some standard enumeration of A. In the construction we will define a ‘speeded-up’ version A of A and build d.r.e. sets B and C to satisfy the requirements PB:
B f @e(A),
P,“:
C # Qe(A),
N,:
Qe(B @A)
= Qe(C @A)
=f
total j
f = T(A).
The strategy for N, is roughly the standard one used in the r.e. case. We measure the length of agreement [(e, s) and allow at most one number into either B or C at expansionary stages, keeping the restraint high otherwise. A positive requirement which assumes the finite outcome of all higher priority negative requirements uses the previous described strategy for working above a low set. Now suppose P,” wants to diagonalize on witness x at stage s and assumes the infinite outcome of some higher priority Ni-strategy. Instead of asking for A-certification of the use $,(A; x), P,” asks whether A, r s is correct; thus we try to A-certify the uses on all of Ni’S current computations also. In case certification is received, we enumerate x into B and restrain C r s. If later some y < s enters A then we may need to remove x from B in order to restore Ni’s computations, and P,” must try to diagonalize on a new witness. In case A-certification is not received we recalculate l(i, s) and repeat the stage. It is essential that the stage be repeated, since otherwise P,” might almost always be refused A-certification and yet still fail to satisfy the requirement. If P,” is almost always refused A-certification, then by repeating the stage we will be able to prove that Ni does not have the infinite outcome. Using Lemma 3.1, we again see that P,” will act at most finitely often to put a witness into B and will eventually be satisfied. The construction takes place on the tree 2’“. We assign node a to P,” if lal=3e,toP,CifJ(yl=3e+l,andtoN,ifIcul=3e+2.Stagesisdividedintoa varying number of substages t 6 s. Let p(s) be the last substage of stage s. We define 6: to be the path followed at substage t of stage s and let 6, = @‘. Substage t of stage s is an a-substage if IXc of. Stage s is an a-stage if a G 6,. When it causes no confusion, we use a variable without the (sub)stage identifier to denote its current value. If t is a substage of s, then we use the notation X, to denote the value of X at the end of substage t. At any stage in the construction we say a node (Y working on P,” requires attention if ~,f, or if x,1, x, 4 B, and @,(B @A;x,)J = 0. We say a: is satisfied if x,J E B and note that if & is satisfied and later initialized, then x, will be
lnfima
declared
undefined.
Node
in the d. r. e. degrees
a: may
need
to
build
219
several
finite
r.e.
sets
. . . ) where each index j(i) such that W,? = Wjcijis given by the recursion as previously described. In order to simplify notation we will refer to To initialize a during the construction means to the current set at any state as W 7 declare x,? and to abandon the current W OI.The definitions for a, working on P,” w;y, w;, Theorem
are similar. on a negative requirement N, define the length of For (YE Sfz: working agreement [(a, s + 1, t + 1) at substage t + 1 of stage s + 1 to be the largest x so that for all y sx
we have
@~(B@A;y,s+l,t)=@c(C@A;y,s+l,t) A, fu=A, u = max{+,(B Define
lu
and
where @A;y,
s + 1, t), $e(CG3A;y,
s + 1, t)}.
l(a, s + 1) = I(a, s + 1, p(s + 1)) and m(cu, s + 1) = max{l(a,
s’): s’ =GS and s’ is an a-stage}.
Substage t of stage s is a-expansionary if t is an a-substage and l(a, s, t) > m(a, s). Stage s is a-expansionary if s is an a-stage and I( a, s) > m( a, s). At each a-expansionary stage s we will define y(x, s) = s for each x s I(cu, s) such that y(x, s) is not defined. When cr is initialized we declare r,(n) and u,(n) to be undefined for all IZ. Now we discuss in more detail the reason and procedure for removing elements from B and C. Let LY and /3 be N,- and Ni-strategies respectively, where a-(O) E p. We outline the sequence of events which can lead to a problem: At stage s, /3 defines yB(x) = s. Later small b and c are put into B and C, so that the uses on both sides of /3’s computation on x are allowed to grow. We refer to the interval between the marker vrs(x) and the smaller of these new uses as a dangerous interval. Still later, at stage V, a number y in the dangerous interval enters A, so that both new uses are affected. At stage w, the computations reconverge with a different outcome. Note that since y was in the dangerous interval, /3 is not allowed to move yO(x). This is the usual sequence which leads to problems in r.e. branching constructions. Here, however, /3 can correct To(x) by removing 6 from B or c from C, thus restoring the earlier use and outcome on one side. Now suppose /3 removes b, and also that there was an n-expansionary stage s’ with u
D. Kaddah
220
To avoid this situation
we give the responsibility
for removing
lead strategy on each path, which we define to be the highest strategy on the path with infinite outcome, if such exists. numbers
enters
A, the lead strategy
elements
to the
priority negative Whenever some
resets B and C at its next expansionary
stage.
numbers unnecessarily from B and C but the above Note that if a positive strategy acts with correct problem is prevented. A-certification and is never injured, then this action will never be erased by a lead This strategy
strategy, To
may remove
so that all strategies
keep
recoverable
track
of this
A-expansionary
can still satisfy their requirements. process
each
stages.
Each
lead
strategy
A keeps
new A-expansionary
stage
a list
LA of
is added
to
the end of the list. In addition, if il resets B = B,. and C = C,. at stage s > s’, then we remove every v > s’ from LA. This is because Y is now a nonrecoverable stage, in the sense that we can never again reset B = BY. Whenever Iz is initialized we reset Ln = 0. Note that by the way we define markers y, the stages on Ln correspond exactly to the A-markers which are defined. The list is nothing more than an extra bookkeeping device, allowing us to refer directly to stages without having to calculate markers. Construction At stage 0 we initialize all nodes. At substage 0 of any stage s we enumerate one new element into a (and thus possibly into A) and proceed to substage 1. Stage s > 0, substage t > 0 Step 1. Let x be the smallest number which entered A after substage 0 of stage s - 1, if such exists. For each negative strategy (Yand each y such that yn(y)l > X, declare y,(y)?. Step 2. Calculate 6: r n for Iz s” from LA. If any numbers were put into B or C after stage sV, then remove them and fix /I of highest priority which enumerated such a number. Declare xpt and initialize all g > p. Note that after this step is completed we will have B 1 s” = B,. 1 s” and C 1 s” = C,.. r s” (although we may not have B 1 s = B,. r s and C 1 s = C,.. 1 s). Go to the next substage. Step 4. If no positive requirement on 6: requires attention then go to the next step. Otherwise fix (Y to be the highest priority such strategy and let it act as
lnfima in the d.r.e.
follows.
If x,t
then
pick a new witness
construction, initialize B (C, respectively),
221
degrees
larger
than
any number
yet used in the
all E > (Y, and go to the next step. If (Ywants to put X, into then let 0, =A 1s and W” = Wi. Put y1 into W” and
Who., and A (and thus A) until either some y w, and go to the next step. Step 5. This is the last step, so we have t = p(s). For each a(O) G 6, such that enumerate
(Yworks
on a negative
requirement
each y < 1( a, s) such that
ya(y)T,
strategy, then add s to L,. Initialize This ends the construction.
and rr was not initialized define
ya(y)
at stage s, and for
= s. If in addition
m is a lead
all E > 6,.
Verification We first note that, at stage s, the number of substages begun during Step 3 is bounded by s, and a new substage is begun in Step 4 only if some y
3.4. Each strategy on 6 is initialized finitely often. Each positive strategy on 6 acts finitely often and its requirement is satisfied. Thus 161= cc).
Proof. We prove the lemma by induction. Fix (Y so that 6 1 n = a E 6 and assume the lemma for all LY’c a. First we show (Yis initialized only finitely often. To see this, use the inductive hypotheses and the fact that (Yis on the true path to find a least stage s(O) so that every positive strategy CY’c a! has finished acting by after stage s(O), and at any stage s 2 s(O) we have 6, 2 (Y. If u is ever initialized s(O), then this must happen in Step 3 where we fix a highest priority /3 G (Y satisfying the conditions and initialize all E > p. Note that we also declare QT, so that p will act again to redefine x6 if given the chance. Thus we must have p
222
D. Kaddah
A-certification A-certification
and puts x, into B, and was incorrect. But then
IWfi,l = w
and
later
x,
is removed
because
the
yf~{(n:&~A}=@,
contradicting Lemma 3.1. So we may assume (Yacts only finitely often and fix a stage s(3) as(2) at which (Ydefines x, for the last time, say (Ydefines x, =x. Suppose @,JA; x)l = 0 but a never puts x into B. Then there is a stage s(4) >s(3) so that, at the final substage p(s) of every a-stage s >s(4), a asks for A-certification and is refused. But if this certification is refused at substage p(s) of stage s then the construction would proceed with substage p(s) + 1, contradicting the definition of p(s). Thus at some a-stage s 2 s(4), CYasks for and receives A-certification and puts x into B, contradicting our assumption that a never acts at a stage ss(3). This shows that P,” is satisfied. Finally we show that 6 1 (n + 1) is defined. If Edis a P-strategy then this is clear from the above arguments. If a is an N-strategy, (YG 6,, and s > layI, then 6,(n) is defined. Thus in either case 6 r (n + 1) is defined, so that, by induction, 161=@J. 0 Lemma
3.5. Each N, is satisfied.
Proof.
Fix a negative strategy a c 6 which works on N, and assume the lemma is true for all a’ c CY.Also suppose a’s hypotheses are satisfied so that it must build P,. By Lemma 3.4, we may fix a least stage s(O) after which (Yis never initialized. First we claim that a-(O) c 6. If this is false then there is a least a-stage s(l) as(O) so that cr^(O) $ & for any stage s as(l). Let m = m(a, s(l)) and note that for every a-stage s 2 s(l) we have m(a, s) = m. Since CY’Shypothesis are satisfied there must be a stage s(2) > s(l) so that for all substages t + 1 of all a-stages s ss(2) we have I(&, s, t + 1) > m. But then for all a-stages s >s(2) we have I(a, s) > m(a, s) = m, and thus there must be an a-expansionary stage s > s(l), contradicting our definition of s(l). Next we show that, for all n, lim, y,(~z, s) exists. Each marker y&n) is eventually defined in Step 5 and, after stage s(O), we only move the marker in Step 1 when some y < y&(n) enters A. Fix 12and an a-expansionary stage s(3) so that m(a; s(3)) > 12. Let /3 work on Pf, where a-(O) E /3 c 6 and i > s(3) is an index so that VX Vx Vs (C&(X; x, s)l = 0 with use &(X; x, s) = 0). Note that p has not yet acted at stage s(3) since i > s(3). By Lemma 3.4, we can find an a-expansionary stage s(4) > s(3) so that p acts for the final time at substage &s(4)) to put x6 into B. Then I(cu, s(4), &s(4))) >n and p’s Acertification is correct, so y&n, s(4)) = ya(n). Finally we show that, for all n, P,(n) is correct. Fix n and a least C-Zexpansionary stage s’ 2 s(0) such that vs 2s
(y&r, s)’ = y&r) = s’).
Injima in the d.r.e.
degrees
223
By the process for moving yLy, we know that r,(n) is correct at stage S’ and that A,. 1 s’ = A 1 s’. Let A be the lead strategy on 6, where possibly il = a. Note that if (Y# J., then a-expansionary
every a-stage is a &expansionary stage, stage is on Ln unless A removes it.
We first claim
that
B,, s B, and
similarly
for
and
C. Suppose
x E B,, - B. At the end of stage s’ all E > 6,, are initialized
in particular
each
this is false
and s’ is added
and
to LA.
Thus no lead strategy to the right of il can remove x from B. No lead strategy to the left of il will remove x from B since otherwise a: would be initialized. Node il will not remove x because by assumption A is fixed on s’. This argument establishes our claim and also shows that s’ is never removed from LA. Define an active substage to be one which ends in Step 3 or Step 5. We show by induction on active A-expansionary substages t of stages s 2,s’ (where t = p(s’) if s = s’), that there is an a-expansionary stage u E Lh.t such that s’-i
ZJGS,
but vfs
if t#p(s),
and either cDe(B@A;n,s,t)=
@e(B@A;n,v)=f(n,s’)
and
or ~~(C~A;n,s,t)=~~(C~A;n,v)=f(n,s’) (C@A),
lv=(CfT+A),
Iv.
and (2)
For each such t let p(t) be the smallest v satisfying the above, where the existence of p(t) is part of the inductive hypothesis. We induct only on active I-expansionary substages because the nonactive substages are those which end in Step 4, when some positive requirement asks for A-certification and is refused because a number is enumerated into A. If this number is small enough to injure both uses in (1) and (2) then possibly neither (1) nor (2) will be true at this nonactive substage. But in this case il will execute Step 3 and correct the uses at its next expansionary substage, and this substage will be active by definition. The inductive hypothesis holds at a-expansionary substage p(s’) because, by the definition of s’, both (1) and (2) hold at the start of substage p(s’) and at most one positive strategy acts during this substage to put a number into either B or C. Now fix an active A-expansionary substage t of stage s >s’ and assume that, in addition to being satisfied at substage p(s’), the inductive hypothesis is satisfied at all active &expansionary substages w where either w is a substage of stage s and w G t, or w is a substage of stage s^ and s’ < s^ dPct) are initialized. remove an element
D. Kaddah
224
below a-(O) will put a number start of substage t+, since
A-expansionary stage after t. First assume that no number
is some
smallest
z which
p(t) E Lh and, by assumption,
entered
A -A,
z 4; y(t),
and is used
B rCtj. Thus (1) is still true at the end of substage suppose
t+ ends
substage
in
Step
in Step 3. We know
so we do not remove 5,
so
that
t+ using p(t’) tf = p(s’).
any numbers
from
= p(t)
is not
Now (Y-
expansionary, then no number =ss+. Finally, assume that some number
Then
B p(p(s’~))1 POW’))
= BY 1 ,~P(s”)> G &+-I
1 ,4W)>.
The equality comes directly from (1) and the last containment fact that s” E LA. at the end of substage t+ we have
because
Blru,(s”jj r P(PW)
= BY 1 Ads”))
Step 3 is executed.
Again
A p(p(s”)) 1 P(P(s”))
= As,, r P(P(~))
Theorem
3.2.
from
the
= &+ 1 PMs”))
using (1) and the fact that s” E LA we get
Thus (1) still holds at the end of substage This establishes
comes
= At+ 1 P(P(s”)). t+ using ,u(t’) = p(p(s”)).
0
0
The next result proves that we can move out of the low r.e. degrees and still find a downward closed interval of r.e. degrees, all of which branch in the d.r.e. degrees. It also shows that Theorems 3.2 and 4.8 cannot be combined to single out the low r.e. degrees in D2 among the r.e. degrees by a definable predicate. Theorem 3.6. There is an r.e. degree branching in D2.
a 4 L1 such that every
r.e. degree
b s a is
lnjima in the d.r.e.
degrees
225
Proof. We will construct an r.e. set A so that deg(A) satisfies the theorem. We first discuss the requirement that A not be low and then add the branching requirements. In order to make
A’ #Tf4’ we will build
a total A-recursive
that, for all X, lim,Y @(A; x, s) exists and all requirements P,:
3x (lip
function
O(A)
so
P, are met:
@(A; x, s) # lim {e}(x, s)).
Note that if all requirements are met, then, by the Limit Lemma, A is not low. A P,-strategy (Y uses a simple version of the technique developed by Lempp and Slaman [14] for controlling the jumps of r.e. sets. First LYpicks a witness x and some large number z and starts setting @(A; x, p) = 1 with use B(A; x, p) = z for larger and larger p . If a ever finds a new u so that {e}(x, V) = 1, then it puts z into A, resets @(A; x, p) = 0 for all previous p, and picks a new larger 2, starting the process over for all new p. Thus cr will possibly put an infinite set into A but has quite a bit of freedom in choosing the members of this set. If cx is ever initialized then all current parameters, including x, are cancelled. In order to satisfy the branching conditions we meet the requirements: R,,;:
w = Qe(A)
j
deg(wi)
is branching
in D,.
An R,,,-strategy LYmeasures the length of agreement for M$ = @?(A). If cv has infinitary outcome then we must build d.r.e. sets D = D, and E = E, to satisfy the subrequirements: N,,i,j:
@j(D @ wl_)= ~j(E @ H$) =f
S,qi,j:
D # Qj(A),
S~i,j:
E # ~j(A).
Note that we are actually proving degrees, of two d.r.e. degrees.
that
total
deg(H$)
+ f = r(w),
is the infimum,
in the Turing
An N,,,i-strategy cx measures the length of agreement of the functionals in its hypothesis and defines more of r as this length increases. The strategy follows the pattern of the infimum strategies in Theorem 3.1, although in this case a will only believe computations which look correct according to higher priority P-strategies with infinite outcome. We delay discussion of the details. An S,,i,j-strategy CYpicks a witness and attempts the usual diagonalization, but first it tries to arrange that no higher priority P-strategy will ever force a higher priority N,,,j,-strategy /3 c cx to remove the witness. This is accomplished by suspending each higher priority requirement, starting with the lowest priority one and moving up. Generally we suspend a strategy by forbidding it to act or extend its definitions. In the construction this is accomplished by ending the stage whenever we reach a suspended strategy, except that the P-strategies are always allowed to extend their definition of 0. A P-strategy E with witness x is
226
D. Kaddah
suspended by setting @(A;x,s)= 0 for larger and larger s. Note that if p has higher priority than E, and E has just enumerated a number into A, then a can suspend E immediately but must wait to see if w changes before suspending p. This it may take several
stages before
cx is ready
Let a and p be as in the above paragraph
to act.
and let E be any P-strategy.
As in the
typical r.e. branching proof, if a: acts, then p can still be injured by ,$ < (Y because, although (Y knows about 5’s witnesses, it cannot stay out of the way of the effect of these
witnesses.
removing
from D and
elements
3.1, we define
For this reason
we may need
to correct
E to restore earlier computations.
a lead N,,,j-strategy
which is in charge
for /3 by
As in Theorem
of this procedure.
The construction takes place on the tree T = 2”“. Let Y be a node on T. We say that requirement P,or R,,i is satisfied OIZ Y if there is a node p c Y assigned to P,or R,,i, respectively. Requirement N,,i,j, Szi,j or S~i,j is satisfied on Y if there is an R,,i-strategy LYsuch that a-( 1) _c Y, or if there is a node p c Y assigned to N,,i,j, S,qi,j or S$j respectively. NOW fix some effective o-ordering of all requirements so that R,,iprecedes all of its subrequirements. For technical reasons we also require the first requirement on the list to be a P-strategy. Requirements are assigned to nodes on T by induction, where node Y E T is assigned the highest priority requirement which is not satisfied on Y. Stage s of the construction proceeds in substages t < s. We define Sf to be the path followed at substage t of stage s and let 6, = l_l, 6:. If a link is traveled we may have 1S:l> t, and if some requirement acts or is suspended we may have fx-stage if it is an 1S,l < s. Stage s is an a-stage if a G 6,. Stage s is a genuine a-stage and a is not suspended at stage s. We assume that at most one element enters any set w at each stage and that this element enters at the beginning of the stage. Let (Y be an R,,i-strategy and suppose at substage t of stage s + 1 we have Wi,,+l(X) = Qe(A;x, s + 1, t). We say w r (x + 1) is correct at s + 1, t if A, / &(A;x, Subrequirements still look correct
s + I, 4 =As+r,t
1 &XA;x,
s + 1, t>.
p of cx will use this definition to check whether at P-stages. We define the length of agreement
cy’s computations for (Yat substage
t of stage s + 1 to be 1(o, s + 1, t) = max{x:
K 1 x is correct
at s + 1, t}.
If stage s + 1 is a genuine a-stage then there is some unique t + 1 so that a, = 6::: and we define the length of agreement and maximum length of agreement at stage s + 1 to be l(a, s + 1) = I(n, s + 1, t), m(a;
s + 1) = max{l(cz,
and
s’): s’ Cs is a genuine
a-stage}.
If I(LY,s + 1) > m((~, s + 1) then stage s + 1 is o-expansionary. for each /3 ZI (Ywhich works on a subrequirement of R,,i.
Define
r(p)
= (Y
lnfima
On each
path
or node
priority N,,,,if-strategy on an &-subrequirement.
in
/3 we define
the d.r.e.
227
degrees
the lead (e, i)-strategy to be the highest
/3’ such that p’-(O) G /3, if such exists. Suppose p works If there is a lead (e, Q-strategy on the path p then we
use the notation J.(p) to designate this strategy. Otherwise Suppose that p is an SE;,j-strategy with currently defined p = 6;::.
we define L(p) = p. witness x(/3) and that
We say that /3 is ready to begin acting if Gj(A; x(/3), s + 1, t) = 0 A, r @j(A;X(P),
and
s + 1, t) =As+~,t 1 Gj(A;X(P),
s + 1, t).
If /3 is allowed to act and n(p) = /I then /3 enumerates x(/3) into D,,,, action is finished. If however 2.(p) # /?, then this becomes a multi-step First /3 establishes a link with top k(p). This link does not behave quite since the link is not traveled until all of p’s suspensions are in place. Note can form a unique longest chain n(p)-(o)
c pi-(o)
E * * * E/k-(o)
and the process. as usual that we
c p,
where each pi works on sone N,,i,j,-requirement. Strategy p begins by suspending all 5 with &( 0) G E c /3 at stage s + 1. Suppose that $’ is a P-strategy. If c-( 1) G /3, then p will be initialized if E ever acts. If E-(O) E /3, then 5 acted at stage s + 1 to put its current z(E) into A, so that z(g)? at the end of stage s + 1, and z(E) will not be redefined until the suspension on 5 is lifted. Also, as mentioned before, Pn only recognizes computations with ‘M/j--use which looks correct according to higher priority strategies. Thus the w-use on all Bncomputations recognize at the first P,-expansionary stage after stage s + 1 will be correct forever unless /3 is injured. At the next genuine /3,-expansionary stage, p suspends all strategies 5 such that pn_1c‘(O) 5 5 c on. This process is repeated until we reach L@?)-(O). For k < n, an argument similar to the one above shows that the w-use on all P,-computations is correct at the stage at which fik is suspended and remains correct unless p is injured. Finally, at the first L(P)-expansionary stage after A(/?)-(0) 1s suspended, we travel the link and p finishes the action by enumerating x(/3) into DXo,. Strategy p is then satisfied unless it is initialized. If p is initialized at any time during this process, or if the link is travelled, then all suspensions are cancelled and the link is destroyed. Also, whenever /3 is initialized we declare x(/3) undefined. The process for an SFi,j-Strategy is similar. Now let p be an N,,i,j-strategy and suppose that p = 6:::. Because of the infinite injury from P-strategies we must be careful in defining the length of agreement for /3. The three paragraphs following are devoted to building up these definitions. We drop the subscriptions on D,,,, and E,,,,. Before /l is allowed to recognize computations, we first make sure that A-use on the I&use for these computations looks correct when we reach p. In addition, if A(P) # /3 then p must coordinate its definitions with those of A(P) since the lead strategy may remove elements from D and E. In this case p will in effect recalculate all of n(p)‘s
228
D. Kaddah
computations
before
making
its own definitions.
Now suppose
Define U(P, X, S + 1) = max{ &(D CBWi; s + 1, t),
@j(E
@
M$;
X,
S
+
1, t)}
and let u = z&l, x, s + 1). First suppose A(p) = /3. If yB(x)l, or if there is no z
+ 1, u(P, x, s + 1)).
In this case we say that x has correct P-computations at s + 1 if w 1 k is correct at s + 1, t, and furthermore, if p is allowed to define the marker in the construction, then it sets yp(x) = k. Note that, using the basic assumption that uses are nondecreasing, we have y@(x) < yP(x + 1) if both yP(x) and yB(x + 1) are defined at stage s + 1. Now suppose A(p) # /3. In this case we say x has correct /?-computuitons at s + 1 if there is some smallest z so that 1. Y*&)L ’ u = 0, x, s + I), 2. K 1 ~~~~(2) is correct at s + 1, t, 3. if yA&z) was defined to its current value at stage s’ c s + 1, then it was the smallest A(/?)-marker defined or redefined at stage s’, and 1 u and Es, 1 u = Es+,,, 1 u. 4. & r 24.= ~,,I,, If the above conditions are satisfied then we say x is associated with z and s’ at s + 1. If yB(x) is undefined, x has correct P-computations and /3 is allowed to define the marker, then we set y@(x) = yh&z). The purpose of these conditions is to ensure that A(p) will not remove any numbers from D or E unless w. condition 3 ensures that if both sides of /3’s changes on y@(x). In addition, computations on x are injured because some number >ys(x) enters A, then one side will be restored to its state at a previous stage. For either lead or non-lead 6 we define the length of agreement and maximum length of agreement at stage s + 1 to be f(p, s + 1) = max{x:
Vy
m(/?, s + 1) = max{l(p, Stage s + 1 is P-expansionary 1). If /3 is ever initialized, arguments.
P-computations
s’): s’ C s is a genuine
if it is a genuine then we declare
at s + l)},
p-stage}.
P-stage and Z(/3, s + 1) > m(j?, s + r, and yg to be undefined on all
lnfima
Now
let
p
expansionary
be stage
an and
in the d.r.e.
N,,,,j-strategy suppose
such there
degrees
that
229
k(p) = p.
is a most
recent
Let
s + 1 be
P-expansionary
a pstage
s’
initialized at any stage v 3 s’. Further suppose that there is some least x E W&+, - w,,, and some y so that yfj(y)J >x. Then j3 wants to reset the branches and performs the following operation when allowed to do so. First,
for each y so that
yO(y)J >x,
declare
the marker
to be undefined.
Next, for each N,,i,j,-strategy p’ so that p-(O) E p’ and for each y so that yO(y)J >x, declare the marker to be undefined. Finally, if there is some P-marker
which is still defined,
this marker
was defined
from D,(p) and &J)
then fix the largest
to its present
position
which were enumerated
such, say yO(z), and suppose
at stage s”. Remove
all numbers
into these sets after stage s”.
Construction At stage 0 we initialize all nodes and define @(A; x, p) = 0 = 0(x, p) for all x and p = 0. Initialize all nodes at stage 0. At stage s + I, proceed with the steps below. Step 1. If t = 0, then define Si,, = 0. Otherwise suppose we have defined &+, = (Y. If t = s + 1 then go to Step 2. If t v(a) so that {e}(x, V) = 1. Then put z(a) into A. Declare z(a)f and set a+ = a-(O). (2D) Otherwise. Set oy+ = a-( 1). Case 3: IX is an Ne,i,j-strategy and A(a) = (Y. Pick the first subcase below which applies. (3A) Stage s + 1 is not a-expansionary. Then set (Y+ = a-( 1). (3B) (3C)
(Ywants to reset the branches. Let it do so and then set (Y+ = (u-( 1). Let (Y define its markers for all x < Z(a, s + 1). If cue(O) is
Otherwise.
not suspended, then set 1y+ = o-(O) _ Otherwise there is a link with top (Y and bottom /3, so travel the link and let 6 finish its action, set S,,, = 0 and initialize all 5 > p, and go to Step 2. Case 4: (Yis an N,,,,j-strategy and ii(m) # (Y. Pick the first subcase below which applies. (4A) Stage s + 1 is not a-expansionary. Then set my+= a-( 1). (4B) Otherwise. Let CYdefine its markers for all x < I(cu, s + 1). If a-(O) is not suspended then set LY+= a-(O). Otherwise let p be the longest N,,i,j,-strategy such that /3-(O) E (Y, suspend all E with p-(O) E g c (Yand set 6 s+l = (Y, and go to Step 2.
230
D. Kaddah
Case 5: (Yis an S$j-strategy. Then pick the first subcase below which applies. (5A) a is satisfied. Then set a+ = (Y-(O). (5B) x(a) is undefined. Choose a large witness x((u) and set IY+= a-( 1). (5C) a is ready to begin acting. Then let it do so. Set a,+, = cr and initialize all E > (Y. Go to Step 2. (5D) Otherwise. Then set (Y+= a-( 1). Case 6: (Yis an SEi,j-strategy. The subcases are analogous to those in Case 5. Step 2. Initialize
all E such that S,,, cL E. Finally, for each x and p = s + 1, 0 as follows. If there is no P-strategy (Y so that x =x(a), then set @(A; x, p) = 0 with use 0(x, p) = 0. Otherwise, fix the (unique) P-strategy a such that x = x(a) and choose the first below which applies. 1. Both z(a) and v(cr) are undefined and (Yis suspended or has not requested that z(a) be defined. Then set @(A; x, p) = 0 with use 0(x, p) = 0. 2. Both z(m) and v(a) are undefined. Then define z(a) large and v((u)=p, and set @(A; x, p) = 1 with use (3(x, p) = z(a). 3. Both ~(a() and v(a) are defined. Then set O(A;x, p) = 1 with use 0(X, p) = z(a). 4. Otherwise. We note that in this case (Yacted at stage s + 1, so that z(a)? but v((Y)~. For each u such that v((u) <.v 6p, set O(A;x, v) = 0 with use 0(x, V) = 0. Declare v((Y) to be undefined. For p
Verification
We define the true path 6 by induction. Let 6 10. Suppose we have defined 6 r n = a. Then let 6(n) be the least k E (0, 1) so that a-(k) G 6, for infinitely many s, if such k exists. Lemma 3.7. Let cx E 6. Then (Y is initialized only finitely often and there are infinitely many genuine a-stages. Zf it is an S-strategy then it acts only finitely often. Thus 161 =m. Proof. The proof is by induction. Assume 6 1 n = (Y, there are infinitely many genuine a-stages, each S-strategy E’ c a acts finitely often, and (Yis initialized only finitely often. Note that, if a is an S-strategy and Q:acts and is never again initialized, then a is satisfied forever. Thus by the inductive hypotheses, (Yacts only finitely often. We must show that 6 r (n + 1) = (Y+ is defined, there are infinitely many genuine a+-stages and (Y+ is initialized only finitely often. First note that, if a = S:,, and s + 1 is a genuine a-stage, then &I: is defined unless t = s + 1, or there is a link with top a’ E LY,or subcase (5C) or (6C) is executed. Now eventually s > Ial, and by induction subcases (5C) or (6C) are
Injima
in the d.r.e.
231
degrees
executed only finitely often by (Y. Thus we may fix a stage s > Jai so that (Ynever again acts via (5C) or (6C) and is never again initialized. Suppose there is a link with top 19 E (Yat genuine a-stage S’ > s. Then this link was established by some S-strategy /I 3 a at a previous stage. If this link is never destroyed then (Y+= a-( 1) c 6 or there is never another genuine a-stage, so assume the link is destroyed at stage s” > s’. When this happens all E > /3 are initialized. Thus, if there is a link with top (Y”c a at the next genuine a-stage >s”, then this link must have been in place before p established its link and must have bottom p’ cLp. Repeated applications of this argument show that there must be some least genuine a-stage u > S” so that at stage v there is no link in place with top sty. Then v is a genuine stage for 6, 1 (n + 1). Note that if 6, 1 (n + 1) = cr-( 1) then no link can be established with bottom /? 2 cum(O) until the next genuine cur‘(0)-stage. Thus (Y+exists, has infinitely many genuine stages, and is initialized only finitely often. q Lemma 3.8. O(A) is a total A-recursive each x. Each P, is satisfied.
function,
and lim, @(A; x, p) exists for
Proof. Fix x. Note that if x =x(a) at some stage and later a abandons x, then we will never again define x = x( cu’) for any a’. Now fix p also. If p = 0 then at all stages we have @(A; x, p) = f3(x, p) = 0. If p > 0 then @(A; x, p) and 0(x, p) are first defined at stages s + 1 =p, and these values are changed at most once, in Step 2.4. In addition, if these values are changed in Step 2.4 at stage s’ >p, then /3(x, p) is enumerated into A at this stage. Thus O(A) is a total A-recursive function. Next we show that lim, @(A; x, p) exists and each P, is satisfied. If there is no (Yso that x =x(a) cofinitely often, then there is some p’ so that, for all p >p’, we define @(A; x, p) = 0 = 6(x, p) in Step 2, and thus lim, @(A; x, p) = 0. So assume there is such an a. Then either Step 2.1 applies cofinitely often, or Step 2.3 applies cofinitely often, or a-(O) c 6 and (Yacts infinitely often. In the first case lim, @(A; x, p) = 0, but (Y cannot be on the true path since otherwise, by Lemma 3.7, we would eventually redefine 2((u) in Step 2.2. In the second case, lim, @(A; x, p) = 1 and lim, {e}(x, p) = 0. Finally, suppose cy works on P,, a-(O) c 6, and & acts infinitely often. Then note that lim, @(A; x, p) = 0 and there are infinitely many v so that {e}(x, v) = 1. Thus P, is satisfied in this case also. 0 Lemma 3.9. Suppose that the hypothesis for requirement requirements S,qi,j and SEi,j are satisfied. Proof.
R,,i is satisfied. Then
Let p c 6 work on SEi,j. We prove this case only since the other is similar. By Lemma 3.7, we may fix a least genuine P-stage s so that /3 is not initialized at
232
any
D. Kaddah
stage
s’ 2 s. Then
6 chooses
a new
final
witness
x = x(/3)
at stage
s. If
@j(A; x) = 0 then /3 will eventually act, so suppose that v + 1 is the least genuine P-stage >s at which /3 begins acting via (5C). Assume that p begins acting at substage t + 1 of stage u + 1 and that we have aj(A; x, v + 1, t) = 0 A,
/ u =&+,,r
1 u,
and where
u = ~j(A; X, u + 1, t).
We first show that A r u = A,,,,, v + 1 either
z(a)
is undefined,
z(a) is defined but is never a will injure U. In addition, since otherwise /3 will be shows that if A(p) = /3, requirement is satisfied. Now suppose A(P) # p. genuine P-stage since it Al’ + 1 > u + 1 be the stage
1 u. Note that for each P-strategy in which case it will be defined larger
enumerated into A because /3 initializes all E > p at stage initialized. So A 1 u = A,+l,t then /I finishes its action
(Yc /3, at than U, or
a^( 1) E p. Thus no such II + 1 and no (Y<= 6 acts r u. This argument also at stage u + 1 and the
Note that p is allowed to finish its action at the is not initialized and is on the true path. at which the link to /3 is travelled in subcase (3C). Y&Y) be the largest A(P)- marker defined at state V’ + 1. We claim that wl: 1 yAtaj(y) = Wi,v’+l 1 yX&y). If this claim is true, note all 5 > A(P) will never remove x from DZo,, and, since p again initializes
next Let Let that /3 at
stage u’ + 1, no other strategy will remove X, so that the lemma is proved. We now prove the claim. Note that stage 2r’ + 1 is ii(P)-expansionary and y = l(Q), n’ + 1) - 1. Thus, if n(p) = a;‘:‘,,, then by definition we have K r ykyncp,(y, U’ + 1) is correct A,.
r Li = A,,,,,,,
1 fi
where
at v’ + 1, t’,
and
~2= @,(A; yACB,(y, v’ + l), v’ + 1, t’).
Now the remarks above on P-strategies WC /3 also apply here. Each such 1y will either choose a new large z(a) > fi or it has a currently defined z(a) G fi but will not enumerate this number into A. Thus A 1 fi =A,,+l,t r fi and the claim is proved.
0
Suppose (Yc 6 works on R,,i and w = @,(A). Then from Lemma 3.7 and the definition of the length of agreement for (II we know a-(O) c 6. We now prove a similar fact for N-strategies and show that their functionals are total when necessary. Lemma 3.10. Let /3 c 6 be an Ne,i,j-strategy and suppose w = QC(A). If k(p) = /3 and /3-(O) c 6, then lim, Y~,~(x) exists for each x and w can compute this limit uniformly in x. Zf N,,i,j’s hypothesis is true then /3-(O) c 6, lim, yB,,(x) exists for each x, and l4$ can compute this limit uniformly in x. Proof. From the remark preceding the lemma we know r(p) has outcome 0 on 6. By Lemma 3.7 we know that there are infinitely many genuine P-stages and
Infima in the d.r.e.
that there is some smallest stages referred
stage so that p is never
to in the rest of the proof
be declared undefined can always compute
again initialized.
are larger
YP(x) is defined at substage t + 1 of stage position unless w changes below the marker
233
degrees
than this stage.
A(P)-expansionary for N,,i,i-strategies
P--(O) C 6. Suppose ;1(/3) = /? and p-( 0) c 6. We show that each P-marker assume
Note that,
all if
s + 1, then the marker stays at this position. In that case the marker will
at the following genuine the marker movemnts
way of contradiction,
We assume
this is false and fix the smallest
stage. Thus I4$ /I such that has a limit.
By
x so that lim, YO,,(x)
does not exist. Fix a stage s such that, for all ys the marker Y[<(x) is (re)defined if necessary. Let j’ >s be an index so that VX Vz (Qj(X;
z)l = 0 with use Qj(X;
z) = 0),
and suppose E c 6 works on Sz,,j,. Then note that 5 has not yet acted at stage s and in addition p-(O) G g because j’ > s. By Lemma 3.9 we may fix a least genuine P-expansionary stage s’ > s so that 5; finishes acting for the last time at stage s’, and the proof of the same lemma shows that w,,. 1 yB(n) = M$ 1 yO(x). Thus this marker is fixed forever at stage s’. Now suppose that h(P) = p and the hypothesis of requirement N,,i,j is satisfied. Then, by the assumption that wj = Gc(A), each x will eventually have correct P-computations at some genuine P-stage. Thus /3-(O) c 6. Finally suppose k(p) #/I and the hypothesis of requirement N,,,, is satisfied. From the above we know all A(p)- markers are eventually defined and reach a limit. Using Lemma 3.7, the fact that W, = @,,(A), and the assumption that the requirement’s hypothesis is satisfied, we see that, for each x, there is eventually a smallest z so that yO(x) is defined to be ynCB)(z) and both markers have reached a limit, so again /3-( 0) c 6. 0 Lemma
3.11. Every requirement N,,i,j is satisfied.
Proof. Let p c 6 be an N,,i,j-strategy. Assume the requirement’s hypothesis is satisfied and I4$ = Qc(A). By Lemma 3.7 there is some smallest stage after which /3 is never again initialized, so we assume all stages in the rest of the proof are larger. We drop the subscripts on D and E and let A = A(p) in the following. From Lemma 3.10 we know that p-(O) c 6 and that each marker eventually reaches a I+computable limit, Fix x and a least genuine P-stage s so that vv 2s
(Y&v(X)J = Y&)d2f
We must show that genuine P-substage. there are parameters y =x and s(O) = s in
Y&)).
h(x) =fa,.y(x). A ssume that substage t + 1 of stage s is a If il # /3 then, from the definition of correct P-computations, y and s(O) with which x is associated at s. If A = /3 then let the following.
234
D. Kaddah
First note that, using the definitions of stage s and of B-correct computations, we have m,,&r) = Ye and wiJ(0) 1 Y*(Y) = K r Y*(Y), so that A.never uses a stage
= Q 1 Y&) G D 1 Y&),
and
E,(o) r Y&) = E, / Y&) GE I Y/~(X). Suppose that A resets the branches at stage s” > s to a stage s’ < s”. By the above, s(O) ss’. Let m(z) be the smallest &marker which was defined or redefined at stage s’. From the conditions for resetting the branches we have (D @ I%),. 1 Y*,&) = (D @ W),, r Y*,&), (E @ J%),v 1 Y&Z) = (E @ K),, 1 m,&), In particular, G
and
Ye,&) = YA,&).
if A # /I and s’ c s then r 4%
x, s) = Q
r u(P, x, 81,
and
Es,, r u(/?, x, s) = Es r ~(6, x, s).
These facts are used in the inductions below. First suppose A.= p. We show by induction on genuine A-expansionary stages s’ >s that for each such s’ thee is a genuine A-expansionary stage u such that s
@
WJx,
v)=f,(x),
(3) 0” r @I, x, v) = Q, 1 u(P, x, u),
and
W,,, r Y~,~(z) = K,,, r y&r);
and
I%,, 1 YA,&) = KS, 1 n,&),
or @j(E @ w; X, v) C_&(X), (4) E, / u(B, x, u) = E,r r u(P, x, u),
where z =x and u = s if s’ = s, and otherwise Y*,~(z) is the smallest &marker defined or redefined at stage u. In this latter case note that z >x and Y~,~,(z)= Y*,~(z) because a marker is only moved if some smaller number enters w. The induction holds at stage s since it is a genuine A-expansionary stage and y*(x) reaches its limit at this stage. Assume the induction holds at each genuine A-expansionary stage w with s c o GS’ and let s+ be the immediate successor of s’ among such stages. Without loss of generality assume (3) holds at stage s’. If K,,+ r Y&U(Z)= wl:,,, 1 Y*,&)
(5)
then (3) still holds at s+ unless some number is enumerated into D at stage s+. If this happens then (4) holds at s+ using v = s+ because s+ is a genuine A-expansionary stage, so that both (3) and (4) hold at the beginning of stage s+, and also because no P-strategies below A are allowed to act. If, on the other
in the d.r.e.
lnfima
235
degrees
hand,
(5) is false, then A. resets the branches at stage S+ to some genuine A-expansionary stage w such that s s w < u, and so, using the comments above on resetting the branches, the induction holds at sc since it holds at w. Now suppose A f p. We show by induction on genuine A-expansionary stages
s’ ss
that for each such S’ there
is a genuine
P-expansionary
stage
u such that
s(0) G ZJs s’ and either @j(D
@ W; x, u, =fs(x), (6)
0,
1 u(P, x, v) = 0,~
1 u(P, x, v),
and
W,,
r ydz)
= JYsf r Ye.&);
and
W,,
1 n,&)
= W,,, 1 m&).
or @j(E
@ W; x, v, =fs(x),
(7) E, 1 u(P, x, u) = &
14,
x, u),
In this case z and u’ are the parameters which were associated with x at stage u in the definition of correct /?-computations. Note that by the conditions on I4$ we is true at s for the same will have Y~,~,(z) = Y*,~(z) = yA,Jz). Th e induction reasons as before, so assume the inductive hypotheses are true for every genuine A-expansionary stage w with s s w SS’ and let s+ be as before. Let v be the smallest stage which satisfies the induction for s’, and without loss of generality assume (6) is true at s’ using stage II. First suppose at stage s+ we have K,s+ 1 Y*,v’(z) # W,,.y’ 1 Y*.v’(z).
(8)
Then h resets the branches to some genuine k-expansionary stage w with s(O) s w /3-(O) were initialized, and so this link has bottom below p-(O). Also the link must have been in place at state s’. Thus /3 was suspended in subcase (4B) at some genuine P-expansionary stage w with s c w CS’. Let z’ and u” be the parameters associated
with x at stage w. We claim that: &
1 u(P, x, w) = D,
ES8 r 44
x, w) = E,
r u(P, x, w), 1 u(P, x, w), and wi,,+ 1 m,&‘)
= W,,
1 Y~,~~(z’).
To see that the claim is true, recall our earlier remarks on the effect of a suspension. When /? is suspended at stage w, all of its computations have correct w-use as long as the suspension remains. Also, as long as this w-use is correct, h will not reset the branches to a stage
Theorem
3.6.
0
236
D. Kaddah
4. Nonbranching In this section
degrees we prove
two nonbranching
theorems
for the d.r.e.
degrees,
both of which were independently proved by Downey and Shore. These results show that the obvious generalizations of Theorem 3.2 fail. We begin by proving: Theorem
4.1. In DZ, for every a > 0 there is a nonbranching
degree b c a.
Proof. Note that we may assume a is r.e. by Lachlan’s result that every properly d.r.e. degree bounds a non-zero r.e. degree. Fix an r.e. set A E a. We will build a d.r.e. set B and ensure B+A by permitting. The construction takes place on a finite-branching tree, and at the end of each stage at which A permits, the highest priority requirement which can use the permission is allowed to act. The discussion of the details of permitting is incorporated into the discussion of the individual requirements below. Before outlining the nonbranching requirements, we introduce the Nrequirements. On each path through the tree there will be infinitely many iv-strategies (Y, each of which works on the requirement: N,:
3 an a-stage
s 2 /myI(B, 1s = B 1 s).
These requirements are not essential but we include them in order to make the proof simpler. An N,-strategy attempts to initialize all lower priority strategies once, at which point (Yis satisfied unless it is initialized. If (Yacts at stage s, then we will show that any higher priority strategy which changes B, 1 s also initializes a. In addition, when (Y acts it will initialize all lower priority strategies, and so these strategies will not change B, r s. Thus, if a is never again initialized after it acts, then we will be able to prove that B r s does not change after stage s. To prove that b is nonbranching, we must show that, for each pair of d.r.e. sets C and D, deg(B) is not the infimum of deg(B @ C) and deg(B @ D). Using Lemma 2.1 we may assume C and D are r.e. in B. Fix some standard list of with 4-tuples (Ee,, Qe,, C,, 0,) where Ee and @= are partial recursive functionals domains U, and V, respectively, and C, and D, are d.r.e. sets. We satisfy the requirements: R,:
C,=
U,(B)
3X,,
c, ,?I=(X, is r.e. and X, = T,(B @ C,), ze(B @ D,)).
and
D,=&(B)
+
A strategy LYworking on requirement R, measures the length of agreement I(cr) of its hypotheses. We will say that an a-stage is a-expansionary if this length of agreement is larger than at any previous a-stage. Using the satisfaction of the N-strategies, we will be able to prove that there are infinitely many LYexpansionary stages if and only if R,‘s hypotheses are true. At each (Yexpansionary stage, (Ywill define more of the functionals r = r, and 2 = 2,. If a: is initialized then both r, and 2, are also initialized, where we initialize a functional by declaring it to be undefined on all arguments.
lnfima in the d.r.e.
We discuss
the construction
of r only,
degrees
237
since that of 2 is similar.
Generally,
if
the marker y(x) is defined and some y
defining the T(C)-use for x to be x and the T(B)-use for x to be y(x) + 1. The T(C) use is kept constant in order to prove that ris total. Note that we do not move y(x) to a new larger position if some y
previous
as discussed smaller
below,
position
we may
in this case.
need
If y
to move enters
then there must (Y’S hypotheses are satisfied, &(B; y)J = 1 for y E C = U(B), and the marker position greater than the verifying use cm(y). If verifying use is destroyed by some small number z
the
C after be
a
marker
back
y(x) is defined
verifying
to a and
computation
y(x) will be moved to a new y later leaves C because the entering B, then LYcan create
a disagreement by removing z from B, so that C(y) = 0 # 1 = U(B ; y). We will call this method the disagreement strategy and show that if cy is able to remove z from B and is never again initialized, then (Y is satisfied forever because its hypotheses are false. When executing the disagreement strategy, (Y must receive A-permission to remove z from B. Associated with a there will be a permitting function pm whose only role is to indicate that a is waiting for permission. To execute the disagreement strategy, o first defines p,(z) = z and initializes r and 2. If A r z later changes, and a is allowed to use this permission, then CYremoves z from B and is satisfied. The permitting function pa will not necessarily be a total function, but we will show that if a: is on the true path and asks for permission infinitely often, so that pa is permanently defined at infinitely many arguments, then A is recursive. There will be some restrictions on when a strategy can remove numbers, and thus LYmay not always be able to execute the disagreement strategy if z enters B and drives y out of C as described above. In that case & may have to move the marker y(x) back to its previous position. Strategy (Y has the three outcomes 0 < 0~< 1 on the tree. The O-outcome is finitary and indicates that LYwins via the disagreement strategy, so that CX’S hypotheses are false. (We note that this outcome can be infinitary if LXasks for infinitely many A-permissions, but in that case we will be able to prove A recursive.) The m-outcome indicates that a has infinitely many a-expansionary stages, and that, in addition, a only asks for finitely any A-permissions and never wins via the disagreement strategy. The l-outcome is finitary and indicates that (Y does not win via the disagreement strategy and also that there are only finitely many a-expansionary stages. As mentioned above, we will use the satisfaction of the N-strategies to prove that if there are only finitely many a-expansionary stages, then (Y’Shypotheses are false. Below the infinitary outcome of a we have the subrequirements:
R,,i: where
(0,:
X, = O,(B) + C,+
i E w} is some
standard
B or D,s,
B,
list of p-r.
functionals.
An
R,,i-strategy
/3
238
D. Kaddah
attempts to diagonalize on 0, = Oi, and, failing that, it tries to prove its conclusion. We give a brief description of one cycle of /3’s strategy, but first we discuss the notation used in this description. The markers rj~”= q; ($J’ = rj~$, respectively) will be used to show C,
B, respectively)
if /3 fails to diagonalize.
Generally
we will not
define
@(x) (q”(x), total functional
respectively) for all X, and so these markers will not determine a Yc such that C, = YJB) (Y d such that 0, = Y(B) respectively), and for this reason we will not mention these functionals again. However, we will show in the proof
that,
if necessary,
the markers
can be extended
to such a total
functional. Note that if LX’S hypotheses are true and y E C,,,, then B can determine whether y E C, by looking for a stage t 2 s such that either there is a B-correct verifying computation on y at stage t, or y 4 C,,,. Similarly, if y E D_ then B can determine whether y E 0,. Strategy p has associated with it the three permitting functions pk = p”p, for k E {c, d, w}, which are used at the three steps in p’s cycle where A-permission is needed to put an element into B. These functions are used in a manner similar to the permitting function pm to prove A recursive if not enough A-permissions are received. However there are several differences which should be mentioned. In contrast to pa, we will usually have p”(x) >x and there will be a use block associated with p”(x). Rather than enumerating x into B, strategy /3 may need to enumerate some number from the use block for p”(x) into B, in which case p waits for A to permit by changing below the smallest element of the use block. To simplify the description of a P-cycle, we delay discussion of the use blocks and assume for now that the strategy simply wants to enumerate p”(x) into B. The following is the description of one cycle for /I. Note that the cycle can be in one of 7 states and will never return to a previous state. (1) Pick a new large witness x. Wait for @(B, x)1 = 0. When this happens we say x has been realized. (c) Define q’(x) > e,(x). (Note that if C r x changes to a new position >rj~“(x).) Wait for some y
then we can move C.
y(x)
(c^) Define p’(x) = w”(x) and then initialize I/J’. (We now want to force markers associated with higher priority strategies to be larger than the verifying use on y.) Restrain B and wait for A 1 p’(x) to change, at which point we enumerate p”(x) into B. (Note that if the verifying use on y is large, then enumerating p”(x) into B may destroy this use.) (d) If y left C, then (Y takes over and executes the disagreement strategy described above. Otherwise we define r&“(x) = y(x) and note that y(x), and thus v~(_x), is larger than both the verifying use on y and e,(x). (Note that if D 1 x changes then we can move o(x) to a new position >vd(x).) Wait for some z
Infima
in the d.r.e.
239
degrees
Restrain B and wait for A 1 p”(x) to change, at which point we enumerate into B. (Note that if the verifying use on z is large, then enumerating p”(x)
p”(x) into B
may destroy this use.) (w) If z left D, then LYtakes over and executes the disagreement strategy. Otherwise, there is a new verifying use for z. Restrain B. (Note that if this restraint previous larger
is not injured, position than
the
because
then
y(x)
both
and o(x)
y(x)
and
will never
a(x)
have
need been
to be returned redefined
to a
by a to be
new
verifying uses on y and 2.) Initialize qd and define e would now like to put x into X, but need to put P”(X) = min{r(x), o(x)>. (W p”(x) into B at the same time in order to keep the functionals r, and 2, correct.) Wait for A 1 p”(x) to change. (0) Put x into X, and p”(x) into B, and restrain B. (We now have R,,; permanently satisfied unless B 1 O,(x) changes.) At first glance Step (2) does not appear to be necessary since we could eliminate r/~“, wait for y(x) to be moved to a position >0(x), and then go directly to Step (d). However, if p had some cycle stalled in Step (c) waiting for y(x) to move, then strategies of lower priority would have no way of guessing the effect of /3’s future action because they would have no way of predicting the future position of y(x). Also, as mentioned in the description of Step (c^) and as discussed below, /3’s actions must be coordinated with those of higher priority strategies. In particular, we need to ensure that if p ever acts to put x into X, in Step (0), then no higher priority strategy will ever destroy the verifying uses on y or z, thus forcing y(x) or a(x) to be returned to a position
(0)< (WI< (a < (4 < (Q < Cc)< (1). We then extend the priority ordering < on T to cycles associated with the same strategy as follows. Let g, and g, be /I-cycles in Steps (k,) and (k2) respectively. Then g, cLg2 if k, < k2, and g, g,. When all cycles are stalled, p starts a new one. We now discuss the outcomes for strategy p. If there are finitely many permanent cycles, then either some cycle reaches Step (0), or no cycle reaches Step (0) but some cycle is permanently stuck in Step (1). These cases represent outcomes 0
240
D. Kaddah
and 1, respectively. R,,i is satisfied.
In both of these
On the other hand,
if there
cases we have X, f O,(B)
are infinitely
many permanent
and requirement
cycles then almost
all
in Step (k), for some k such that (0) < (k) < (1).
of these cycles are permanently
If k = w, (i, or 2, then we will be able to prove A recursive since p asks for infinitely many A-permissions and receives at most finitely many of them. If k = d then q”(x)
we will be able to show D cT B by using is permanently
be able to prove
defined
that C+
for infinitely
(Y’Shypotheses
many x. Similarly,
and the fact that if k = c then we will
B.
There are no outcomes on the tree for the permitting steps since, if A is recursive, then we need not carry out the rest of the construction. Thus the infinitary outcomes of the strategy on the tree are c and d, and they are considered injurious to R,. of lower priority than R,. Below these outcomes we eliminate all other R,,i.-strategies since we will be able to prove C, +- B or De G,. B. We note that the N-strategies below the c-outcome are crucial in allowing us to prove that there are infinitely many cycles which are permanently in Step (c), and similarly for the N-strategies below the d-outcome. The definition of the priority ordering is completed as follows. Suppose that p # /3’ and /3’ assumes that p has outcome kl, where 0 is an R,,i-strategy and p’ is any strategy. Let g be a P-cycle. If g is in Step (k2) c (k,) then g < p’, and otherwise g > p’. If /3’ is an R,,,,-strategy and g’ is a /3’-cycle, then g g’. Step &) s (U We now discuss the main conflict between strategies. Let p be an R,,i-strategy below an infinitary outcome of R,‘,i,- strategy p’, where R,-strategy (Y has higher priority than Rep-strategy a’. Suppose /3 acts successfully to put x into X,, where y entered C, to allow yol(x) to be moved. Further suppose that later p’ puts some w into B, injuring p’s verifying use for y so that y leaves C, and LY’Sfunctionals are incorrect. One solution to this problem is to allow (Y to execute the disagreement strategy by removing w from B. In the full construction this method becomes more complicated because a’s action may conflict with higher priority strategies. We solve the problem in a different way by adding some extra machinery which prevents the original conflict from occurring. This approach will be needed in Theorem 4.8 because in that theorem B must be an r.e. set. To avoid the conflict between /3 and /I’, we require /l to coordinate its markers with p’. For example, when p defines r/$(x) in Step (c), it makes sure it places this marker on some $$(x’). Note that Steps (c) and (2) ensure that I&(X’) is larger than any of its own verifying uses, and recall that we require r/&x’) = ya(x’). If p acts to put r@(x) into B (because some y entered C,), then we require it to define r&(x) = ya(x) in Step (d), but we can also move r~$,(x’) at the same time since these two markers share a position. In the construction, strategy cr will take control and redefine Y,(X) G=r&x’) Since
a’ has lower
= 9$(x)*
priority
than
a, strategy
a’ will redefine
ym(x’) Z= &.(x’)
at
Infima in the d.r.e.
degrees
241
its next expansionary stage. Furthermore, we see that, after the markers have been redefined, the x’-cycle cannot injure the x-cycle if its advances, and vice versa. In particular, if /3 later wins with witness x, then /3’ will never injure r,. In the full construction we require each strategy to coordinate its markers with all infinitary strategies of higher priority and define several marker rules to ensure this. The construction takes place on a tree T c A<@, where A = (0 < 03< d < c < l}. Fix an effective o-ordering of all requirements such that 1. R, precedes R,. if e < e’, 2. R, precedes all R,,i, and 3. R,,i precedes R,,ir if i e, i’ E o}.
Otherwise define L,+ = L, U {R,,i}. (iii) If R = N, t hen (Yhas outcome 0. Define L,+ = L, U {N,}. Assign to node LL+E T the highest priority requirement which is not on the list L (Y+. Note that, for each e and each infinite path f on T, there is a longest R,-strategy (Yc f,which we call the final R,-strategy on f. For an R,i-strategy /I define r(p) to be the R,-strategy c p of longest length. For any strategy /3 define J-C(~)to be the longest strategy /3’ so that p’-(c) c_ /3 or P’-(d) c p, if such a p’ exists. If /3’ does not exist then n(p) is undefined. We note that the definition n(p) is never used if /3 is an N-strategy. For all strategies /3 which have markers associated with them we enforce the First Marker Rule. If E(P)J then all p markers must be placed on t&B)-marker positions, where k = c if J@)-(c) c p and k = d if Q)-(d) G /3. This sharing of marker positions presents a minor technical problem: It is possible that some position may be used up by being put into and taken out of B. Thus we assign use blocks to the markers. In fact we only assign these blocks to primary q-markers, which are defined to be those W-markers associated with strategies /3 for which n(P) 1s undefined. Note that these are the strategies which do not have to obey the first marker rule, and thus do not have to coordinate their markers with those of any higher priority requirement. All other R,,;strategies’ markers will inherit their use blocks from the markers whose positions they share, and we enforce the
D. Kaddah
242
Second Marker Rule. If rc(/?) is defined, where IpI = IZ, and /3 wants to place a P-marker on a n(P) marker position, then the n(P)-marker must have a use block
of size at least s - 4”+‘, where
s is the most
recent
stage
at which
p was
initialized.
The purpose marker
of this rule is to ensure
are large enough
to allow
that the use blocks
all strategies
to act which
associated need
with each
to act. In the
construction only an &-strategy will want to remove an element from B, and, once it does this, it is satisfied and does not act again unless initialized. Furthermore, no two markers attached to the same strategy will ever share a position. Thus, since there are <4”+’ strategies of length
Third Marker Rule. If a primary marker W(x) is to be (re)defined, then it is assigned a use block size of n, where n is larger than ever before. Its new position must be greater than its essential use plus n.
If we are able to (re)define q(x) without violating any of the first three rules, then the use block for q(x) is defined to be the interval (I/J(X) -n,
marker q(x)].
lnfima in the d.r.e. degrees
Now suppose
a non-primary
Then we enforce
marker
G(x)
243
is to be (re)defined
on a @position.
the
Fourth Marker Rule.
If $(x)
essential
must be less than the smallest
use for 4(x)
If such a y exists,
then choose
is defined the largest
to the same position element
or node
on the tree
then
the
use block.
such.
If we are able to (re)define I&X) = r/~(y) without violating marker rules, then both markers share the same use block. Let cy be a path
as q(y), in q(y)‘s
any of the other
and let 6 be an Q-strategy
such that
p^( k) c my,where k E {c, d}. Then we say t& is a marker on (Y. Let /? be an &-strategy and suppose that Q(B; x)J = 0 for /3’s current unrealized witness. We say the x-cycle is ready to progress to Step (c) if /3 is able to define the marker I/$(X) without violating the marker rules. Let k E {c, d}. When an &i-strategy p puts a number 2 from the use block for p”p(x) into B, it will request that t(P) take control of the markers. This is a two p’ such that step process. First r(p) corrects the t/~~.-markers for all &;atrategies /3’ c r(p). It does this by moving all such markers which share a position with &(x) together to some new large position. Also, I/J~,(Y) is declared to be undefined for all y such that .f is smaller than any number in the use block for r/~~(y). The second step of the process corrects the VP.-markers for all R,.,i.-strategies /3’ such that t(P)“(m) E p’ G p. This step must wait until the next @)-expansionary stage since, when /? acts, it generally destroys one of t(P)‘s verifying uses and t(P) must wait to see whether a new verifying computation converges. When carrying out the second step, r(p) moves all such p’-markers which share a position with p#c) together to some new large number m. If 3t(r(p))i then the marker rules require that m be a n(z(/3))-marker position. If t(P) moves racy,, and 2 is smaller than any number in the use block for r#~~.(z), then qs(z) is now undefined (as expected it turns out that these are the z > y). Finally, if k = c then r(p) defines I&X) = m, and if k = d then it defines p;(x) then,
= m. We note that if some marker I&(Z) is moved in either of these steps, at the next @‘)-expansionary stage, yrta.,(z) will be redefined larger by
x(/3’). Similarly, if r&x) is defined by r(p), then at the same stage it will also redefine yT&) larger. Lastly, if r(p) defines pi(x) = m then at the same stage it will redefine both yrCp,(x) and a,&~) larger. When an &-strategy < removes a number from B it will request that all strategies above it immediately reset their markers. Suppose 5 is removing 2, which was put into B at stage s. Then all of the markers on the node c which were moved or declared undefined at s are redefined to their positions at stage s, and any new markers which were not defined at stage s are now permanently undefined. Note that any such new marker was originally defined larger than i, and so we are allowed to do this. Markers ya and oa associated with a c 5 will automatically be reset by LYat its next expansionary stage, if such exists.
244
D. Kaddah
We point out a potential problem with these marker redefinitions and explain why this problem in fact never occurs. Suppose that q&x) G yT(&) and both markers depend on y staying in C. Further suppose that later some lower priority strategy acts to put a small w into B, driving y out of C and causing I&X) to be declared undefined, and that still later w is removed from B, so that I#$(x) is returned to its previous position. Since y left C we may now have Y&X) < r&x), so that, if /l acts to put x into X, then r(p)‘s functional Tis incorrect. But note that, when w is removed from B, the verifying use on y is restored but y $ C since C is a d.r.e. set. Thus r(p)‘s length of agreement drops and it will not have an expansionary stage unless the verifying use on y is again destroyed. But in order for this to happen, another small number would have to enter B, at which point we would again have r/$(x)?. Thus /3 will never again have a chance to act on I&X) and the problem is avoided. At stage s of the construction we will define the approximation to the true path S,, where 1S,l s s. Stage s is an o-stage if (Yc S,. At times we will use a parameter without its stage subscript to denote its current value. Also we may omit a strategy subscript when it causes no confusion. We assume that at each stage at most one element is enumerated in any set which we do not control. For a working on R, we define the length of agreement l(a, s) at stage s to be the greatest x so that for all y
and
D,(Y) =
UB;y),
and (Yis allowed to define y(y) and o(y) without violating the marker rules. The maximum length of agreement at stage s is defined to be m(cx, s) = max{Z(a; t): t
Stage s + 1 is an cY-expansionary stage if it is an a-stage and l(a, s) > m(a, s). For (Y on the true path, the N-strategies will allow us to prove that if x E U,,, at infinitely many a-stages, then x E U,, and similarly for V,. In addition we will show in the proof that & will always eventually be able to define markers without violating the marker rules. Thus we will also be able to prove that if (Yis on the true path and a’s hypotheses are true, then there are infinitely many aexpansionary stages. At the end of each stage we allow the highest priority strategy ~6, which can use an A-permission to act. Suppose y enters A at stage s. An R,-strategy a: is able to use the permission if y < p&x)1 for some x. An R,,i-strategy /3 is able to use the permission if p$(x)l for some x and some k E {c, d, w}, and y is less than the smallest element in this marker’s use block. To initialize a function or functional means to declare it to be undefined on all arguments. In addition, when a p-marker is initialized, its requests are cancelled. To initialize a strategy means to cancel its witnesses and cycles and to initialize its functions and functionals. If p is an R,,i-strategy, /3-(k) G (Y, and a: initializes all 5 such that LY<,_c, then all P-cycles in any state (k’) such that (k) CL(k)) are
Infima in the d.r.e.
initialized define
also.
A satisfied
a parameter
strategy
large means
which
to define
245
degrees
is initialized it bigger
becomes
than
the construction. At each a-expansionary stage s + 1, for &-strategy markers and note that (Y must also follow the marker
unsatisfied.
any number (Y, we rules
yet used in
(re)define
given
To
above.
the We
describe the process for r only since it is similar for E. First, if there is no a-expansionary state t < s + 1 such that r has not been initialized at any stage at, then,
for each x < I(a, s) in order,
the smallest
define
y(x, s + 1) large.
such stage. For each x < E(cu, s) determine
cu-expansionary
Otherwise
whether
fix t to be
there is a smallest
stage u 2 f such that y(x, v)J and
B, r y(x, v) = B, r y (x, V)
and
C, r x = C, r x.
If so redefine y(x, s + 1) = y(x, v). Finally, for each x < l(cu, s) in order such that y(x) has not yet been redefined, define the marker large. Suppose (Y and s + 1 are as in the previous paragraph and let t + 1 be the a-expansionary stage immediately preceding stage s + 1. Further suppose some /3 with r(p) = (Ywas permitted to put a number x in the use block for pi(y) into B at a stage ~3 t + 1, where p;(y) was defined at stage s’ + 1 =Zt + 1. If pi(y) has not been initialized since stage s’ + 1 and (i) k = c and there is some z
Construction Stage 0. Initialize
all nodes.
State s + 1. Define S,,, by induction until either /&+,I = s or some strategy ends the induction. Then check whether some strategy wants to act on A-permission before ending the stage. Let S,,, 1 II = (II. We define (Y+ = &+, r (n + 1) as follows. If (Y is satisfied then set (Y+ = cy-(0). Otherwise define (Y+, if possible, using the appropriate case below and execute the actions mentioned, else end the definition of hs+ 1. Case 1: (Yis an R,-strategy. Pick the first subcase below which applies. (1.1) Stage s + 1 is not a-expansionary. Then a+ = a-( 1). (1.2) a is ready to execute the disagreement strategy for some x. Choose the smallest such x and define pm(x) =x. Set S,Y+,= (Y. Initialize r,, &, and all 5‘> a: (1.3) Some highest priority /? has requested that LY= r(p) take control of the markers. Let (Y carry out the second step of this process and then redefine (Y’S markers. Cancel p’s request. Set LY+= a-(m).
246
D. Kaddah
(1.4)
Otherwise.
Redefine
Case 2: 1y is an R,,i-strategy. (2.1)
For some
(necessarily
(Y’Smarkers Then
and set (Y+ = (u”(oo).
pick the first subcase
unique)
x, the x-cycle
below which applies. has progressed
to Step
(w) since the last a-stage (i.e., p”(x) has been defined by r(a) in subcase since the last a-stage). Set Ss+, = a. Initialize all 5 > (Yand a-cycles >(w). (2.2)
Some x-cycle
is ready
to progress
Let the cycle progress for the largest 6 s+l = cy, Initialize all 5; 3 a-(d). (2.3)
to Step (2) (i.e.,
such x (i.e.,
define
@(x)J p”(x)
(1.3)
< o,(,)(x)). = qd(x)).
Set
For some
(necessarily unique) x, the x-cycle has progressed to Step (i.e., q”(x) has been defined by r(a) in subcase (1.3) since the last a-stage.) Then set a+ = (u-(d).
(d) since the last (Y stage.
(2.4) Some x-cycle is ready to progress to Step (e) (i.e., I&,“(x)J < rF(&x)). Let the cycle progress for the largest such x (i.e., define p’(x) = r/~‘(x)). Set S,,, = cy, and initialize all c 2 a-(c). (2.5) For some (necessarily unique) x, the x-cycle is ready to progress to Step (c). Let the cycle progress (i.e., let a define r/~‘(x)), and set (Y+ = a-(c). (2.6) Otherwise. If there is no cycle in Step (l), then pick a new large witness x and say the x-cycle is now in Step (1). Let LY+= a-( 1). Case 3: a is an unsatisfied NC-strategy. Initialize all c> a: (thus in effect restraining B r (s + 1)). Strategy a: is now satisjied. Finally, if A,+1 -A,=0, orifyEA,+i-A,but no a~&+, isabletouse this permission, then initialize all c such that as+, <,_ Z; and end the stage. Otherwise fix the highest priority node or cycle which is able to use the permission. Let a denote this node and, if appropriate, let the associated cycle have witness x. Choose the first case below which applies. (i) (Y is an R,-strategy. Then there is some least x such that y < p(x)j,. Remove p(x) from B and initialize all c > (Y. (Note that we always have p(x) =x.) Request that all strategies above a: immediately reset their markers. Strategy (Yis now satisfied. (ii) The x-cycle is in Step (w) (i.e., p”(x) can use the permission). Put x into X,, and the largest fresh number in p”(x)‘s use block into B. Request that r(a) take control of the markers. Initialize all f > (Y. Cancel all of LY’Scycles and declare cy satisfied. (iii) The x-cycle is in Step (2) (i.e., p*(x) can use the permission). Put the largest fresh number in pd(x)‘s use block into B. Request that t(a) take control of the markers. Initialize all c > (Y and cu-cycles ~(2). Put the (iv) The x-cycle is in Step (c^) (i.e., p’( x ) can use the permission). largest fresh number in p’(x)‘s use block into B. Request that t(a) take control of the markers. Initialize all f 3 a-(c) and &-cycles a(?). This ends the construction. Verification Assume A is nonrecursive. We define the true path 6 by induction. Let 6 r 0 = 0. Suppose we have defined 6 1 n = a. Then let 6(n) be the least k so that
Infima in the d.r.e.
a-(k)
E 13, for infinitely
path to be a marker Lemma
degrees
many s, if such k exists.
on the path
247
We define
a marker on the true
6.
4.2. Each node a E 6 is initialized only finitely often and initializes nodes
below it on 6 only finitely often. If a = 6 r n then 6(n) is defined. Thus 16 1= 03. If (Y is an N,-strategy,
then a is satisfied.
Proof. We prove the lemma
by induction,
for all LY’c CY.By the inductive all t a s, a is not initialized
hypotheses
so assume
LY5 6 and the lemma
is true
we can can fix a stage s so that,
by any a’ c a at stage t and 6, # (Y. Thus
for
any node
which initializes (Y after stage s must be to the left of a and must have been waiting for A-permission at stage s, and there are only finitely many of these. Once such a node receives permission it will act at most one more time via permission, so we may fix a stage s’ > s so that (Y is not initialized after state s’. The lemma is now clear if a: is an N,-strategy. Let (Y be an &-strategy. If permitting case (i) is executed by (Y after stage s’ then LYis satisfied forever and S(n) = 0, so assume this does not happen. If (Y executes subcase (1.2) infinitely often then we claim A is recursive: To determine whether n E A, find a stage t b-s’ and an x such that p,(x)J > n, and then n E a e n E A,. Thus LYinitializes nodes below it at most finitely often and 6(n) exists. Now let a be an &-strategy. If a executes permitting case (ii) at any stage t >s’ then it is satisfied forever and 6(n) = 0, so assume this does not happen. Strategy a cannot execute subcases (2.1) or (2.2) (and thus permitting case (iii)) infinitely often, since then A would be recursive. To see this, use the same argument as in the previous paragraph but also require that n is less than the p-marker’s use block. So assume permitting cases (ii) and (iii) and subcases (2.1) and (2.2) are only executed finitely often for (Y. Now if LYexecutes subcase (2.3) infinitely often then 6(n) = d and the lemma is proved, so assume this does not happen. Finally, many permissions
if (Y executes are received
subcase (2.4) infinitely often, then for pc and A is recursive as before.
only finitely Thus in any
case 6(n) is defined. Node (Ymay initialize nodes below a-(c) infinitely subcase (2.4), but then s(n) G d. Thus the lemma is true for a. Cl
often via
Lemma 4.3. Suppose V(X) is a VW-marker on the true path, or V(X) = ye(x) o,(x) where a-(m) c 6. Then Y(X) is moved finitely often and 3”s (Y(X, s)?)
+
or
3s Vt 2s (Y(X, t)J,).
Proof. We drop the subscript LYin the following. Note that Y can only be initialized by nodes or cycles to its left. Using Lemma 4.2 and the fact that Y is a marker on the true path, we may fix a stage s so that Y is never again initialized and, if Y is y or CJ, then
D. Kaddah
248
Also we may fix a stage t > s so that Y(X, t)i, Now let 5 be some N,-strategy Again
by Lemma
since otherwise the lemma is clear. so that (Yc c c 6 and 5 has not acted by stage t.
4.2, we may fix a stage n > t so that
stage r~ and is satisfied
forever.
Then
no element
5; acts for the final time at
or leave
B after
stage V. This is because c initializes all nodes and cycles > < and if any node or cycle
or undefined
after stage u if B changes
on some element
less than the
marker, and this will not happen. If Y(X, u)? then the marker can only be redefined if it is returned to some previous position, and this will only happen if some element less than the marker’s last defined position leaves of these previous positions are
B. Note that all will
never
be
0
Let (Y be an &-strategy with infinite outcome on the true path. Note that (Y redefines yIy and a, at each a-expansionary stage if necessary. From Lemma 4.3 we know that these markers reach a limit for each x. The rules for redefining markers
automatically lim y(-,
s
Next we prove Lemma
guarantee
s) Gr B G3C,
similar
and
lim a(-,
s
s) +. B 63 0,.
facts for q-markers.
4.4. Let /3 be an R,,i-strategy
with outcome k E {c, d} on the true path.
Then lim #‘(-,
s
s) =+ B
and
3”~ 3s Vt > s @(x, t)J.
Proof. We only prove the lemma for the case k = c, since the other case is similar. By Lemma 4.2 we may fix a least stage s such that 1~,= q’p is not initialized at any stage t 2 s. Fix X. First we prove that lim, I/J(X, s) + B. Note that V(X) is a marker on the true path so it satisfies the conditions in Lemma 4.3. Now B can determine whether I/J(X) is undefined at all stages t s s by searching for any stage t s s such that V(Y, t)l f or some y > X. If v(x) has not yet been defined by stage t, then q(x) will never be defined, so assume there is some least stage t s s such that V(x, 94. Note that no node or cycle
lnjima
q-markers
to their
positions
in the d.r.e.
degrees
249
at stage s’ and declares
defined at stages 3s ‘. Thus, if ZJ< n’ are p-( c)-stages
undefined
all markers
first
u’>J=
~1,
such that t s u, then
(r/G,
~J)L and (B,,
1 144~ ~1) = (B, 1 VG
(r/G,
v)? and r/G,
u’)l)
~1)) +
Vk
VG
and 3
3~ < 21(Y E B, - B,,),
and so lim, r/~(x, s) + B. We also note limit using a B-oracle is uniform in n. Finally
we show that there
that
are infinitely
this procedure
many
+-markers
for determining
the
which are eventually
always defined. Suppose this is false and fix a stage s after which the finitely many permanently defined markers are fixed. Now let x be the largest number so that the marker I/J(X) is permanently defined. Let < be any N,-strategy on the true path such that P-(c) E 5; and 5 1s . not yet satisfied at stage s, and let t’ > s be the stage at which c acts for the last time. We know that stage t’ exists by Lemma 4.2. Then there must be some y >x such that 1+9(y, t’)L since 5 is below the c-outcome of /3. As in the proof of Lemma 4.3 we see that rj~(y, t’) is fixed at its present position because no node or cycle of higher priority than 5 will ever act and I$’initializes all nodes and cycles of lower priority. 0 Lemma
4.5.
Let
LX be an R,-strategy
on the true path.
If a’s hypotheses
are
satisfied then a-( 1) $6. Proof. We drop the subscripts on C and D in the following. Assume (Y’S hypotheses are satisfied and suppose, by way of contradiction, that cu-( 1) c 6. Note that, if x E C, then there is some stage se(x) such that K(B;x,
sc(x))l
and
Vt 2s (B, 1 &(B;x,
se(x))
= B 1 &(B;x,
We similarly define sD(x) for x E D. Fix the largest x so that at infinitely a-stages s we have I( a; s) = x. Finally fix a smallest stage s such that 1. (Yis not initialized at any stage t 2 s,
se(x))). many
2. for all y 6x, if y E C, then y E C, and s > s”(y), and similarly for D, and 3. for all y =Gx and all t 3 s, if y 4 C then y 4 C,, and similarly for D. Let g be an N-strategy on the true path so that a-( 1) G < and < has not acted by stage s. Fix an CY-( 1)-stage t > s so that c acts for the final time at stage t. Since l(c~, t) SX, there must be some y GX such that y E (U,JB) - Ct) or y E (V&(B) - 0,). Now, as in the proof of Lemma 4.3, we have B, 1 t = B 1 t. Thus in the first case we have y E U, and in the second we have y E V,. In either case a’s hypotheses are false, a contradiction. Finally, if X(LY)J then, by Lemma 4.4, (Y will always eventually be able to define its markers without violating the marker rules. Thus LY-( 1) 4 6. 0 Lemma 4.6. Let p c 6 be an Reli-strategy /3-(O) # 6, then R,,i is satisfied.
and suppose
R,‘s hypotheses
are true.
If
250
D. Kaddah
Proof.
Note that for this lemma we do not need to assume that r(p) is the final &-strategy on 6. First suppose p-( 1) c 6. Then R,,i is clearly satisfied unless p has some permanently uncancelled witness x so that @(B;x)J = 0, but the x-cycle is never able to progress to Step (c). If qi is a primary marker then this cannot happen, so assume n(p)J. By Lemma 4.4, there are infinitely many permanently defined qZurJ-markers on 6 so the x-cycle will eventually be able to progress. Thus R,,i is satisfied since its hypothesis is not. Next suppose /3-(c) c 6. (The argument for P-(d) c 6 is similar.) Fix a stage s so that I&; is never again initialized. To determine whether IZE C,, use a B-oracle to find a P-stage t >s and an x > n so that ~Cp(x,t)l = lim, q@). Lemma 4.4 ensures that this is possible. Using the fact that R,‘s hypotheses are true, it is clear that, if IZE C,,,, then B can determine whether n E C,. On the other hand, if 12$ C,,, then it 4 C,, since otherwise we would define pi(y) for some y ax and initialize $$. Thus C, + B and R,,i is satisfied. 0 Lemma m-(m)
4.7. Let cz be the final R,-strategy c 6 and p c S is an R,,i-strategy
on 6. Then R, is satisjied.
Zf in addition
with t(P) = a; then R,,i is satisfied.
Proof.
Using Lemma 4.5, the first part of this lemma is clear if (u-( 1) c 6. Using Lemma 4.2 and the definition of the disagreement strategy, it is also clear if cum(O) c 6, because if a executes this strategy, then its disagreement will not be injured unless a is also initialized. So assume (u-(~) c 6 and fix a least stage s such that, for all stages t 2 s, we have 6, +,_ a-(m) and 1~ is not initialized at stage t. We drop most strategy subscripts in the following. Suppose that p acts to put x into X, where r(p) = a, but do not necessarily assume that /3 c 6. We now examine in some detail the sequence of events leading up to this action. Note that x’s cycle cannot be initialized at any time during this sequence since otherwise x would be cancelled and p would never enumerate x into X. Also recall that a strategy always defines a marker position as large as possible. Let s(O) >s be the stage at which p picks witness x, and s(l) the stage at which it defines v’(x) > B(B;x, s(l)). Also let s(2) be the stage at which /3 defines p’(x) = V(x, s(2)) < Y(X, s(2)). Th en, because the x-cycle has not been initialized by stage s(2), we have B,(l) r B(B; x, s(l)) = R(z) r o(B; x, 4)). At stages s(2) all nodes and cycles >fi-(c) are initialized. Let s(3) be the stage at which p”(x) receives A-permission to put some element n from its use block into B. Then, because of the initialization at stage s(2) and the fact that the x-cycle remains uncancelled at stage s(3), we have (K(Z) r Y(X, 42)))
U {n> = K(X) 1 vk
s(2)).
Note that this implies B,(i) 1 o(B; x, s(l)) = K(3) 1 o(B; x, s(l)).
lnjima in the d.r.e.
251
degrees
are again initialized and p requests At stage s(3) all nodes and cycles ~/3-(c) that a take control of the markers. At the next a-expansionary stage s(4), if (Y does not act via the disagreement strategy, then it moves all markers on the path between greater
cu”(oo) and /? which shared than
a position
with v’(x,
s(2))
to a new position
the new verifying
use on y(x). Also LYassigns r@‘(x) to this same position and redefines y(x, s(4)) even larger. Note that any new cycles which are started at stages >s(4) by strategies p’ such that (Yc j3’ c p will pick witnesses larger than this new position. We claim that if y >x and y E CSc2) then executed
the disagreement (R,,,,
1 Ye> 43))
strategy, u W(x>>
y E CSc4). Otherwise
(Y would
have
unless #K(4)
r Yk
(9)
m).
But if equation (9) is true, then some node or cycles(4) and the x-cycle is uncancelled stage t. This same basic reasoning can be applied to the second half of the cycle to show that, if cr defines p(x) = p;(x) at stage s(5) and p enumerates x into X at stage s(6),
then
p(x) G y(x, s(6)), Also note that p(x)
o(x, s(6)),
and
R,+) 1 e(B; x, s(l))
= Kclj
1 @B;x,
4)).
is defined
larger than the essential uses for y(x) and a(x) at we will say that y is critical. We claim that stage s(5). If x Gy G p(x) R.&Y) = R(Y) f or all critical y. Note that if this claim is true then T(x) and Z(x) are correct because witnesses are always chosen large, so that y(x) and a(x) have not yet been defined at stage s(O) and will always be defined to be larger than if the claim is true and /3 c 6 then requirement R,,iis s(0). Furthermore, satisfied, because if /3 ever acts on a witness x and is never initialized afterwards, then Rsc6) r p(x) = B 1 p(x) and thus O,(B; x) = 0 while x E X. We assume in the rest of the proof that B 1 x = Bsc6) 1 x, because otherwise it is clear that T(x) and Z(x) are correct. To prove the claim we first note that at stage s(6) all nodes and cycles >p are initialized so that no such node or cycle will ever again change Bsc6) r s(6). Also, because the x-cycle is not cancelled at any stage as(6), we know that if any node of cycle CL p changes B(y) at at stage >s(6), then either y s(6), and so y is not critical since s(0) s(6) for any critical y. Let LY’c 6 be an R,,-strategy ZCU-(~) and assume, by way of contradiction, that a’ removes some critical y E Bsc6) from B at a stage >s(6). Then y must have been enumerated into B at some stage
D. Kaddah
252
after
this initialization.
when
/3’ defines
critical.
p;,(z),
On the other
Thus y E B for each critical
Finally
we show
then there
that
hand,
and thus y
if P’-(k)
contradicting
G j3, then
c 6 and
t)i and w >s(6)
t > s(6)
is some
z so that
t&(z)J
= p(x).
is a P’-(k)-stage,
for each w in this marker’s
use block. Note that this will finish the proof of the claim chooses the largest r&-marker when defining a &-marker. there
that y is
y E BsC6). E /3’-(k)
is some z such that t&z,
stage s(6)
J3 is initialized
our assumption
because /3’ always First note that at
At the next
a-expansionary
stage s(7) >s(6), strategy a, moves I&(Z) and all other markers along /3 below (u”(w) which share a position with p(x) together to a new large position. Now at each p’-( k)-stage as(7) we will have q$(z) 2 $$,(z, s(7)) unless I&(Z) has been declared undefined, so suppose this happens at stage u 2 s(7). If Q!&(Z) is declared undefined because all /3’-cycles in Step (k) are initialized then the result is clear, so assume that the marker is declared undefined because enumerates a number y into B, where y is in the use some node or cycle ~/3’-(k) block for &(z’) < q;.(z). Note that as long as I&(x)~, it is the smallest &.-marker which is >s(6). Thus the node or cycle which enumerates y must be -CUP and we must have r&(z’)s(6), there is some z such that rj~$,(z, t)J and w > s(6) for each w in this marker’s use block. This proves the claim and the lemma.
0
This establishes
Theorem
4.1.
0
Next we show that not all r.e. degrees are the infima of two d.r.e. degrees. In the following theorem we will build a low, r.e. degree which is nonbranching in D2. The construction and proof for this theorem are similar to those of Theorem 4.1, and we will use some of the same notation and strategies. Theorem
4.8. There is a low, r.e. degree b which is nonbranching
in the d.r.e.
degrees. Proof. We will build an r.e. set B whose degree satisfies the theorem. Below we indicate the similarities to the construction for Theorem 4.1 and discuss in detail the differences. We will use N-requirements to force B to be of low, degree. These requirements are essential to the proof and they are of a different form than those in Theorem 4.1. Suppose a, has been assigned to requirement N,. Roughly, we will say that an a-stage is expansionary if 1We(B)1 is larger than ever before. Then requirement N, is as follows: N,:
3” a-expansionary
stages
e
1W,(B)1 = 00.
Injima
in the d.r.e.
253
degrees
At each a-expansionary stage there will be some new x in W,(B) with verifying use cpe(B;x). Strategy a tries to keep x in the set by defining a marker qIol(-lc) larger than the verifying use and restraining B. These markers will never be enumerated
into B, but otherwise
markers
in Theorem
markers
associated
4.1.
they function
There
with different
will
strategies
in the end, we will be able to prove
very much
be no
actual
like the R,,j-strategy
a-restraint
will be coordinated
that there
are infinitely
stages if and only if there are infinitely many permanently An N-strategy a has the two outcomes 00< 1. Suppose
B.
Instead
as before
on
so that,
many
a-expansionary
defined a-markers. that each N-strategy
on
the true path satisfies its requirement. Then we claim that B is low,. To see this note that there will be a unique strategy on the true path associated with W,(B). Thus we will have FinE ef {e: Wt
is finite} =+ 0”.
Now FinB is 2:-complete, so B” c1 FinB and thus B” ST@“. We next turn to the nonbranching requirements. Let (Y work on requirement R,. In Theorem 4.1, the strength of the N-requirements allowed us to prove that if (Yc 6 and a?s hypotheses were true, then a-( 1) # 6. The N- strategies in this proof do not guarantee the same result and so we weaken R,‘s hypotheses: R,:
C, E U,(B) 3X,,
and
0, c V,(B)
+
&, ,I$ (Xc is r.e. and X, = T,(B 63 C,), ,Ze(B 43 D,)).
The requirement’s hypotheses are used to define l(a; s), m(cu, s), and (Yexpansionary stages much as before, although we delay the formal definitions until after the marker rules are given. Care must be taken to ensure that the functionals r = r, and 2 = Z, are well-defined. The marker y(x) is moved whenever some y < x enters C = C,. It is also moved if some y s y(x) enters B and no substrategy of a: has an uncancelled request that the marker not be moved. We will be able to show that if (Y’S hypotheses are true then there are infinitely many such uncancelled requests, so that y(x) reaches a limit. The rules for moving a(x) are analogous. Strategy a may still need to redefine these markers to earlier positions when numbers leave C or D. Since a cannot remove numbers from B, it has no disagreement strategy. Thus the permitting function pa is no longer needed and a has only the two outcomes m < 1. Below the infinitary outcome of an R,-strategy a: we again have subrequirements: R,,i:
X, = O;(B)
3
C,+
B
or
D,+B.
Let /I be an R,,;-substrategy of R,-strategy (Y. The basic cycle from Theorem 4.1 must be altered in several ways. We eliminate the permitting functions pk and the permitting steps (2) and (2). Previously, if the x-cycle executed a permitting step, say (e), then either the cycle advanced (if a new verifying use was found) or (Y
254
took
D. Kaddah
over
and
construction disagreements.
tried
to create
the x-cycle
a disagreement
may act and achieve
However,
(if some neither
each time this happens
result
y
left
C).
since a: cannot
some y
must
leave
In this create C, and
so we will be able to prove that no marker is enumerated into B infinitely often. Here the secondary function of the P-markers is to drive numbers out of C or D, enabling
us to prove
We now describe
C+
B or D +
B if p fails to diagonalize.
the basic cycle for p. Note
states and will never return Step (d).
to a previous
(1) Pick a new large witness say x has been realized.
state,
that the cycle can be in one of 4 although
it may repeat
x. Wait for O@(B, x)s = 0. When
Step (c) or
this happens
we
(c) Define q”(x) > e,(x). (Note that if C 1 x changes then we can move y(x) to a new position >$J’(x).) Wait for some y@(x).) Wait for some z
then y(x) and a(x) will never need to be returned both y(x) and a(x) have been redefined by a uses on y and z.) We say /3 is satisfied. (We now unless B r e,(x) changes.) Theorem 4.1 the outcomes of the cycle are 0
to a previous position to be larger than the have R,,i permanently < d < 1. The priority
ordering is extended to cycles as before. The tree T is defined as in the previous proof, except that now R,- and N,-strategies both have the two outcomes 00< 1. We also use the same definition of a marker on path or node a, with the addition that, if /3 is an NC-strategy and p-(w) E (Y, then $J~ is a marker on (Y. The essential use for markers r/~,“,qd, y, and o is defined as before. We note that, if p is an N,-strategy, then r/~~(x) will only be defined if n E W,(B), in which case the essential use for VP(x) is defined to be the verifying use on x. For R,,i-strategies p, the function r(p) is defined as before. For any strategy (Y we define n(a) to be the longest R,,i- or N,-strategy /3 c cy which a guesses to have infinite outcome, if such exists. We eliminate the use blocks from Theorem 4.1 because elements can no
lnfima in the d. r. e. degrees
255
longer be extracted from B. This makes the marker rules considerably simpler. The first marker rule is still enforced, namely if 3t(p)l, then P-markers must share positions with n(p)-markers on the path /3. To ensure no marker is moved infinitely often we enforce two additional marker rules: If I/31= IZ, then p must avoid the first IZmarkers associated with n(p) on the path /3, and if /3 is initialized at stage s or if some p-marker is moved at stage S, then all P-markers first defined after stage s must avoid the first s markers associated with J@) on the path p. For a working on R, we define the length of agreement I(&, s) to be the largest x so that for every y
Y e 4,
+ Y e K,(B),
and a is allowed to define y(y) and a(y) larger than their essential uses without violating the marker rules. We define m(a, s) and cu-expansionary stages as before. Let s + 1 be an a-expansionary stage. The strategy (re)defines its markers for each x < l((~, s) as follows. First, if a has been initialized since the most recent a-expansionary stage then, for each x in order, define y(x, s + 1) large. Otherwise fix t to be the smallest a-expansionary stage such that LXhas not been initialized at any stage at. For each x determine whether there is a smallest a-expansionary stage v 2 t such that y(x, v)J and B,, /y(x,v)=B,
ry(x,~)
and
C,, rx=C,
rx.
If so redefine y(x, s + 1) = y(x, v). Now let s’ + 1x in W’(B) so that /3 can define q(y) larger than both q’(x) and the essential use for y without violating the marker rules. If allowed to act, p chooses the largest such y and defines r/~(y) as large as possible. Let p be an R,,i-strategy and suppose that O,(B; x)J = 0 for p’s current unrealized witness. We say that the x-cycle is ready to progress to Step (c) if /3 is able to define the marker t&(x) without violating the marker rules. Now suppose that R,,j-strategy p puts the marker &j(x) into B at stage s + 1 and let a = t(P). If n(n)4 then let Y(W) be the Jc(cy)-marker with which t&(x) shares a position at the beginning of stage s + 1. As in Theorem 4.1, /3 requests that (Ytake control of the markers, and this procedure is unchanged if the x-cycle
256
D. Kaddah
advances. However, in this construction the procedure becomes more complicated if the x-cycle does not advance because the markers associated with each 5 2 /3-(k) must be corrected. In particular we must ensure that if r(E) c a and (Y moves t&y), then at the end of the procedure we still have r/.$(y) G yrYrts,(y).If n(a)l this is accomplished by linking the new position of the &markers to the position of Y(W). There is the danger that Y(W) may become undefined before a is able to complete the procedure, but we will show that (Y must have been initalized in that case. If ~(a)? then a simply moves the E-markers to a position which is not too large, for example the position s + 1. We give the details of the procedure in the following two paragraphs. Whenever some marker Y(Z) is moved, all Y(W) are declared to be undefined for all w > z. Also, if (Yis initialized before it has completed all steps, then the rest of the procedure is cancelled. To execute the first step, a immediately corrects all markers 1+9~.(y)on the path a which share a position with r&(x) by moving them together to some new large position. We note that t(j3’) will redefine yscBrJ(y) large at the next @‘)-expansionary stage if r&(y) is moved. Also note that if n( (Y)Tthen there are no such markers qa(y), and thus this step is not needed. Strategy cr must wait for its next expansionary stage t + 1 > s + 1 to execute the second step. We give the details in outline form in order to make the cases clearer. First assume k = c. Form the set S of all VP.-markers on the path p, where (Yc /3’ G 0 and the marker shares a position with r/$(x). Case (I): C, r x = C, 1 x. In this case the x-cycle is ready to progress to Step (d) and we may ignore all nodes 5 1 p-(c) . M ove all of the markers in the set S to some new large position and define r/$(x) to be this number. Note that this position will be 2 the position used in the first step. Case (II): otherwise. In this case the x-cycle is not ready to advance and thus we must correct all markers for E 2 /3-(c). Form the set S’ of all such vE-markers which share a position with r#i(x). Subcase (i): PDF. Move all markers in S U S’ to position s + 1. Note that, for any 5 and z such that I&(Z) ES’, the marker y&z) will already have been moved by r(E) to a position 2s + 1. Subcase (ii): otherwise. Move all markers in S U S’ to the current position of Y(W). We will show in the proof that, for any g and z such that r&(z) ES’, we must have ‘y&z) > Y(W) else /3 would have been initialized.
The procedure is analogous for k = d. In this case, (I) may be executed twice for the x-cycle: once after the cycle enumerates I&X) into B in Step (d), and again after the cycle enumerates r/&x) into B in Step (0). When (I) is executed for the first time for the x-cycle, we say that the cycle is ready to execute Step (0). When (I) is executed for the second time for the x-cycle, j3 requests that for all y GX, y,(y) and o,(x) not be moved, and we say that p is satisfied.
Infima in the d.r.e.
257
degrees
Construction Stage 0. Initialize
all nodes.
until either 16,+11 = s or some strategy ends Stage s + 1. Define &+i by induction the induction. Let 6,+i 10 = 0. If S,,, 1 IZ = a, then define S,,, 1 (n + 1) = a+ as follows. If a is satisfied then set ay+ = a-(O). Otherwise define my+, if possible, using the appropriate
case below
definition. Case 1: (Yis un R,-strategy.
and execute
the actions
Pick the first subcase
mentioned,
else end the
below which applies.
(1.1) Stage s + 1 is not a-expansionary. Then a(+ = (w-( 1). (1.2) Some highest priority 6 has requested that (Y= r(p) take control of the markers. Let o carry out the second step of this process and then redefine a’s markers. Cancel p’s request. Set cy+ = LY-( m). (1.3) Otherwise. Redefine a’s markers and set (Y+ = a-(m). Case 2: IXis an R,,i-strategy. Pick the first subcase below which applies. (2.1) For some (necessarily unique) x, the x-cycle is ready to execute Step (0). Let the cycle act (i.e., enumerate x into XZca, and q”(x) into B, and request that t(m) take control of the markers). Strategy (Y is not satisfied. Set S,,, = a. Initialize all 5‘> (Yand cancel all a-cycles. (2.2) Some x-cycle is ready to execute Step (d) (i.e., @(x)l< r+,(x)). Let the cycle act for the largest such x (i.e., enumerate r/~“(x) into B, and request that r(a) take control of the markers). Set S,,, = (Yand initialize all 5 2 a-(c). (2.3) For some (necessarily unique) x, the x-cycle is ready to progress to Step (d) (i.e., $J~(x) has been defined by r(a) in subcase (1.2) since the most recent LYstage). Set a+ = a-(d). (2.4) Some x-cycle is ready to execute Step (c) (i.e., IJJ’(X)J < JJ~(~,(x)). Let the cycle act for the largest such x (i.e., enumerate q’(x) into B, and request that r(a) take control of the markers). Set S,,, = (Yand initialize all 5‘2 a-( 1). (2.5) For some (necessarily unique) x, the x-cycle is ready to progress to Step (c). Let the cycle progress (i.e., let (Ydefine q’“(x)). Set LY+= a-(c). (2.6) Otherwise. If there is no cycle in Step (l), then pick a new large witness x and say the x-cycle is now in Step (1). Set &+ = a-( 1). Case 3: (Yis an NC-strategy. Pick the first subcase below which applies. (3.1) Stage s + 1 is a-expansionary. Let (Y define the new marker and set Ly+= a_(m). (3.2) Otherwise. Set my+= a^( 1). Finally, initialize all 5‘ such that asfl < L 5‘ and end the stage. This ends the construction. Verification We define the true path 6 by induction. Let 6 10 = 0. Suppose we have defined 6 112 = (Y. Then let 6(n) be the least k so that a-(k) G 6, for infinitely many S, if such k exists. We define a marker on the true path to be a marker on the path 6.
258
Lemma
D. Kaddah
4.9. Zf LY= 6 1 n then 6(n)
is defined and (Y initializes each /3 2 a-(n)
at
most finitely often. Thus 161 = w and each node on 6 is initialized at most finitely often. Proof. We prove the lemma by induction, so assume a: E 6 and the lemma is true for each a’ c LY.By the inductive hypotheses we can can fix a stage s so that, for all t 2 s, a is not initialized at stage t. If LYis an N,- or &-strategy then the lemma is clear. Suppose o! is an &-strategy. If a executes subcase (2.1) at any stage t > s then it is satisfied forever and 6(n) = 0, so assume this does not happen. Suppose (Y executes subcase (2.2) infinitely often. Then we claim a also executes subcase (2.3) infinitely often. If the claim is false then there are only finitely many x so that @(x) is ever defined, and thus a must execute subcase (2.2) infinitely often for some x. But, since a! never advances to subcase (2.1), some y
(we say such a
Fix a P-marker r+!~ on 6. Note that, by the procedure for taking control of markers, q(x) is only moved when the current position of v(x) is enumerated in /3, and if some y < I&(X)is enumerated in B then q(x) is undefined from then on. We first claim that any &-strategy I; will only enumerate finitely many numbers into B for any fixed cycle. To see this, note that c enumerates a marker ~~(2) into B only if I#~(z) = $$(z) < y&z) or r&(z) = r&z) < o,o,(z). In either case the z-cycle advances unless some z’
Injima in the d.r.e.
degrees
259
on q(x), and from above we know that /I’ will only enumerate V(X) into B finitely often for the z-cycle. Thus we may fix a stage s so that no p’ 2 p enumerates V(X) into B at any stage 2s. Finally, for /3’ c /I, if p’ enumerates q(x) into B after stage s, then the cycle for /I’ does not advance else /3 would be initialized. Thus r(p’) moves all markers which share ~(x)‘s position together to some new position. By the inductive hypotheses, v(x) will only be moved finitely often. Now we show that (i) implies (ii). If (ii) is false then, by (i), we may fix a least stage s so that the finitely many permanently defined q-markers have finished moving. Suppose p is an &-strategy and qj = v’ (the other two cases are similar). By assumption /3-(c) c 6 and at the next /I-( c)-stage after s some new 1+9(y)will be defined. Note that all 0’ such that P-(c) c/I’ place their markers on r/$-markers. By definition of stage s, no such p’ will enumerate a marker position
4.11. Every
N, is satisjied.
Proof.
Fix 6 c 6 so that /3 works on N,. Suppose W,(B) is an infinite set. Then there are infinitely many P-stages at which IW,(B)I is larger than ever before. At infinitely many of these stages /3 will be able to define a new marker, where we use Lemma 4.10 to prove this if n(P)j,. Thus ,6-(m) c 6. Now assume /I-( co) c 6. Again by Lemma 4.10, there are infinitely many permanently defined P-markers. Note that when We is first defined we have q&) ’ $e(B; .x). Th us if r+!+r(x)is a permanently defined marker then x E W,(B) and so IW,(B)I = 03. q Lemma 4.12. Suppose R,‘s hypotheses are satisfied. such that p-(O) # 6. Then R,,i is satisjied.
Let /3 c 6 be an R,,i-strategy
Proof. Let /I’ be the immediate successor of j? on 6. Fix a stage s so that for all t 3 s, /3 is not initialized at stage t and 6, #,_ B’. Let C = C,. Suppose /I’ = /I-( 1). If R,,i is not satisfied then there must be some realized
witness x so that /3 is never able able to progress to Step (c). But by Lemma 4.10 this cannot happen. Now suppose /-I’ = /l-(c). (Th e case where /3 has outcome d is similar.) We claim C cT B. By Lemma 4.10 there are infinitely many permanently defined t/$-markers. From the method for moving these markers it is clear that
260
D. Kaddah
lim, I#‘;(-, s) + B. Then
to determine
whether
n E C, use a B-oracle
to find an
x > n and a stage t so that r,!$j(x, t)J and
Note that if n E C,, then p can determine
whether
n E C because
R,‘s hypotheses
are satisfied. On the other hand, if n E C - C,, then some y-cycle, where y ax, would advance to Step (d), say at stage u > t, and we would have 6,
proving
R,-markers, similarly
the next lemma
we automatically
we remark
have lim, y(-,
that,
from the method
for moving
B @ C, if this limit exists,
s) +
and
for u and B G3D.
Lemma 4.13. Let LXbe the final R,-strategy on 6. Zf (Y+ = a-( 1) c 6 then R, is satisfied. Zf R,‘s hypotheses are satisfied, then a+ = (u-(m) c 6 and (Y’S markers are moved finitely often. Proof. Note that if R,‘s hypotheses are satisfied then a will always eventually be able to define markers by Lemma 4.10. Thus if (Y+ = a-( 1)) then R, is satisfied since its hypotheses are not. Now assume that R,‘s hypotheses are satisfied, show that y(x) and a(x) are only moved finitely every stage t 2 s, cx is not initialized, C,,, 1 x = C,,, 1 x There
are infinitely @(B;y)J
many =O
and indices
o,,,
1 x = Q,,
and so (Y+ = cr-( m) . Fix x. We often. Fix a stage s so that, at
1x.
i so that for all y,
with use 8i(B;y)
~0.
Fix such an index i >s and an R,,i-strategy #I c 6 so that r(p) = LYand p has not yet acted at stage s. Then each p-witness will be >x. Now /3 will want to act on each of its witnesses y and, by Lemma 4.10, the y-cycle will advance to Step (c). Thus /3-( 1) # 6. By Lemma 4.12 and our assumptions on R,, /3-(c) # 6 and /T(d) $6, and so we may fix a stage t > s so that j3 acts for the final time to put witness y >x into B. At the next a-expansionary stage >t, all p’-markers on the path /?, where a-(m) c p’, are moved together to a new large n(m)-marker position m. Also, /3 requests that the markers y(z) and a(z) not be moved for all z < y, so in particular m > y(x), a(x). Note that this request will not be cancelled since otherwise j? would be initialized. By the definition of stage t and the fact that each strategy chooses the largest possible marker position when it acts, we know that no p’ G 6 will act again using a position/3 at stage t, so these nodes and cycles will not cause y(x) or a(x) to be moved. Thus y(x) and a(x) are not moved after stage t. 0
in the d.r.e.
Infima
261
degrees
Lemma 4.14. Let a, be the final R,-strategy on 6 and suppose the requirement’s hypotheses are true. Then R, is satisfied. Let 0 be an R,,i-strategy such that t(P) = cx and /3-(O) c 6. Then R,.i is satisfied. Proof. By Lemma We must prove
4.13 we know
a-(m)
that these functionals
c 6 and r, and 2, are total functionals.
are correct.
Fix a smallest
stage s so that a
is not initialized at any stage t 2 s. We drop most subscripts in the following. Let /3 be any R,,i-strategy with r(p) = (Y, but do not yet assume that /3 c 6. Let x be a witness and
a(x)
for /3. We first prove
are moved
when
necessary.
a useful
fact, namely
Suppose
I/&X)
that the markers
is enumerated
into
Y(x) B at
stage s’ and the x-cycle is not initialized. Then we must show that, at stage s’, no a-substrategy p’ has an uncancelled request that the a-markers y(x) and a(x) not be moved. Suppose otherwise and note that p’ must have made this request at a stage t when it executed subcase (2.1) for some y-cycle, where y > x. From the definition of the tree and the fact that y > x we must have /3-( 1) =Gp’. Now if r&(x) is enumerated into B by p or by some 5 2 /3-(d), then p’ would be initialized and its request cancelled. The only other possiblity is that r&(x) is enumerated into B by some Z+cycle in Step (k), where c-(k) G p. Suppose this is true and let r&(z) be the marker which shares a position with r&x). We claim that in this case /I?’ was initialized at a stage IJ such that t < u s s’. This is clear if c-(k) cL/3’, so assume g-(k) c p’. Then at stage t there was some t&z’) which shared a position with +$(y). By the construction we have z < z’. Then r&z’, s’)f because otherwise E would have chosen this marker for its action at stage s ’ . But r+!$(z;, s’)? implies that /3’ has been initialized at some stage u with t < v 6s’. Thus the m-markers will be moved whenever I&(X) is enumerated into B. Now suppose /3 picks witness x at stage s(O) > s and enumerates x into X, at stage s(6). We examine in detail the sequence of events leading up to this action. Fix a stage s(l) >s(O) at which p defines v”(x), so that $“(x, s(l))
’ oi(R; x, s(l)).
Then we have
B~(I, I ei(B; XT s(l)) = Bs(6)I ei(B; x, s(l)) since otherwise some higher priority node or cycle must x-cycle would have been cancelled. Now fix stages s(4) > s(3) >s(2) > s(1) so that /3 executes final time at stage s(2), (Y defines vd(x, s(3)) at the next s(3), and p executes subcase (2.2) for the final time at stage rI%
s(2)) < Y(X) s(2))
and
have
acted
and
the
subcase (2.4) for the a-expansionary stage s(4). Then
qd(x, s(4)) < a@, s(4)).
From the rules for taking control of the markers, Cs(2j r x = Csc3) r x. Also (Y redefines I/J’(x, s(3)) < ym(x, s(3)) an d makes sure that each marker is larger than
D. Kaddah
262
both of their verifying uses. We then have r,!~“(x,s(4)) < y&, s(4)) because, as was noted at the beginning of this proof, y(x) is moved whenever I/J’(X) is moved. At the next a-expansionary stage s(5) >s(4), a will redefine vd(x, s(5)) smaller than both y&, s(5)) and a&, s(5)) and will also make sure that each marker is larger than all three of their verifying uses. Strategy /3 is now ready to execute subcase (2.1) for the x-cycle, and at the next P-stage s(6) it does so by enumerating x into X, and Wd(x, s(6)) into B, and requesting that CYtake control of the markers. At the following o-expansionary stage s(7), a: executes the second step of this procedure. Arguments similar to the above show that, for all t as(6) Y&G 9, 4~
0 ’ v%,
s(6)),
and
B, r &(B;-G s(l)) = Bsc6) 1 f%(B;x, s(l))
unless some y < IJJ~(X,s(6)) is enumerated into B. We claim that if y is enumerated into B after stage s(6), then either ys(6) > $J’(x, s(6)). Note that if the claim is true then T(x) and Z(x) are correct. To prove the claim, first note that nodes or cycles
207
Logic 62 (1993) 207-263
Infima in the d.r.e. degrees D. Kaddah* SPL,
113 Steuart St., San Francisco,
CA 94105, USA
Communicated by A. Nerode Received 3 December 1992
Abstract Kaddah, 207-263.
D.,
Infima
in the
d.r.e.
degrees,
Annals
of Pure
and
Applied
Logic
62 (1993)
This paper analyzes several properties of infima in D,, the n-r.e. degrees. We first show that, for every n > 1, there are n-r.e. degrees a, b, and c, and an (n + 1)-r.e. degree x such that a < x < b, c and, in D,, b A e = a. We also prove a related result, namely that there are two d.r.e. degrees that form a minimal pair in D,, for each n < w, but that do not form a minimal pair in D,. Next, we show that every low r.e. degree branches in the d.r.e. degrees. This result does not extend to the low, r.e. degrees. We also construct a non-low r.e. degree a such that every r.e. degree b c a branches in the d.r.e. degrees. Finally we prove that the nonbranching degrees are downward dense in the d.r.e. degrees.
1. Introduction For n 2 1, a set D c w is defined to be (n + 1)-r.e. if there is an r.e. set A and an n-r.e. set B so that D =A - B, where a 1-r.e. set is defined to be an r.e. set. A 2-r.e. set is also called a difieerence of r.e. sets, abbreviated d.r.e. Equivalently we can define an n-r.e. set D as one for which there is a recursive function f so that for every x, D(x) = limf(x, S The definition
s),
f(x,
0) = 0,
of an co-r.e. set is obtained
and
I{s:f(x,
by replacing
S) #tf(x,
s + l)}/ c 12.
the third condition
above
by I{s:f(x,
S) #f(x,
S + l)}[ Gx.
The set of all n-r.e. Turing degrees Turing degrees is denoted by D,.
is denoted
by D, and the set of all w-r.e.
Correspondence to: D. Kaddah, SPL, 113 Steuart St., San Francisco, * The author would like to thank S. Lempp for his many helpful selection of the author’s Ph. D. thesis. 016%0072/93/$06.00
0
1993 -
Elsevier
Science
Publishers
CA 94105, comments.
B.V. All rights reserved
USA. This paper
forms
a
208
D. Kaddah
The d.r.e.
sets, and more
generally
the difference
hierarchy,
were first studied
by Putnam [15] and Ershov [7]. Cooper [2] showed that R# D, by building a properly d.r.e. degree (i.e., a d.r.e. degree which contains no r.e. set). In fact, this result easily extends to show D, # D, for nz # II. Much recent work in this area has compared and contrasted the structures (R, G) and (D2, c). The elementary theories of these partial orders were shown to be undecidable by Harrington more,
and
Shelah
the .X,-elementary
[lo]
and
Slaman
equivalence
(unpublished),
of the partial
orders
respectively.
Further-
is immediate
from the
proof of the decidability of the &-theory of R. Differences between the partial orders first appeared at the &level. Arslanov [l] proved that 0’ does not have the anticupping property in D2, in contrast to the independent results of Cooper and Yates (unpublished) that 0’ does have this property in R. Cooper, Lempp, and Watson [5] extended Arslanov’s result to show that every nonrecursive d.r.e. degree cups to any high r.e. degree above it by a low d.r.e. degree. Then a Z2 difference was demonstrated by the Diamond Theorem (Downey [6]), which says that the diamond (and more generally the l-n-l) lattice can be embedded in D2 preserving 0 and 1. This contrasts with Lachlan’s Nondiamond Theorem for R. Jiang [ll] has claimed that Downey’s theorem can be extended by replacing 0’ with any high r.e. degree. Slaman (unpublished) further sharpened the comparison between the partial orders by showing that R is not a Z,-substructure of DZ. One of the most basic results in r.e. degree theory is the Sacks Density Theorem [la], and researchers naturally tried to extend this to Dz. Partial positive results included Lachlan’s unpublished proof that the d.r.e. degrees are downward dense and Arslanov’s proof [l] that there is a properly d.r.e. degree above every incomplete r.e. degree. Cooper, Lempp, and Watson [5] extended these results to show that the properly d.r.e. degrees are dense among the r.e. degrees in D2. Then Cooper [3] proved the density of the low, d.r.e. degrees as a corollary to his splitting theorem. The general density problem was finally solved by Cooper, Harrington, Lachlan, Lempp, and Soare [4], who constructed a maximal incomplete d.r.e. degree. Jiang [ll] showed that this proof works above any low r.e. degree. In this paper we extend the study of the d.r.e. degrees by examining infima in Dz. Although the r.e. degrees form a proper upper semilattice, Lachlan proved that if two r.e. degrees have a meet in the Turing degrees, then they have the same meet in R. This is a corollary of his stronger result. Theorem (Lachlan [13]). Zf b, c are r.e. degrees and d is any degree sb then there is an r.e. degree a such that d s a < b, c. In Section
2 we prove
that this result
does not extend
to the n-r.e.
and ~c,
degrees:
Theorem 2.6. For all n > 1 there are n-r.e. degrees a, b, and c, and an (n + 1)-r.e. degreexsuchthata
209
lnjima in the d. r. e. degrees
Theorem 2.7. There are d.r.e. O
use the method
In Section
degrees b and c and an mr.e.
developed
3 we look at branching
by Downey in the d.r.e.
both the branching and the nonbranching degrees Fejer [9] respectively. In contrast to Fejer’s result Theorem
degree x such that
for his Diamond are dense, we prove;
3.2. If a is a low r.e. degree then there are d.r.e.
Theorem.
In R it is known
degrees.
by Slaman
that
[19] and
degrees b and c so that
a
of Theorem
3.2 we note that:
Corollary 3.3.
The lattice S, is embeddable
in D,.
This contrasts with the proof by Lachlan and Soare [12] that S, cannot be embedded in the r.e. degrees. A related result, the existence of a properly d.r.e. branching degree, was shown by Jiang [ll]. Because the nondensity of the d.r.e. degrees precludes any sort of full density result, we instead examine downward density for two properties. Theorem 3.2 provides one method for showing that the d.r.e. branching degrees are downard dense. In Section 4 we prove that the d.r.e. nonbranching degrees are downward dense. Theorem
4.1. In D2, for every a > 0 there is a nonbranching
From the partial degrees Theorem degrees.
evidence
are branching
of Theorems
in D,. However
3.2 and 3.6 it appears this is shown
degree b 6 a. that possibly
all r.e.
to be false in:
4.8. There is a low, r.e. degree b which is nonbranching
in the d.r.e.
Recently our Theorems 3.2, 4.1, and 4.8 were independently obtained by Downey and Shore. Our notation is standard and follows Soare [21]. Capital Greek letters are used to denote partial recursive functionals, with associated uses denoted by the corresponding lowercase Greek letters. In addition the notation @(X;x, s, t) represents the value of the functional @ on argument x at substage t of stage s. A parameter defined at stage s is assumed to have the same value at stage s + 1 unless explicitly redefined.
210
D. Kaddah
2. Infima in D, We begin with a lemma which is used extensively in the rest of this paper. proof of this lemma follows from the proof of Lachlan’s result that there nonrecursive Lemma
r.e. degree
below every properly
2.1. Let a be a d.r.e.
d.r.e.
The is a
degree.
degree and d a properly
d.r.e.
degree with a < d.
Then there is a d.r.e. degree e which is r.e. in a and which satifises a < e G d. Proof.
Fix d.r.e.
sets A E a and D Ed. Let B be the r.e. set
B={(x,s):x~D,
andx$D}.
Then B s,D and D is r.e. in B. If B +A, then D is r.e. in A. Otherwise, A -+ B @A s,. D and B @A is r.e. in A. In the first case let e = deg(D) and in the second let e = deg(B @A). 0 Note that Lemma 2.1 can easily be generalized to show that there is a nonrecursive n-r.e. degree below every properly (n + 1)-r.e. degree, for n 2 1. Instead of proving Theorems 2.6 and 2.7 directly, we prove the special case n = 2 and then indicate how the proof can be extended. Theorem 2.2. There are d.r.e. a
degrees a, b, and c and a 3-r.e. degree x so that
Proof. Note that we cannot have a = 0 by the generalized version of Lemma 2.1. We will build d.r.e. sets A, B, and C, a 3-r.e. set X, and functionals 4, &, and 0 satisfying the requirements: R:
X=T,(B@A),
P,:
X # @,z(A),
N,:
@JB @A)
where
G(C@A),
= @+(C @A)
= Dj j
Dj = O(A)
e = (i,j).
Here {Dj: j E o} is some standard listing of all d.r.e. sets. The strategy for N, is a slight variant of that introduced by Downey in his Diamond Theorem [6]. Instead of trying to preserve one side of the computation as in a typical r.e. minimal pair argument, we allow both sides to be injured at once. This happens when some Pi acts by putting x into X, y,(x) into B, and yZ(x) into C. If this causes a change, say z 4 Dj before Pi acts and z E Dj afterwards, then we force a disagreement or show Dj is not d.r.e. as follows. First remove x from X and y,(x) from B, thus restoring the (B @ A)-use to what it was when z $ Dj. In order to keep r, correct we must also put the new yZ(x) into C @A. If all of N,‘s computations again agree on z, then remove y*(x) from C @ A and
Infima in the d.r.e.
degrees
211
put x into X. In order to keep 4 correct we must also put some number less than y,(x) into B. These actions restore the (C @ A)-use to what it was when z E Dj, but since Dj must be d.r.e. we have achieved our goal. Now suppose N, is below the infinite outcome of N,,. Then there may be N,,-expansionary stages between the stage s at which & acts and the stage t at which N, acts for the first time. The N,,-computations which are recognized at these intermediate stages may count on y,(x) remaining in B. Then both sides of such a computation will be injured when N, acts for the first time and, in addition, the B-side will be irrevocably injured. In order to keep 0,. correct, at stage t we will require N, to put s into A instead of putting y&x) into C. Then r, is still correct since yZJx) > s, and 0,. is correct since 0,.(z’) >s if the N,.-computation on z’ was first recognized after stage s. If N, acts again at stage u to remove s from A and put some small number into B, then the C @A-side of N,,‘s computation on z’ is restored. Finally, note that if N,, first recognizes a computation on w at some stage between stages t and u, then e,.(w) >s, so that 0 is correct on w when s is removed from A. The construction takes place on the tree 2’“, where we interpret 0 as the infinite and 1 as the finite outcome. To node (Y we assign requirement N, if la1 = 2e and P, if Ial = 2e + 1. Each node a: working on N, builds its own version of the functional 0,. At stage s + 1 of the construction we will define the approximation to the true path 6s+l, where I&+ll
m(a; s) = max{l(ct; t): t
stage if it is an a-stage and I(cu, s) > m(a(, s).
Construction Stage 0. Initialize all nodes. Stage s + 1 > 0. First update each c as follows: For each x =Zs in increasing order
such that ~&)~, redefine y,(x) large. Next define a,+, by induction until either 16s+11= s or some strategy ends the induction. Let S,,, 10 = 0. Now suppose we
D. Kaddah
212
have
defined
S,,,
r (n) = a. Define
(Y+ = S,,,
1 (n + 1) using
the
appropriate
x,,
initialize
case below and execute the actions mentioned. If Q works on P,, then pick the case below which applies. (i) (Y has no uncancelled 5 > a: and end the stage. (ii) cy has uncancelled ix_(O). (iii) a has uncancelled yl(xn) Initialize
into
B, and
witness. witness
witness
y&,)
into
Pick a new large
x,, x,,
@_(A,; @_(A,;
C. For
xa)l
witness
= 0, and x, E X.
Let
all
my+=
xol)l = 0, and x, $ X. Put x, into X,
all y ?=x,
and
i = 1, 2 declare
X(Y)?.
all 5 > (Yand end the stage.
(iv) Otherwise. Let LY+= In cases (i) and (iii) above If (Yworks on N,, where e a-expansionary stages
a-( 1). we say CYacted. = (i, j) , then let t + 1 < u + 1 be the two most recent 1 (if these exist), and pick the first case below ‘which
applies. (i) Stage s + 1 is not cu-expansionary. Let a+ = a-( 1). (ii) Some (unique) P-strategy /3 with a-(O) c /3 and witness xg executed (iii) above at stage u + 1, (Y has not been initialized since, and there is a z < l(cu, U) such that Oj,s(Z) # Dj,u(Z). Then remove xp from X and Y~,~+~(x~) from B. For all y 2x6, declare yl(y)T if y > u and otherwise reset y,(y) = ~~,~+i(y). Put u + 1 into A and declare yZ( y)t for all y 3 x6. For all N-strategies (Y’ with a’-(O) s (Y U) declare e,,(y)t. Initialize all E > (Y and end the stage. and for all y sI(a’, Note that if &(z) = 0 then LYsatisfied by this action. (iii) w executed (ii) above at stage u + 1 and has not been initialized since. Then we must have 4,U(z) = 1. Remove t + 1 from A. For all y 3xB, declare yz(y)T if y >u and otherwise reset y2(y) = ~~++~(y). Put xB into X and yl,u+l(~a) - 1 into B. Declare y,(y)? for all y 3 xp. For all N-strategies (Y’ with a’-(O) E LYand for all y L I(&, t), declare e,,(y)T if y 2 I(&, U) and otherwise reset 8,(y) = ea,,t+l ( y ). Initialize all c > LYand end the stage. (iv) Otherwise. For each y < Z(a, s) in increasing order such that define 8,(y) large. Let c~+ = cu-( 0). In cases (ii) and (iii) we say (Y acted.
6(Y,S+l(y)t,
At the end of stage s + 1 initialize
all c
with 6s+l
Injima in the d.r.e. degrees
6 10 = 0. We show by induction
Proof. By definition at a = 6 1 rz and 6(n) a-stage infinitely
is defined.
s so that (Yis never
First suppose many
213
IZ is even,
Assume
cr-expansionary
stages,
holds
this is true for all 12’< n and fix the least
again initialized say cy works
on n that the lemma
after stage s.
on N, = Nci,jj. We may assume since
otherwise
the inductive
there
are
hypotheses
are satisfied. We show that if LYacts after stage s then it has the finite outcome and satisfies N,. So assume there is such a stage and fix t + 1 > s to be the least one. Then at the previous n-expansionary stage s’ + 1< t + 1 some P-strategy /3 acted,
and at stage
t + 1 there
stage t + 1, (Yexecutes
is some
(ii) and restores
z < Z((Y,s’) so that
If Oj,s,(Z) = 0 then, at the next a-expansionary and restores the use
So if Dj,sj(z) = 0 then,
Dj,t(~) # Dj,,Y(z). At
the use
stage u + 1 > t + 1, a executes
for all u > U, Qj,,(CU @A,;
(iii)
z) = 1 and Dj,v(z) = 0, while if
Dj,s,(Z) = 1 then, for all u > t, ~i,v(B” CBA,; Z) = 1 and Oj,v(Z) = 0. In either case 6 1 (n + 1) = a_(l), N, IS satisfied, and a never acts again. Now suppose n is odd, say (Yworks on PC. At the first a-stage t B s, cx picks a new witness X, in case (i) which is never cancelled. If a: never executes case (iii) then the inductive hypotheses are satisfied for n, so assume a: does so at stage u > t. Then a: is satisfied since X, E X and @,(A; xn)j, = 0. By assumption no higher priority strategy will later act and destroy a’s diagonalization, and (Y initializes all 5 > a! at stage U, so P, is satisfied. Also, at all a-stages TV> U, a is in case (iv) and 6 r (n + 1) = cu-(0). 0 Lemma 2.4. Let cu-(0) c 6, where LXworks on N,. Then N, is satisfied. Proof. By Lemma 2.3 we may fix a least a-stage s so that cr is never again initialized, and in addition we assume a never acts after stage s, since otherwise the proof of Lemma 2.3 shows that a-( 1) c 6. We show that every p =I o! keeps 0, properly defined. First note that, if a-( 1) c /3 of if /I is a P-strategy, then /3 does not affect the correctness of O,, in the latter case because otherwise & would have acted. Fix an N-strategy p and a P-strategy /I’, with a-(O) c P-(O) E P’, so that /I’ acts at stage t + 1, causing /3 to execute case (ii) at stage u+l>t+l.Ifz
On the other hand, if z 2 /(a, t) then t + l< f3_+i(z), allowing us to correct O,(z). If /3 executes case (iii) at stage ZJ+ 1 > u + 1 then,
and p puts t + 1 into A, for z < I(a, u), p restores
214
D. Kaddah
the use
If z 2 I((Y, u) then t + 1 < Oru,v+l(z) and p removes t + 1 from A, allowing us to correct O,(z). Thus in any case O,(z) is correct at stages t + 1, u + 1, and ?J + 1. Finally we show that V, (limS e,,,(z) exists). Fix z and a smallest stage s’ >S such that 13,(z) has been defined by stage s’. Note that if e,(z) is moved by /3 after stage s’, and later /3 is initialized, where c_z~(O)G /3, then /3 will not move 0,(z) at any future stage. By Lemma 2.3, every requirement on the true path acts at most finitely often. Thus, if there is some /lo a-(O) which moves 0,(z) after stage s’, then we may fix the highest priority such and the largest stage t > s’ at which it does so. Otherwise set t = s’. Then at the next a-expansionary stage u > t, 0,(z) is (re)defined for the last time. 0 Lemma
2.5. Requirement
R is satisfied.
First we note that the functionals c for i = 1, 2 are well-defined. The uses vi(x) are defined for the first time at stage x and are only redefined when some strategy acts on a witness y Sx. By Lemma 2.3 and the procedures for initialization and witness selection, each such witness y sx will be acted on at most finitely often. Thus r, and r, are total. We now show that whenever a strategy acts it keeps q and G correctly defined. If a P-strategy executes case (iii) at stage s + 1 by putting x into X, then it also puts the markers y,(x) into the appropriate sets, allowing us to redefine them large (in particular larger than s + 1). If an iv-strategy then executes case (ii) at stage t + 1> s + 1 by removing x, then it also moves y,(x) back to Y~,~+~(x)and puts s + 1 into A, which allows us to redefine y2(x) large. If this same strategy executes case (iii) at stage u + 1 > t + 1 by putting x back into X, then it puts Y~,~+~(X)- 1 into B, removes s + 1 from A, and moves y*(x) back to its position stage t + 1. Thus the c are correctly defined at each stage. q Proof.
This concludes the proof of Theorem Using similar methods, Theorem 2.2:
2.2.
it is possible
0
to prove
the following
extensions
of
Theorem 2.6. For all n > 1 there are n-r.e. degrees a, b and c, and an (n + 1)-r.e. degreexsuchthata
Note that, as in Theorem 2.2, we cannot make a=0 in Theorem generalized version of Lemma 2.1. However we can make a = 0 in: Theorem 2.7. There are d.r.e. x
degrees b and c and a nonrecursive
2.6 by the
wr.e.
degree
Injima in the d.r.e.
The proofs
of these
theorems
only give a brief description Proof of Theorem change
we make
are similar
is to the actions
215
to that
of the necessary
2.6. We use the same
degrees
of Theorem
2.2, and so we
changes. setup
as for Theroem
of N-strategies,
where
extra
2.2. The
only
steps are needed.
Fix an Nci,jj-strategy a and assume some P-strategy 0 c a-(O) put x into X at stage s(0) + 1, forcing (Yto act at stage s(l) + 1, as in (ii) of Theorem 2.2. Assume also that Dj,,(o,(z) = 0 f Dj,,cr,(z) (th e other case is similar) and let s(0) + 1 < a-expansionary stages. Then LY executes the S(1) + 1 < * * - be consecutive following steps until there are no more cY-expansionary stages. Step 1. At stage s(1) + 1, put s(0) + 1 into A, and remove
yl,S(Oj+l(x) from B and
x from X, thus restoring the (B @ A)-use on z to what it was at stage s(0) and forcing Dj,s(*)(Z) = 0. Step 2. At stage s(2) + 1, remove s(0) + 1 from A, and put yl,S(Oj+l(x) into B and x into X, thus restoring the (C @ A)-use to what it was at stage s(1) and forcing Dj,s(3)(z) = l.
Step k > 1 odd. At stage s(k) + 1, put s(0) + restoring the (C @A)-use to what it was at 0. Step k > 2 even. At stage s(k) + 1, remove thus restoring the (C @A)-use to what
1 into A and remove x from X, thus stage s(2) and forcing Dj,s(k+r)(z) = s(0) + 1 from A and put x into X, it was at stage s(1) and forcing
Dj,s(k+l)(Z) = l.
If all IZ steps are (n + 1)-r.e. set, and earlier construction, yi in order to keep construct B to be a
executed, then (Y has forced a disagreement since Di is an thus there will be no more cu-expansionary stages. As in the a also resets (or declares t) the appropriate markers 8 and the higher priority strategies satisfied. Note that we actually 3-r.e. set and C to be a d.r.e. set, although A is n-r.e.
Proof of Theorem 2.7. Note that, by the generalized version of Lemma 2.1, we need only prove that there is no nonrecursive r.e. degree a below both b and c. Thus we will requirements:
build
d.r.e.
R:
X = T;(B),
P,:
X#{e},
N,:
pi
where
sets
B and
C and
an
w-r.e.
set X to satisfy
the
T,(C),
= @i(C) = Wj 3
y
is recursive
e = (i, j).
We use the same tree and assign nodes to here A cannot be used to reflect X-changes as still need to change the value of X(n) several each IZ. To (re)define the markers vi(n) for i =
requirements as before, although in the previous theorems. We will times so we assign use blocks to 1, 2, we search for a large unused
216
block entire
D. Kaddah
of numbers [x, x + n] in both B and C, set x(X) =x + n and reserve the block for these markers. A P-strategy (1: acts as before except that if
]cr] = 2n + 1 then LYpicks all witnesses greater than IZ. This will ensure there is enough room in the witness’s use block for a! and all higher priority N-strategies to act. Fix an N,-strategy a-(O)
(Y with ]m] = 2n, where
c p. Suppose
/3 acts at m-expansionary
into B, and y*(x) into C, and at the next
e = (i, j),
and a P-strategy
p, with
stage s(O) + 1 to put x into X, y,(x) a-expansionary
stage s(l)
+ 1 we have
x E wj,s(l) - wj,s(0). Strategy (Yremoves x from X and y,,s(0)+,(x) from B and puts yZ,so)+l(x) into C, thus restoring the B-use to what it was at stage s(O). Since M$ is an r.e. set, these will be no more a-expansionary stages. Next we examine a’s effect on higher priority N-strategies. Fix g so that g-(O) E (Yand E works on N,,, where e’ = (i’, j’). Assume s(l) + 1
at
c * * * c a,-(~)
= cy(m)
c qj=
a
s(k) + 1< s(k + 1) + 1 are consecutive most n, where stages and (yk acts at stage s(k) + 1 to create a disagreement
itself. If k is even,
cu,for
‘%kremoves x from X and yl,s(o)+,(x) - (k/2) from B and into C. If k is odd, then ak puts x into X and puts YZ,~(~)+&) - (k/2) yl,s(o)+l(x) - ((k + 1)/2) into B and removes yl,sclj+L(x) - ((k - 1)/2) from C. A use block of size at most n is needed to accomodate all of these strategies, and thus if /3 chooses a witness x > II, then the construction succeeds and X is an w-r.e.
set.
then
0
3. Branching
degrees
In this
the
and
next
section
we will
look
at branching
and
nonbranching
properties of the d.r.e. degrees. These properties have been thoroughly studied in the r.e. degrees, starting with the proof by Lachlan [13] (independently Yates also constructed a [22]) that 0 branches in R. In the same paper Lachlan nonrecursive r . e . branching degree. Next Fejer showed the density of the nonbranching r.e. degrees in [9] and the existence of an r.e. branching degree above any low r.e. degree in [8]. Finally Slaman [19] completed this analysis by proving that the r.e. branching degrees are dense. The situation in D2 is not so well-studied. We cannot hope to prove density for any property since the d.r.e. degrees themselves are not dense. Also, in contrast with R, care must be taken to specify the relevant degree structure when discussing infima. For example, by Lemma 2.2, it is possible to have two d.r.e.
Infima
in
the d.r.e.
217
degrees
degrees with a meet in Dz which does not carry over to D3. On the other hand any r.e. degree which is branching in R remains branching in D,. In [ll], Jiang proved
there
is a properly
d.r.e.
branching
degree,
and,
as his construction
follows that of the r.e. case, the meet may be taken in the Turing roundabout way we can also conclude that there is a properly d.r.e. degree
since there is a maximal
d.r.e.
degree.
In fact the maximal
[4] is maximal in D, and thus nonbranching in the o-r.e. In this section we prove two results about r.e. degrees
degrees. In a nonbranching d.r.e.
degree
degrees. which are branching
of in
the d.r.e. degrees. We begin by showing that every low r.e. degree a branches in D2. More specifically we show that a is the meet in the Turing degree of two incomparable d.r.e. degrees. This result was independently proved by Downey and Shore. Our proof draws on Fejer’s proof in [S], which in turn used a technique developed by Robinson [16] for working above a low r.e. degree. The following lemma is the basis for Robinson’s method. Lemma 3.1 (Soare such that for all j
[20]).
y fl {n: D, GA}
If A
. a low LS
r.e. set then there is a recursive function h
= Who.,fl {n: D,, GA},
M’jf~ {n: D, E A} finite 3
and
Who.,finite.
Here D,, is the finite set with canonical index n. We will briefly outline Robinson’s method below. Suppose we want to build a set B satisfying the requirement B # @=(A), where A is a given low r.e. set. The strategy selects a witness x and waits for a stage such that Qe(A; x)j, = 0. Then it tries to A-certify the use $,(A; x) by enumerating it into Wj, where D, = A r #,(A ; x) and the index i is given by the Recursion Theorem. We now enumerate Who., and A until either n enters Who., or some y < c$~(A; x) enters A. By Lemma 3.1, one of these must occur. In the first case the strategy acts by putting x into B, and in the second case it waits for a new stage such that @JA; x)l = 0 and tries again. In addition, if action is taken at stage s and later some y < $,(A; x, s) enters A, then we abandon witness x and start over. Again using Lemma 3.1 we see that the strategy acts at most finitely often and is eventually satisfied. This is the standard finite injury technique for working above a low r.e. degree. The next theorem contrasts with Fejer’s result that the nonbranching degrees are dense in R. Theorem 3.2. If a is a low r.e. degree then there are d.r.e. a
degress b and c so that
in the d.r.e. degrees using the method Theorem [6]. The same result can be
218
D. Kaddah
obtained as a corollary to Theorem 3.2 by noting that the lattice MS can be embedded in the low r.e. degrees. This embedding contrasts with the proof of Lachlan and Soare [12] that S, cannot be embedded in the r.e. degrees. Corollary 3.3. The lattice S, can be embedded Proof
of
Theorem
in the d.r.e. degrees.
3.2. Fix
an r.e. set A E a, along with some standard enumeration of A. In the construction we will define a ‘speeded-up’ version A of A and build d.r.e. sets B and C to satisfy the requirements PB:
B f @e(A),
P,“:
C # Qe(A),
N,:
Qe(B @A)
= Qe(C @A)
=f
total j
f = T(A).
The strategy for N, is roughly the standard one used in the r.e. case. We measure the length of agreement [(e, s) and allow at most one number into either B or C at expansionary stages, keeping the restraint high otherwise. A positive requirement which assumes the finite outcome of all higher priority negative requirements uses the previous described strategy for working above a low set. Now suppose P,” wants to diagonalize on witness x at stage s and assumes the infinite outcome of some higher priority Ni-strategy. Instead of asking for A-certification of the use $,(A; x), P,” asks whether A, r s is correct; thus we try to A-certify the uses on all of Ni’S current computations also. In case certification is received, we enumerate x into B and restrain C r s. If later some y < s enters A then we may need to remove x from B in order to restore Ni’s computations, and P,” must try to diagonalize on a new witness. In case A-certification is not received we recalculate l(i, s) and repeat the stage. It is essential that the stage be repeated, since otherwise P,” might almost always be refused A-certification and yet still fail to satisfy the requirement. If P,” is almost always refused A-certification, then by repeating the stage we will be able to prove that Ni does not have the infinite outcome. Using Lemma 3.1, we again see that P,” will act at most finitely often to put a witness into B and will eventually be satisfied. The construction takes place on the tree 2’“. We assign node a to P,” if lal=3e,toP,CifJ(yl=3e+l,andtoN,ifIcul=3e+2.Stagesisdividedintoa varying number of substages t 6 s. Let p(s) be the last substage of stage s. We define 6: to be the path followed at substage t of stage s and let 6, = @‘. Substage t of stage s is an a-substage if IXc of. Stage s is an a-stage if a G 6,. When it causes no confusion, we use a variable without the (sub)stage identifier to denote its current value. If t is a substage of s, then we use the notation X, to denote the value of X at the end of substage t. At any stage in the construction we say a node (Y working on P,” requires attention if ~,f, or if x,1, x, 4 B, and @,(B @A;x,)J = 0. We say a: is satisfied if x,J E B and note that if & is satisfied and later initialized, then x, will be
lnfima
declared
undefined.
Node
in the d. r. e. degrees
a: may
need
to
build
219
several
finite
r.e.
sets
. . . ) where each index j(i) such that W,? = Wjcijis given by the recursion as previously described. In order to simplify notation we will refer to To initialize a during the construction means to the current set at any state as W 7 declare x,? and to abandon the current W OI.The definitions for a, working on P,” w;y, w;, Theorem
are similar. on a negative requirement N, define the length of For (YE Sfz: working agreement [(a, s + 1, t + 1) at substage t + 1 of stage s + 1 to be the largest x so that for all y sx
we have
@~(B@A;y,s+l,t)=@c(C@A;y,s+l,t) A, fu=A, u = max{+,(B Define
lu
and
where @A;y,
s + 1, t), $e(CG3A;y,
s + 1, t)}.
l(a, s + 1) = I(a, s + 1, p(s + 1)) and m(cu, s + 1) = max{l(a,
s’): s’ =GS and s’ is an a-stage}.
Substage t of stage s is a-expansionary if t is an a-substage and l(a, s, t) > m(a, s). Stage s is a-expansionary if s is an a-stage and I( a, s) > m( a, s). At each a-expansionary stage s we will define y(x, s) = s for each x s I(cu, s) such that y(x, s) is not defined. When cr is initialized we declare r,(n) and u,(n) to be undefined for all IZ. Now we discuss in more detail the reason and procedure for removing elements from B and C. Let LY and /3 be N,- and Ni-strategies respectively, where a-(O) E p. We outline the sequence of events which can lead to a problem: At stage s, /3 defines yB(x) = s. Later small b and c are put into B and C, so that the uses on both sides of /3’s computation on x are allowed to grow. We refer to the interval between the marker vrs(x) and the smaller of these new uses as a dangerous interval. Still later, at stage V, a number y in the dangerous interval enters A, so that both new uses are affected. At stage w, the computations reconverge with a different outcome. Note that since y was in the dangerous interval, /3 is not allowed to move yO(x). This is the usual sequence which leads to problems in r.e. branching constructions. Here, however, /3 can correct To(x) by removing 6 from B or c from C, thus restoring the earlier use and outcome on one side. Now suppose /3 removes b, and also that there was an n-expansionary stage s’ with u
D. Kaddah
220
To avoid this situation
we give the responsibility
for removing
lead strategy on each path, which we define to be the highest strategy on the path with infinite outcome, if such exists. numbers
enters
A, the lead strategy
elements
to the
priority negative Whenever some
resets B and C at its next expansionary
stage.
numbers unnecessarily from B and C but the above Note that if a positive strategy acts with correct problem is prevented. A-certification and is never injured, then this action will never be erased by a lead This strategy
strategy, To
may remove
so that all strategies
keep
recoverable
track
of this
A-expansionary
can still satisfy their requirements. process
each
stages.
Each
lead
strategy
A keeps
new A-expansionary
stage
a list
LA of
is added
to
the end of the list. In addition, if il resets B = B,. and C = C,. at stage s > s’, then we remove every v > s’ from LA. This is because Y is now a nonrecoverable stage, in the sense that we can never again reset B = BY. Whenever Iz is initialized we reset Ln = 0. Note that by the way we define markers y, the stages on Ln correspond exactly to the A-markers which are defined. The list is nothing more than an extra bookkeeping device, allowing us to refer directly to stages without having to calculate markers. Construction At stage 0 we initialize all nodes. At substage 0 of any stage s we enumerate one new element into a (and thus possibly into A) and proceed to substage 1. Stage s > 0, substage t > 0 Step 1. Let x be the smallest number which entered A after substage 0 of stage s - 1, if such exists. For each negative strategy (Yand each y such that yn(y)l > X, declare y,(y)?. Step 2. Calculate 6: r n for Iz
lnfima in the d.r.e.
follows.
If x,t
then
pick a new witness
construction, initialize B (C, respectively),
221
degrees
larger
than
any number
yet used in the
all E > (Y, and go to the next step. If (Ywants to put X, into then let 0, =A 1s and W” = Wi. Put y1 into W” and
Who., and A (and thus A) until either some y
(Yworks
on a negative
requirement
each y < 1( a, s) such that
ya(y)T,
strategy, then add s to L,. Initialize This ends the construction.
and rr was not initialized define
ya(y)
at stage s, and for
= s. If in addition
m is a lead
all E > 6,.
Verification We first note that, at stage s, the number of substages begun during Step 3 is bounded by s, and a new substage is begun in Step 4 only if some y
3.4. Each strategy on 6 is initialized finitely often. Each positive strategy on 6 acts finitely often and its requirement is satisfied. Thus 161= cc).
Proof. We prove the lemma by induction. Fix (Y so that 6 1 n = a E 6 and assume the lemma for all LY’c a. First we show (Yis initialized only finitely often. To see this, use the inductive hypotheses and the fact that (Yis on the true path to find a least stage s(O) so that every positive strategy CY’c a! has finished acting by after stage s(O), and at any stage s 2 s(O) we have 6, 2 (Y. If u is ever initialized s(O), then this must happen in Step 3 where we fix a highest priority /3 G (Y satisfying the conditions and initialize all E > p. Note that we also declare QT, so that p will act again to redefine x6 if given the chance. Thus we must have p
222
D. Kaddah
A-certification A-certification
and puts x, into B, and was incorrect. But then
IWfi,l = w
and
later
x,
is removed
because
the
yf~{(n:&~A}=@,
contradicting Lemma 3.1. So we may assume (Yacts only finitely often and fix a stage s(3) as(2) at which (Ydefines x, for the last time, say (Ydefines x, =x. Suppose @,JA; x)l = 0 but a never puts x into B. Then there is a stage s(4) >s(3) so that, at the final substage p(s) of every a-stage s >s(4), a asks for A-certification and is refused. But if this certification is refused at substage p(s) of stage s then the construction would proceed with substage p(s) + 1, contradicting the definition of p(s). Thus at some a-stage s 2 s(4), CYasks for and receives A-certification and puts x into B, contradicting our assumption that a never acts at a stage ss(3). This shows that P,” is satisfied. Finally we show that 6 1 (n + 1) is defined. If Edis a P-strategy then this is clear from the above arguments. If a is an N-strategy, (YG 6,, and s > layI, then 6,(n) is defined. Thus in either case 6 r (n + 1) is defined, so that, by induction, 161=@J. 0 Lemma
3.5. Each N, is satisfied.
Proof.
Fix a negative strategy a c 6 which works on N, and assume the lemma is true for all a’ c CY.Also suppose a’s hypotheses are satisfied so that it must build P,. By Lemma 3.4, we may fix a least stage s(O) after which (Yis never initialized. First we claim that a-(O) c 6. If this is false then there is a least a-stage s(l) as(O) so that cr^(O) $ & for any stage s as(l). Let m = m(a, s(l)) and note that for every a-stage s 2 s(l) we have m(a, s) = m. Since CY’Shypothesis are satisfied there must be a stage s(2) > s(l) so that for all substages t + 1 of all a-stages s ss(2) we have I(&, s, t + 1) > m. But then for all a-stages s >s(2) we have I(a, s) > m(a, s) = m, and thus there must be an a-expansionary stage s > s(l), contradicting our definition of s(l). Next we show that, for all n, lim, y,(~z, s) exists. Each marker y&n) is eventually defined in Step 5 and, after stage s(O), we only move the marker in Step 1 when some y < y&(n) enters A. Fix 12and an a-expansionary stage s(3) so that m(a; s(3)) > 12. Let /3 work on Pf, where a-(O) E /3 c 6 and i > s(3) is an index so that VX Vx Vs (C&(X; x, s)l = 0 with use &(X; x, s) = 0). Note that p has not yet acted at stage s(3) since i > s(3). By Lemma 3.4, we can find an a-expansionary stage s(4) > s(3) so that p acts for the final time at substage &s(4)) to put x6 into B. Then I(cu, s(4), &s(4))) >n and p’s Acertification is correct, so y&n, s(4)) = ya(n). Finally we show that, for all n, P,(n) is correct. Fix n and a least C-Zexpansionary stage s’ 2 s(0) such that vs 2s
(y&r, s)’ = y&r) = s’).
Injima in the d.r.e.
degrees
223
By the process for moving yLy, we know that r,(n) is correct at stage S’ and that A,. 1 s’ = A 1 s’. Let A be the lead strategy on 6, where possibly il = a. Note that if (Y# J., then a-expansionary
every a-stage is a &expansionary stage, stage is on Ln unless A removes it.
We first claim
that
B,, s B, and
similarly
for
and
C. Suppose
x E B,, - B. At the end of stage s’ all E > 6,, are initialized
in particular
each
this is false
and s’ is added
and
to LA.
Thus no lead strategy to the right of il can remove x from B. No lead strategy to the left of il will remove x from B since otherwise a: would be initialized. Node il will not remove x because by assumption A is fixed on s’. This argument establishes our claim and also shows that s’ is never removed from LA. Define an active substage to be one which ends in Step 3 or Step 5. We show by induction on active A-expansionary substages t of stages s 2,s’ (where t = p(s’) if s = s’), that there is an a-expansionary stage u E Lh.t such that s’-i
ZJGS,
but vfs
if t#p(s),
and either cDe(B@A;n,s,t)=
@e(B@A;n,v)=f(n,s’)
and
or ~~(C~A;n,s,t)=~~(C~A;n,v)=f(n,s’) (C@A),
lv=(CfT+A),
Iv.
and (2)
For each such t let p(t) be the smallest v satisfying the above, where the existence of p(t) is part of the inductive hypothesis. We induct only on active I-expansionary substages because the nonactive substages are those which end in Step 4, when some positive requirement asks for A-certification and is refused because a number is enumerated into A. If this number is small enough to injure both uses in (1) and (2) then possibly neither (1) nor (2) will be true at this nonactive substage. But in this case il will execute Step 3 and correct the uses at its next expansionary substage, and this substage will be active by definition. The inductive hypothesis holds at a-expansionary substage p(s’) because, by the definition of s’, both (1) and (2) hold at the start of substage p(s’) and at most one positive strategy acts during this substage to put a number into either B or C. Now fix an active A-expansionary substage t of stage s >s’ and assume that, in addition to being satisfied at substage p(s’), the inductive hypothesis is satisfied at all active &expansionary substages w where either w is a substage of stage s and w G t, or w is a substage of stage s^ and s’ < s^
D. Kaddah
224
below a-(O) will put a number start of substage t+, since
A-expansionary stage after t. First assume that no number
is some
smallest
z which
p(t) E Lh and, by assumption,
entered
A -A,
z 4; y(t),
and is used
B rCtj. Thus (1) is still true at the end of substage suppose
t+ ends
substage
in
Step
in Step 3. We know
so we do not remove 5,
so
that
t+ using p(t’) tf = p(s’).
any numbers
from
= p(t)
is not
Now (Y-
expansionary, then no number
Then
B p(p(s’~))1 POW’))
= BY 1 ,~P(s”)> G &+-I
1 ,4W)>.
The equality comes directly from (1) and the last containment fact that s” E LA. at the end of substage t+ we have
because
Blru,(s”jj r P(PW)
= BY 1 Ads”))
Step 3 is executed.
Again
A p(p(s”)) 1 P(P(s”))
= As,, r P(P(~))
Theorem
3.2.
from
the
= &+ 1 PMs”))
using (1) and the fact that s” E LA we get
Thus (1) still holds at the end of substage This establishes
comes
= At+ 1 P(P(s”)). t+ using ,u(t’) = p(p(s”)).
0
0
The next result proves that we can move out of the low r.e. degrees and still find a downward closed interval of r.e. degrees, all of which branch in the d.r.e. degrees. It also shows that Theorems 3.2 and 4.8 cannot be combined to single out the low r.e. degrees in D2 among the r.e. degrees by a definable predicate. Theorem 3.6. There is an r.e. degree branching in D2.
a 4 L1 such that every
r.e. degree
b s a is
lnjima in the d.r.e.
degrees
225
Proof. We will construct an r.e. set A so that deg(A) satisfies the theorem. We first discuss the requirement that A not be low and then add the branching requirements. In order to make
A’ #Tf4’ we will build
a total A-recursive
that, for all X, lim,Y @(A; x, s) exists and all requirements P,:
3x (lip
function
O(A)
so
P, are met:
@(A; x, s) # lim {e}(x, s)).
Note that if all requirements are met, then, by the Limit Lemma, A is not low. A P,-strategy (Y uses a simple version of the technique developed by Lempp and Slaman [14] for controlling the jumps of r.e. sets. First LYpicks a witness x and some large number z and starts setting @(A; x, p) = 1 with use B(A; x, p) = z for larger and larger p . If a ever finds a new u so that {e}(x, V) = 1, then it puts z into A, resets @(A; x, p) = 0 for all previous p, and picks a new larger 2, starting the process over for all new p. Thus cr will possibly put an infinite set into A but has quite a bit of freedom in choosing the members of this set. If cx is ever initialized then all current parameters, including x, are cancelled. In order to satisfy the branching conditions we meet the requirements: R,,;:
w = Qe(A)
j
deg(wi)
is branching
in D,.
An R,,,-strategy LYmeasures the length of agreement for M$ = @?(A). If cv has infinitary outcome then we must build d.r.e. sets D = D, and E = E, to satisfy the subrequirements: N,,i,j:
@j(D @ wl_)= ~j(E @ H$) =f
S,qi,j:
D # Qj(A),
S~i,j:
E # ~j(A).
Note that we are actually proving degrees, of two d.r.e. degrees.
that
total
deg(H$)
+ f = r(w),
is the infimum,
in the Turing
An N,,,i-strategy cx measures the length of agreement of the functionals in its hypothesis and defines more of r as this length increases. The strategy follows the pattern of the infimum strategies in Theorem 3.1, although in this case a will only believe computations which look correct according to higher priority P-strategies with infinite outcome. We delay discussion of the details. An S,,i,j-strategy CYpicks a witness and attempts the usual diagonalization, but first it tries to arrange that no higher priority P-strategy will ever force a higher priority N,,,j,-strategy /3 c cx to remove the witness. This is accomplished by suspending each higher priority requirement, starting with the lowest priority one and moving up. Generally we suspend a strategy by forbidding it to act or extend its definitions. In the construction this is accomplished by ending the stage whenever we reach a suspended strategy, except that the P-strategies are always allowed to extend their definition of 0. A P-strategy E with witness x is
226
D. Kaddah
suspended by setting @(A;x,s)= 0 for larger and larger s. Note that if p has higher priority than E, and E has just enumerated a number into A, then a can suspend E immediately but must wait to see if w changes before suspending p. This it may take several
stages before
cx is ready
Let a and p be as in the above paragraph
to act.
and let E be any P-strategy.
As in the
typical r.e. branching proof, if a: acts, then p can still be injured by ,$ < (Y because, although (Y knows about 5’s witnesses, it cannot stay out of the way of the effect of these
witnesses.
removing
from D and
elements
3.1, we define
For this reason
we may need
to correct
E to restore earlier computations.
a lead N,,,j-strategy
which is in charge
for /3 by
As in Theorem
of this procedure.
The construction takes place on the tree T = 2”“. Let Y be a node on T. We say that requirement P,or R,,i is satisfied OIZ Y if there is a node p c Y assigned to P,or R,,i, respectively. Requirement N,,i,j, Szi,j or S~i,j is satisfied on Y if there is an R,,i-strategy LYsuch that a-( 1) _c Y, or if there is a node p c Y assigned to N,,i,j, S,qi,j or S$j respectively. NOW fix some effective o-ordering of all requirements so that R,,iprecedes all of its subrequirements. For technical reasons we also require the first requirement on the list to be a P-strategy. Requirements are assigned to nodes on T by induction, where node Y E T is assigned the highest priority requirement which is not satisfied on Y. Stage s of the construction proceeds in substages t < s. We define Sf to be the path followed at substage t of stage s and let 6, = l_l, 6:. If a link is traveled we may have 1S:l> t, and if some requirement acts or is suspended we may have fx-stage if it is an 1S,l < s. Stage s is an a-stage if a G 6,. Stage s is a genuine a-stage and a is not suspended at stage s. We assume that at most one element enters any set w at each stage and that this element enters at the beginning of the stage. Let (Y be an R,,i-strategy and suppose at substage t of stage s + 1 we have Wi,,+l(X) = Qe(A;x, s + 1, t). We say w r (x + 1) is correct at s + 1, t if A, / &(A;x, Subrequirements still look correct
s + I, 4 =As+r,t
1 &XA;x,
s + 1, t>.
p of cx will use this definition to check whether at P-stages. We define the length of agreement
cy’s computations for (Yat substage
t of stage s + 1 to be 1(o, s + 1, t) = max{x:
K 1 x is correct
at s + 1, t}.
If stage s + 1 is a genuine a-stage then there is some unique t + 1 so that a, = 6::: and we define the length of agreement and maximum length of agreement at stage s + 1 to be l(a, s + 1) = I(n, s + 1, t), m(a;
s + 1) = max{l(cz,
and
s’): s’ Cs is a genuine
a-stage}.
If I(LY,s + 1) > m((~, s + 1) then stage s + 1 is o-expansionary. for each /3 ZI (Ywhich works on a subrequirement of R,,i.
Define
r(p)
= (Y
lnfima
On each
path
or node
priority N,,,,if-strategy on an &-subrequirement.
in
/3 we define
the d.r.e.
227
degrees
the lead (e, i)-strategy to be the highest
/3’ such that p’-(O) G /3, if such exists. Suppose p works If there is a lead (e, Q-strategy on the path p then we
use the notation J.(p) to designate this strategy. Otherwise Suppose that p is an SE;,j-strategy with currently defined p = 6;::.
we define L(p) = p. witness x(/3) and that
We say that /3 is ready to begin acting if Gj(A; x(/3), s + 1, t) = 0 A, r @j(A;X(P),
and
s + 1, t) =As+~,t 1 Gj(A;X(P),
s + 1, t).
If /3 is allowed to act and n(p) = /I then /3 enumerates x(/3) into D,,,, action is finished. If however 2.(p) # /?, then this becomes a multi-step First /3 establishes a link with top k(p). This link does not behave quite since the link is not traveled until all of p’s suspensions are in place. Note can form a unique longest chain n(p)-(o)
c pi-(o)
E * * * E/k-(o)
and the process. as usual that we
c p,
where each pi works on sone N,,i,j,-requirement. Strategy p begins by suspending all 5 with &( 0) G E c /3 at stage s + 1. Suppose that $’ is a P-strategy. If c-( 1) G /3, then p will be initialized if E ever acts. If E-(O) E /3, then 5 acted at stage s + 1 to put its current z(E) into A, so that z(g)? at the end of stage s + 1, and z(E) will not be redefined until the suspension on 5 is lifted. Also, as mentioned before, Pn only recognizes computations with ‘M/j--use which looks correct according to higher priority strategies. Thus the w-use on all Bncomputations recognize at the first P,-expansionary stage after stage s + 1 will be correct forever unless /3 is injured. At the next genuine /3,-expansionary stage, p suspends all strategies 5 such that pn_1c‘(O) 5 5 c on. This process is repeated until we reach L@?)-(O). For k < n, an argument similar to the one above shows that the w-use on all P,-computations is correct at the stage at which fik is suspended and remains correct unless p is injured. Finally, at the first L(P)-expansionary stage after A(/?)-(0) 1s suspended, we travel the link and p finishes the action by enumerating x(/3) into DXo,. Strategy p is then satisfied unless it is initialized. If p is initialized at any time during this process, or if the link is travelled, then all suspensions are cancelled and the link is destroyed. Also, whenever /3 is initialized we declare x(/3) undefined. The process for an SFi,j-Strategy is similar. Now let p be an N,,i,j-strategy and suppose that p = 6:::. Because of the infinite injury from P-strategies we must be careful in defining the length of agreement for /3. The three paragraphs following are devoted to building up these definitions. We drop the subscriptions on D,,,, and E,,,,. Before /l is allowed to recognize computations, we first make sure that A-use on the I&use for these computations looks correct when we reach p. In addition, if A(P) # /3 then p must coordinate its definitions with those of A(P) since the lead strategy may remove elements from D and E. In this case p will in effect recalculate all of n(p)‘s
228
D. Kaddah
computations
before
making
its own definitions.
Now suppose
Define U(P, X, S + 1) = max{ &(D CBWi; s + 1, t),
@j(E
@
M$;
X,
S
+
1, t)}
and let u = z&l, x, s + 1). First suppose A(p) = /3. If yB(x)l, or if there is no z
+ 1, u(P, x, s + 1)).
In this case we say that x has correct P-computations at s + 1 if w 1 k is correct at s + 1, t, and furthermore, if p is allowed to define the marker in the construction, then it sets yp(x) = k. Note that, using the basic assumption that uses are nondecreasing, we have y@(x) < yP(x + 1) if both yP(x) and yB(x + 1) are defined at stage s + 1. Now suppose A(p) # /3. In this case we say x has correct /?-computuitons at s + 1 if there is some smallest z so that 1. Y*&)L ’ u = 0, x, s + I), 2. K 1 ~~~~(2) is correct at s + 1, t, 3. if yA&z) was defined to its current value at stage s’ c s + 1, then it was the smallest A(/?)-marker defined or redefined at stage s’, and 1 u and Es, 1 u = Es+,,, 1 u. 4. & r 24.= ~,,I,, If the above conditions are satisfied then we say x is associated with z and s’ at s + 1. If yB(x) is undefined, x has correct P-computations and /3 is allowed to define the marker, then we set y@(x) = yh&z). The purpose of these conditions is to ensure that A(p) will not remove any numbers from D or E unless w. condition 3 ensures that if both sides of /3’s changes on y@(x). In addition, computations on x are injured because some number >ys(x) enters A, then one side will be restored to its state at a previous stage. For either lead or non-lead 6 we define the length of agreement and maximum length of agreement at stage s + 1 to be f(p, s + 1) = max{x:
Vy
m(/?, s + 1) = max{l(p, Stage s + 1 is P-expansionary 1). If /3 is ever initialized, arguments.
P-computations
s’): s’ C s is a genuine
if it is a genuine then we declare
at s + l)},
p-stage}.
P-stage and Z(/3, s + 1) > m(j?, s + r, and yg to be undefined on all
lnfima
Now
let
p
expansionary
be stage
an and
in the d.r.e.
N,,,,j-strategy suppose
such there
degrees
that
229
k(p) = p.
is a most
recent
Let
s + 1 be
P-expansionary
a pstage
s’
initialized at any stage v 3 s’. Further suppose that there is some least x E W&+, - w,,, and some y so that yfj(y)J >x. Then j3 wants to reset the branches and performs the following operation when allowed to do so. First,
for each y so that
yO(y)J >x,
declare
the marker
to be undefined.
Next, for each N,,i,j,-strategy p’ so that p-(O) E p’ and for each y so that yO(y)J >x, declare the marker to be undefined. Finally, if there is some P-marker
which is still defined,
this marker
was defined
from D,(p) and &J)
then fix the largest
to its present
position
which were enumerated
such, say yO(z), and suppose
at stage s”. Remove
all numbers
into these sets after stage s”.
Construction At stage 0 we initialize all nodes and define @(A; x, p) = 0 = 0(x, p) for all x and p = 0. Initialize all nodes at stage 0. At stage s + I, proceed with the steps below. Step 1. If t = 0, then define Si,, = 0. Otherwise suppose we have defined &+, = (Y. If t = s + 1 then go to Step 2. If t
(Ywants to reset the branches. Let it do so and then set (Y+ = (u-( 1). Let (Y define its markers for all x < Z(a, s + 1). If cue(O) is
Otherwise.
not suspended, then set 1y+ = o-(O) _ Otherwise there is a link with top (Y and bottom /3, so travel the link and let 6 finish its action, set S,,, = 0 and initialize all 5 > p, and go to Step 2. Case 4: (Yis an N,,,,j-strategy and ii(m) # (Y. Pick the first subcase below which applies. (4A) Stage s + 1 is not a-expansionary. Then set my+= a-( 1). (4B) Otherwise. Let CYdefine its markers for all x < I(cu, s + 1). If a-(O) is not suspended then set LY+= a-(O). Otherwise let p be the longest N,,i,j,-strategy such that /3-(O) E (Y, suspend all E with p-(O) E g c (Yand set 6 s+l = (Y, and go to Step 2.
230
D. Kaddah
Case 5: (Yis an S$j-strategy. Then pick the first subcase below which applies. (5A) a is satisfied. Then set a+ = (Y-(O). (5B) x(a) is undefined. Choose a large witness x((u) and set IY+= a-( 1). (5C) a is ready to begin acting. Then let it do so. Set a,+, = cr and initialize all E > (Y. Go to Step 2. (5D) Otherwise. Then set (Y+= a-( 1). Case 6: (Yis an SEi,j-strategy. The subcases are analogous to those in Case 5. Step 2. Initialize
all E such that S,,, cL E. Finally, for each x and p = s + 1, 0 as follows. If there is no P-strategy (Y so that x =x(a), then set @(A; x, p) = 0 with use 0(x, p) = 0. Otherwise, fix the (unique) P-strategy a such that x = x(a) and choose the first below which applies. 1. Both z(a) and v(cr) are undefined and (Yis suspended or has not requested that z(a) be defined. Then set @(A; x, p) = 0 with use 0(x, p) = 0. 2. Both z(m) and v(a) are undefined. Then define z(a) large and v((u)=p, and set @(A; x, p) = 1 with use (3(x, p) = z(a). 3. Both ~(a() and v(a) are defined. Then set O(A;x, p) = 1 with use 0(X, p) = z(a). 4. Otherwise. We note that in this case (Yacted at stage s + 1, so that z(a)? but v((Y)~. For each u such that v((u) <.v 6p, set O(A;x, v) = 0 with use 0(x, V) = 0. Declare v((Y) to be undefined. For p
Verification
We define the true path 6 by induction. Let 6 10. Suppose we have defined 6 r n = a. Then let 6(n) be the least k E (0, 1) so that a-(k) G 6, for infinitely many s, if such k exists. Lemma 3.7. Let cx E 6. Then (Y is initialized only finitely often and there are infinitely many genuine a-stages. Zf it is an S-strategy then it acts only finitely often. Thus 161 =m. Proof. The proof is by induction. Assume 6 1 n = (Y, there are infinitely many genuine a-stages, each S-strategy E’ c a acts finitely often, and (Yis initialized only finitely often. Note that, if a is an S-strategy and Q:acts and is never again initialized, then a is satisfied forever. Thus by the inductive hypotheses, (Yacts only finitely often. We must show that 6 r (n + 1) = (Y+ is defined, there are infinitely many genuine a+-stages and (Y+ is initialized only finitely often. First note that, if a = S:,, and s + 1 is a genuine a-stage, then &I: is defined unless t = s + 1, or there is a link with top a’ E LY,or subcase (5C) or (6C) is executed. Now eventually s > Ial, and by induction subcases (5C) or (6C) are
Injima
in the d.r.e.
231
degrees
executed only finitely often by (Y. Thus we may fix a stage s > Jai so that (Ynever again acts via (5C) or (6C) and is never again initialized. Suppose there is a link with top 19 E (Yat genuine a-stage S’ > s. Then this link was established by some S-strategy /I 3 a at a previous stage. If this link is never destroyed then (Y+= a-( 1) c 6 or there is never another genuine a-stage, so assume the link is destroyed at stage s” > s’. When this happens all E > /3 are initialized. Thus, if there is a link with top (Y”c a at the next genuine a-stage >s”, then this link must have been in place before p established its link and must have bottom p’ cLp. Repeated applications of this argument show that there must be some least genuine a-stage u > S” so that at stage v there is no link in place with top sty. Then v is a genuine stage for 6, 1 (n + 1). Note that if 6, 1 (n + 1) = cr-( 1) then no link can be established with bottom /? 2 cum(O) until the next genuine cur‘(0)-stage. Thus (Y+exists, has infinitely many genuine stages, and is initialized only finitely often. q Lemma 3.8. O(A) is a total A-recursive each x. Each P, is satisfied.
function,
and lim, @(A; x, p) exists for
Proof. Fix x. Note that if x =x(a) at some stage and later a abandons x, then we will never again define x = x( cu’) for any a’. Now fix p also. If p = 0 then at all stages we have @(A; x, p) = f3(x, p) = 0. If p > 0 then @(A; x, p) and 0(x, p) are first defined at stages s + 1 =p, and these values are changed at most once, in Step 2.4. In addition, if these values are changed in Step 2.4 at stage s’ >p, then /3(x, p) is enumerated into A at this stage. Thus O(A) is a total A-recursive function. Next we show that lim, @(A; x, p) exists and each P, is satisfied. If there is no (Yso that x =x(a) cofinitely often, then there is some p’ so that, for all p >p’, we define @(A; x, p) = 0 = 6(x, p) in Step 2, and thus lim, @(A; x, p) = 0. So assume there is such an a. Then either Step 2.1 applies cofinitely often, or Step 2.3 applies cofinitely often, or a-(O) c 6 and (Yacts infinitely often. In the first case lim, @(A; x, p) = 0, but (Y cannot be on the true path since otherwise, by Lemma 3.7, we would eventually redefine 2((u) in Step 2.2. In the second case, lim, @(A; x, p) = 1 and lim, {e}(x, p) = 0. Finally, suppose cy works on P,, a-(O) c 6, and & acts infinitely often. Then note that lim, @(A; x, p) = 0 and there are infinitely many v so that {e}(x, v) = 1. Thus P, is satisfied in this case also. 0 Lemma 3.9. Suppose that the hypothesis for requirement requirements S,qi,j and SEi,j are satisfied. Proof.
R,,i is satisfied. Then
Let p c 6 work on SEi,j. We prove this case only since the other is similar. By Lemma 3.7, we may fix a least genuine P-stage s so that /3 is not initialized at
232
any
D. Kaddah
stage
s’ 2 s. Then
6 chooses
a new
final
witness
x = x(/3)
at stage
s. If
@j(A; x) = 0 then /3 will eventually act, so suppose that v + 1 is the least genuine P-stage >s at which /3 begins acting via (5C). Assume that p begins acting at substage t + 1 of stage u + 1 and that we have aj(A; x, v + 1, t) = 0 A,
/ u =&+,,r
1 u,
and where
u = ~j(A; X, u + 1, t).
We first show that A r u = A,,,,, v + 1 either
z(a)
is undefined,
z(a) is defined but is never a will injure U. In addition, since otherwise /3 will be shows that if A(p) = /3, requirement is satisfied. Now suppose A(P) # p. genuine P-stage since it Al’ + 1 > u + 1 be the stage
1 u. Note that for each P-strategy in which case it will be defined larger
enumerated into A because /3 initializes all E > p at stage initialized. So A 1 u = A,+l,t then /I finishes its action
(Yc /3, at than U, or
a^( 1) E p. Thus no such II + 1 and no (Y<= 6 acts r u. This argument also at stage u + 1 and the
Note that p is allowed to finish its action at the is not initialized and is on the true path. at which the link to /3 is travelled in subcase (3C). Y&Y) be the largest A(P)- marker defined at state V’ + 1. We claim that wl: 1 yAtaj(y) = Wi,v’+l 1 yX&y). If this claim is true, note all 5 > A(P) will never remove x from DZo,, and, since p again initializes
next Let Let that /3 at
stage u’ + 1, no other strategy will remove X, so that the lemma is proved. We now prove the claim. Note that stage 2r’ + 1 is ii(P)-expansionary and y = l(Q), n’ + 1) - 1. Thus, if n(p) = a;‘:‘,,, then by definition we have K r ykyncp,(y, U’ + 1) is correct A,.
r Li = A,,,,,,,
1 fi
where
at v’ + 1, t’,
and
~2= @,(A; yACB,(y, v’ + l), v’ + 1, t’).
Now the remarks above on P-strategies WC /3 also apply here. Each such 1y will either choose a new large z(a) > fi or it has a currently defined z(a) G fi but will not enumerate this number into A. Thus A 1 fi =A,,+l,t r fi and the claim is proved.
0
Suppose (Yc 6 works on R,,i and w = @,(A). Then from Lemma 3.7 and the definition of the length of agreement for (II we know a-(O) c 6. We now prove a similar fact for N-strategies and show that their functionals are total when necessary. Lemma 3.10. Let /3 c 6 be an Ne,i,j-strategy and suppose w = QC(A). If k(p) = /3 and /3-(O) c 6, then lim, Y~,~(x) exists for each x and w can compute this limit uniformly in x. Zf N,,i,j’s hypothesis is true then /3-(O) c 6, lim, yB,,(x) exists for each x, and l4$ can compute this limit uniformly in x. Proof. From the remark preceding the lemma we know r(p) has outcome 0 on 6. By Lemma 3.7 we know that there are infinitely many genuine P-stages and
Infima in the d.r.e.
that there is some smallest stages referred
stage so that p is never
to in the rest of the proof
be declared undefined can always compute
again initialized.
are larger
YP(x) is defined at substage t + 1 of stage position unless w changes below the marker
233
degrees
than this stage.
A(P)-expansionary for N,,i,i-strategies
P--(O) C 6. Suppose ;1(/3) = /? and p-( 0) c 6. We show that each P-marker assume
Note that,
all if
s + 1, then the marker stays at this position. In that case the marker will
at the following genuine the marker movemnts
way of contradiction,
We assume
this is false and fix the smallest
stage. Thus I4$ /I such that has a limit.
By
x so that lim, YO,,(x)
does not exist. Fix a stage s such that, for all y
z)l = 0 with use Qj(X;
z) = 0),
and suppose E c 6 works on Sz,,j,. Then note that 5 has not yet acted at stage s and in addition p-(O) G g because j’ > s. By Lemma 3.9 we may fix a least genuine P-expansionary stage s’ > s so that 5; finishes acting for the last time at stage s’, and the proof of the same lemma shows that w,,. 1 yB(n) = M$ 1 yO(x). Thus this marker is fixed forever at stage s’. Now suppose that h(P) = p and the hypothesis of requirement N,,i,j is satisfied. Then, by the assumption that wj = Gc(A), each x will eventually have correct P-computations at some genuine P-stage. Thus /3-(O) c 6. Finally suppose k(p) #/I and the hypothesis of requirement N,,,, is satisfied. From the above we know all A(p)- markers are eventually defined and reach a limit. Using Lemma 3.7, the fact that W, = @,,(A), and the assumption that the requirement’s hypothesis is satisfied, we see that, for each x, there is eventually a smallest z so that yO(x) is defined to be ynCB)(z) and both markers have reached a limit, so again /3-( 0) c 6. 0 Lemma
3.11. Every requirement N,,i,j is satisfied.
Proof. Let p c 6 be an N,,i,j-strategy. Assume the requirement’s hypothesis is satisfied and I4$ = Qc(A). By Lemma 3.7 there is some smallest stage after which /3 is never again initialized, so we assume all stages in the rest of the proof are larger. We drop the subscripts on D and E and let A = A(p) in the following. From Lemma 3.10 we know that p-(O) c 6 and that each marker eventually reaches a I+computable limit, Fix x and a least genuine P-stage s so that vv 2s
(Y&v(X)J = Y&)d2f
We must show that genuine P-substage. there are parameters y =x and s(O) = s in
Y&)).
h(x) =fa,.y(x). A ssume that substage t + 1 of stage s is a If il # /3 then, from the definition of correct P-computations, y and s(O) with which x is associated at s. If A = /3 then let the following.
234
D. Kaddah
First note that, using the definitions of stage s and of B-correct computations, we have m,,&r) = Ye and wiJ(0) 1 Y*(Y) = K r Y*(Y), so that A.never uses a stage
= Q 1 Y&) G D 1 Y&),
and
E,(o) r Y&) = E, / Y&) GE I Y/~(X). Suppose that A resets the branches at stage s” > s to a stage s’ < s”. By the above, s(O) ss’. Let m(z) be the smallest &marker which was defined or redefined at stage s’. From the conditions for resetting the branches we have (D @ I%),. 1 Y*,&) = (D @ W),, r Y*,&), (E @ J%),v 1 Y&Z) = (E @ K),, 1 m,&), In particular, G
and
Ye,&) = YA,&).
if A # /I and s’ c s then r 4%
x, s) = Q
r u(P, x, 81,
and
Es,, r u(/?, x, s) = Es r ~(6, x, s).
These facts are used in the inductions below. First suppose A.= p. We show by induction on genuine A-expansionary stages s’ >s that for each such s’ thee is a genuine A-expansionary stage u such that s
@
WJx,
v)=f,(x),
(3) 0” r @I, x, v) = Q, 1 u(P, x, u),
and
W,,, r Y~,~(z) = K,,, r y&r);
and
I%,, 1 YA,&) = KS, 1 n,&),
or @j(E @ w; X, v) C_&(X), (4) E, / u(B, x, u) = E,r r u(P, x, u),
where z =x and u = s if s’ = s, and otherwise Y*,~(z) is the smallest &marker defined or redefined at stage u. In this latter case note that z >x and Y~,~,(z)= Y*,~(z) because a marker is only moved if some smaller number enters w. The induction holds at stage s since it is a genuine A-expansionary stage and y*(x) reaches its limit at this stage. Assume the induction holds at each genuine A-expansionary stage w with s c o GS’ and let s+ be the immediate successor of s’ among such stages. Without loss of generality assume (3) holds at stage s’. If K,,+ r Y&U(Z)= wl:,,, 1 Y*,&)
(5)
then (3) still holds at s+ unless some number is enumerated into D at stage s+. If this happens then (4) holds at s+ using v = s+ because s+ is a genuine A-expansionary stage, so that both (3) and (4) hold at the beginning of stage s+, and also because no P-strategies below A are allowed to act. If, on the other
in the d.r.e.
lnfima
235
degrees
hand,
(5) is false, then A. resets the branches at stage S+ to some genuine A-expansionary stage w such that s s w < u, and so, using the comments above on resetting the branches, the induction holds at sc since it holds at w. Now suppose A f p. We show by induction on genuine A-expansionary stages
s’ ss
that for each such S’ there
is a genuine
P-expansionary
stage
u such that
s(0) G ZJs s’ and either @j(D
@ W; x, u, =fs(x), (6)
0,
1 u(P, x, v) = 0,~
1 u(P, x, v),
and
W,,
r ydz)
= JYsf r Ye.&);
and
W,,
1 n,&)
= W,,, 1 m&).
or @j(E
@ W; x, v, =fs(x),
(7) E, 1 u(P, x, u) = &
14,
x, u),
In this case z and u’ are the parameters which were associated with x at stage u in the definition of correct /?-computations. Note that by the conditions on I4$ we is true at s for the same will have Y~,~,(z) = Y*,~(z) = yA,Jz). Th e induction reasons as before, so assume the inductive hypotheses are true for every genuine A-expansionary stage w with s s w SS’ and let s+ be as before. Let v be the smallest stage which satisfies the induction for s’, and without loss of generality assume (6) is true at s’ using stage II. First suppose at stage s+ we have K,s+ 1 Y*,v’(z) # W,,.y’ 1 Y*.v’(z).
(8)
Then h resets the branches to some genuine k-expansionary stage w with s(O) s w
with x at stage w. We claim that: &
1 u(P, x, w) = D,
ES8 r 44
x, w) = E,
r u(P, x, w), 1 u(P, x, w), and wi,,+ 1 m,&‘)
= W,,
1 Y~,~~(z’).
To see that the claim is true, recall our earlier remarks on the effect of a suspension. When /? is suspended at stage w, all of its computations have correct w-use as long as the suspension remains. Also, as long as this w-use is correct, h will not reset the branches to a stage
Theorem
3.6.
0
236
D. Kaddah
4. Nonbranching In this section
degrees we prove
two nonbranching
theorems
for the d.r.e.
degrees,
both of which were independently proved by Downey and Shore. These results show that the obvious generalizations of Theorem 3.2 fail. We begin by proving: Theorem
4.1. In DZ, for every a > 0 there is a nonbranching
degree b c a.
Proof. Note that we may assume a is r.e. by Lachlan’s result that every properly d.r.e. degree bounds a non-zero r.e. degree. Fix an r.e. set A E a. We will build a d.r.e. set B and ensure B+A by permitting. The construction takes place on a finite-branching tree, and at the end of each stage at which A permits, the highest priority requirement which can use the permission is allowed to act. The discussion of the details of permitting is incorporated into the discussion of the individual requirements below. Before outlining the nonbranching requirements, we introduce the Nrequirements. On each path through the tree there will be infinitely many iv-strategies (Y, each of which works on the requirement: N,:
3 an a-stage
s 2 /myI(B, 1s = B 1 s).
These requirements are not essential but we include them in order to make the proof simpler. An N,-strategy attempts to initialize all lower priority strategies once, at which point (Yis satisfied unless it is initialized. If (Yacts at stage s, then we will show that any higher priority strategy which changes B, 1 s also initializes a. In addition, when (Y acts it will initialize all lower priority strategies, and so these strategies will not change B, r s. Thus, if a is never again initialized after it acts, then we will be able to prove that B r s does not change after stage s. To prove that b is nonbranching, we must show that, for each pair of d.r.e. sets C and D, deg(B) is not the infimum of deg(B @ C) and deg(B @ D). Using Lemma 2.1 we may assume C and D are r.e. in B. Fix some standard list of with 4-tuples (Ee,, Qe,, C,, 0,) where Ee and @= are partial recursive functionals domains U, and V, respectively, and C, and D, are d.r.e. sets. We satisfy the requirements: R,:
C,=
U,(B)
3X,,
c, ,?I=(X, is r.e. and X, = T,(B @ C,), ze(B @ D,)).
and
D,=&(B)
+
A strategy LYworking on requirement R, measures the length of agreement I(cr) of its hypotheses. We will say that an a-stage is a-expansionary if this length of agreement is larger than at any previous a-stage. Using the satisfaction of the N-strategies, we will be able to prove that there are infinitely many LYexpansionary stages if and only if R,‘s hypotheses are true. At each (Yexpansionary stage, (Ywill define more of the functionals r = r, and 2 = 2,. If a: is initialized then both r, and 2, are also initialized, where we initialize a functional by declaring it to be undefined on all arguments.
lnfima in the d.r.e.
We discuss
the construction
of r only,
degrees
237
since that of 2 is similar.
Generally,
if
the marker y(x) is defined and some y
defining the T(C)-use for x to be x and the T(B)-use for x to be y(x) + 1. The T(C) use is kept constant in order to prove that ris total. Note that we do not move y(x) to a new larger position if some y
previous
as discussed smaller
below,
position
we may
in this case.
need
If y
to move enters
then there must (Y’S hypotheses are satisfied, &(B; y)J = 1 for y E C = U(B), and the marker position greater than the verifying use cm(y). If verifying use is destroyed by some small number z
the
C after be
a
marker
back
y(x) is defined
verifying
to a and
computation
y(x) will be moved to a new y later leaves C because the entering B, then LYcan create
a disagreement by removing z from B, so that C(y) = 0 # 1 = U(B ; y). We will call this method the disagreement strategy and show that if cy is able to remove z from B and is never again initialized, then (Y is satisfied forever because its hypotheses are false. When executing the disagreement strategy, (Y must receive A-permission to remove z from B. Associated with a there will be a permitting function pm whose only role is to indicate that a is waiting for permission. To execute the disagreement strategy, o first defines p,(z) = z and initializes r and 2. If A r z later changes, and a is allowed to use this permission, then CYremoves z from B and is satisfied. The permitting function pa will not necessarily be a total function, but we will show that if a: is on the true path and asks for permission infinitely often, so that pa is permanently defined at infinitely many arguments, then A is recursive. There will be some restrictions on when a strategy can remove numbers, and thus LYmay not always be able to execute the disagreement strategy if z enters B and drives y out of C as described above. In that case & may have to move the marker y(x) back to its previous position. Strategy (Y has the three outcomes 0 < 0~< 1 on the tree. The O-outcome is finitary and indicates that LYwins via the disagreement strategy, so that CX’S hypotheses are false. (We note that this outcome can be infinitary if LXasks for infinitely many A-permissions, but in that case we will be able to prove A recursive.) The m-outcome indicates that a has infinitely many a-expansionary stages, and that, in addition, a only asks for finitely any A-permissions and never wins via the disagreement strategy. The l-outcome is finitary and indicates that (Y does not win via the disagreement strategy and also that there are only finitely many a-expansionary stages. As mentioned above, we will use the satisfaction of the N-strategies to prove that if there are only finitely many a-expansionary stages, then (Y’Shypotheses are false. Below the infinitary outcome of a we have the subrequirements:
R,,i: where
(0,:
X, = O,(B) + C,+
i E w} is some
standard
B or D,s,
B,
list of p-r.
functionals.
An
R,,i-strategy
/3
238
D. Kaddah
attempts to diagonalize on 0, = Oi, and, failing that, it tries to prove its conclusion. We give a brief description of one cycle of /3’s strategy, but first we discuss the notation used in this description. The markers rj~”= q; ($J’ = rj~$, respectively) will be used to show C,
B, respectively)
if /3 fails to diagonalize.
Generally
we will not
define
@(x) (q”(x), total functional
respectively) for all X, and so these markers will not determine a Yc such that C, = YJB) (Y d such that 0, = Y(B) respectively), and for this reason we will not mention these functionals again. However, we will show in the proof
that,
if necessary,
the markers
can be extended
to such a total
functional. Note that if LX’S hypotheses are true and y E C,,,, then B can determine whether y E C, by looking for a stage t 2 s such that either there is a B-correct verifying computation on y at stage t, or y 4 C,,,. Similarly, if y E D_ then B can determine whether y E 0,. Strategy p has associated with it the three permitting functions pk = p”p, for k E {c, d, w}, which are used at the three steps in p’s cycle where A-permission is needed to put an element into B. These functions are used in a manner similar to the permitting function pm to prove A recursive if not enough A-permissions are received. However there are several differences which should be mentioned. In contrast to pa, we will usually have p”(x) >x and there will be a use block associated with p”(x). Rather than enumerating x into B, strategy /3 may need to enumerate some number from the use block for p”(x) into B, in which case p waits for A to permit by changing below the smallest element of the use block. To simplify the description of a P-cycle, we delay discussion of the use blocks and assume for now that the strategy simply wants to enumerate p”(x) into B. The following is the description of one cycle for /I. Note that the cycle can be in one of 7 states and will never return to a previous state. (1) Pick a new large witness x. Wait for @(B, x)1 = 0. When this happens we say x has been realized. (c) Define q’(x) > e,(x). (Note that if C r x changes to a new position >rj~“(x).) Wait for some y
then we can move C.
y(x)
(c^) Define p’(x) = w”(x) and then initialize I/J’. (We now want to force markers associated with higher priority strategies to be larger than the verifying use on y.) Restrain B and wait for A 1 p’(x) to change, at which point we enumerate p”(x) into B. (Note that if the verifying use on y is large, then enumerating p”(x) into B may destroy this use.) (d) If y left C, then (Y takes over and executes the disagreement strategy described above. Otherwise we define r&“(x) = y(x) and note that y(x), and thus v~(_x), is larger than both the verifying use on y and e,(x). (Note that if D 1 x changes then we can move o(x) to a new position >vd(x).) Wait for some z
Infima
in the d.r.e.
239
degrees
Restrain B and wait for A 1 p”(x) to change, at which point we enumerate into B. (Note that if the verifying use on z is large, then enumerating p”(x)
p”(x) into B
may destroy this use.) (w) If z left D, then LYtakes over and executes the disagreement strategy. Otherwise, there is a new verifying use for z. Restrain B. (Note that if this restraint previous larger
is not injured, position than
the
because
then
y(x)
both
and o(x)
y(x)
and
will never
a(x)
have
need been
to be returned redefined
to a
by a to be
new
verifying uses on y and 2.) Initialize qd and define e would now like to put x into X, but need to put P”(X) = min{r(x), o(x)>. (W p”(x) into B at the same time in order to keep the functionals r, and 2, correct.) Wait for A 1 p”(x) to change. (0) Put x into X, and p”(x) into B, and restrain B. (We now have R,,; permanently satisfied unless B 1 O,(x) changes.) At first glance Step (2) does not appear to be necessary since we could eliminate r/~“, wait for y(x) to be moved to a position >0(x), and then go directly to Step (d). However, if p had some cycle stalled in Step (c) waiting for y(x) to move, then strategies of lower priority would have no way of guessing the effect of /3’s future action because they would have no way of predicting the future position of y(x). Also, as mentioned in the description of Step (c^) and as discussed below, /3’s actions must be coordinated with those of higher priority strategies. In particular, we need to ensure that if p ever acts to put x into X, in Step (0), then no higher priority strategy will ever destroy the verifying uses on y or z, thus forcing y(x) or a(x) to be returned to a position
(0)< (WI< (a < (4 < (Q < Cc)< (1). We then extend the priority ordering < on T to cycles associated with the same strategy as follows. Let g, and g, be /I-cycles in Steps (k,) and (k2) respectively. Then g, cLg2 if k, < k2, and g,
240
D. Kaddah
and 1, respectively. R,,i is satisfied.
In both of these
On the other hand,
if there
cases we have X, f O,(B)
are infinitely
many permanent
and requirement
cycles then almost
all
in Step (k), for some k such that (0) < (k) < (1).
of these cycles are permanently
If k = w, (i, or 2, then we will be able to prove A recursive since p asks for infinitely many A-permissions and receives at most finitely many of them. If k = d then q”(x)
we will be able to show D cT B by using is permanently
be able to prove
defined
that C+
for infinitely
(Y’Shypotheses
many x. Similarly,
and the fact that if k = c then we will
B.
There are no outcomes on the tree for the permitting steps since, if A is recursive, then we need not carry out the rest of the construction. Thus the infinitary outcomes of the strategy on the tree are c and d, and they are considered injurious to R,. of lower priority than R,. Below these outcomes we eliminate all other R,,i.-strategies since we will be able to prove C, +- B or De G,. B. We note that the N-strategies below the c-outcome are crucial in allowing us to prove that there are infinitely many cycles which are permanently in Step (c), and similarly for the N-strategies below the d-outcome. The definition of the priority ordering is completed as follows. Suppose that p # /3’ and /3’ assumes that p has outcome kl, where 0 is an R,,i-strategy and p’ is any strategy. Let g be a P-cycle. If g is in Step (k2) c (k,) then g < p’, and otherwise g > p’. If /3’ is an R,,,,-strategy and g’ is a /3’-cycle, then g
a’ has lower
= 9$(x)*
priority
than
a, strategy
a’ will redefine
ym(x’) Z= &.(x’)
at
Infima in the d.r.e.
degrees
241
its next expansionary stage. Furthermore, we see that, after the markers have been redefined, the x’-cycle cannot injure the x-cycle if its advances, and vice versa. In particular, if /3 later wins with witness x, then /3’ will never injure r,. In the full construction we require each strategy to coordinate its markers with all infinitary strategies of higher priority and define several marker rules to ensure this. The construction takes place on a tree T c A<@, where A = (0 < 03< d < c < l}. Fix an effective o-ordering of all requirements such that 1. R, precedes R,. if e < e’, 2. R, precedes all R,,i, and 3. R,,i precedes R,,ir if i
Otherwise define L,+ = L, U {R,,i}. (iii) If R = N, t hen (Yhas outcome 0. Define L,+ = L, U {N,}. Assign to node LL+E T the highest priority requirement which is not on the list L (Y+. Note that, for each e and each infinite path f on T, there is a longest R,-strategy (Yc f,which we call the final R,-strategy on f. For an R,i-strategy /I define r(p) to be the R,-strategy c p of longest length. For any strategy /3 define J-C(~)to be the longest strategy /3’ so that p’-(c) c_ /3 or P’-(d) c p, if such a p’ exists. If /3’ does not exist then n(p) is undefined. We note that the definition n(p) is never used if /3 is an N-strategy. For all strategies /3 which have markers associated with them we enforce the First Marker Rule. If E(P)J then all p markers must be placed on t&B)-marker positions, where k = c if J@)-(c) c p and k = d if Q)-(d) G /3. This sharing of marker positions presents a minor technical problem: It is possible that some position may be used up by being put into and taken out of B. Thus we assign use blocks to the markers. In fact we only assign these blocks to primary q-markers, which are defined to be those W-markers associated with strategies /3 for which n(P) 1s undefined. Note that these are the strategies which do not have to obey the first marker rule, and thus do not have to coordinate their markers with those of any higher priority requirement. All other R,,;strategies’ markers will inherit their use blocks from the markers whose positions they share, and we enforce the
D. Kaddah
242
Second Marker Rule. If rc(/?) is defined, where IpI = IZ, and /3 wants to place a P-marker on a n(P) marker position, then the n(P)-marker must have a use block
of size at least s - 4”+‘, where
s is the most
recent
stage
at which
p was
initialized.
The purpose marker
of this rule is to ensure
are large enough
to allow
that the use blocks
all strategies
to act which
associated need
with each
to act. In the
construction only an &-strategy will want to remove an element from B, and, once it does this, it is satisfied and does not act again unless initialized. Furthermore, no two markers attached to the same strategy will ever share a position. Thus, since there are <4”+’ strategies of length
Third Marker Rule. If a primary marker W(x) is to be (re)defined, then it is assigned a use block size of n, where n is larger than ever before. Its new position must be greater than its essential use plus n.
If we are able to (re)define q(x) without violating any of the first three rules, then the use block for q(x) is defined to be the interval (I/J(X) -n,
marker q(x)].
lnfima in the d.r.e. degrees
Now suppose
a non-primary
Then we enforce
marker
G(x)
243
is to be (re)defined
on a @position.
the
Fourth Marker Rule.
If $(x)
essential
must be less than the smallest
use for 4(x)
If such a y exists,
then choose
is defined the largest
to the same position element
or node
on the tree
then
the
use block.
such.
If we are able to (re)define I&X) = r/~(y) without violating marker rules, then both markers share the same use block. Let cy be a path
as q(y), in q(y)‘s
any of the other
and let 6 be an Q-strategy
such that
p^( k) c my,where k E {c, d}. Then we say t& is a marker on (Y. Let /? be an &-strategy and suppose that Q(B; x)J = 0 for /3’s current unrealized witness. We say the x-cycle is ready to progress to Step (c) if /3 is able to define the marker I/$(X) without violating the marker rules. Let k E {c, d}. When an &i-strategy p puts a number 2 from the use block for p”p(x) into B, it will request that t(P) take control of the markers. This is a two p’ such that step process. First r(p) corrects the t/~~.-markers for all &;atrategies /3’ c r(p). It does this by moving all such markers which share a position with &(x) together to some new large position. Also, I/J~,(Y) is declared to be undefined for all y such that .f is smaller than any number in the use block for r/~~(y). The second step of the process corrects the VP.-markers for all R,.,i.-strategies /3’ such that t(P)“(m) E p’ G p. This step must wait until the next @)-expansionary stage since, when /? acts, it generally destroys one of t(P)‘s verifying uses and t(P) must wait to see whether a new verifying computation converges. When carrying out the second step, r(p) moves all such p’-markers which share a position with p#c) together to some new large number m. If 3t(r(p))i then the marker rules require that m be a n(z(/3))-marker position. If t(P) moves racy,, and 2 is smaller than any number in the use block for r#~~.(z), then qs(z) is now undefined (as expected it turns out that these are the z > y). Finally, if k = c then r(p) defines I&X) = m, and if k = d then it defines p;(x) then,
= m. We note that if some marker I&(Z) is moved in either of these steps, at the next @‘)-expansionary stage, yrta.,(z) will be redefined larger by
x(/3’). Similarly, if r&x) is defined by r(p), then at the same stage it will also redefine yT&) larger. Lastly, if r(p) defines pi(x) = m then at the same stage it will redefine both yrCp,(x) and a,&~) larger. When an &-strategy < removes a number from B it will request that all strategies above it immediately reset their markers. Suppose 5 is removing 2, which was put into B at stage s. Then all of the markers on the node c which were moved or declared undefined at s are redefined to their positions at stage s, and any new markers which were not defined at stage s are now permanently undefined. Note that any such new marker was originally defined larger than i, and so we are allowed to do this. Markers ya and oa associated with a c 5 will automatically be reset by LYat its next expansionary stage, if such exists.
244
D. Kaddah
We point out a potential problem with these marker redefinitions and explain why this problem in fact never occurs. Suppose that q&x) G yT(&) and both markers depend on y staying in C. Further suppose that later some lower priority strategy acts to put a small w into B, driving y out of C and causing I&X) to be declared undefined, and that still later w is removed from B, so that I#$(x) is returned to its previous position. Since y left C we may now have Y&X) < r&x), so that, if /l acts to put x into X, then r(p)‘s functional Tis incorrect. But note that, when w is removed from B, the verifying use on y is restored but y $ C since C is a d.r.e. set. Thus r(p)‘s length of agreement drops and it will not have an expansionary stage unless the verifying use on y is again destroyed. But in order for this to happen, another small number would have to enter B, at which point we would again have r/$(x)?. Thus /3 will never again have a chance to act on I&X) and the problem is avoided. At stage s of the construction we will define the approximation to the true path S,, where 1S,l s s. Stage s is an o-stage if (Yc S,. At times we will use a parameter without its stage subscript to denote its current value. Also we may omit a strategy subscript when it causes no confusion. We assume that at each stage at most one element is enumerated in any set which we do not control. For a working on R, we define the length of agreement l(a, s) at stage s to be the greatest x so that for all y
and
D,(Y) =
UB;y),
and (Yis allowed to define y(y) and o(y) without violating the marker rules. The maximum length of agreement at stage s is defined to be m(cx, s) = max{Z(a; t): t
Stage s + 1 is an cY-expansionary stage if it is an a-stage and l(a, s) > m(a, s). For (Y on the true path, the N-strategies will allow us to prove that if x E U,,, at infinitely many a-stages, then x E U,, and similarly for V,. In addition we will show in the proof that & will always eventually be able to define markers without violating the marker rules. Thus we will also be able to prove that if (Yis on the true path and a’s hypotheses are true, then there are infinitely many aexpansionary stages. At the end of each stage we allow the highest priority strategy ~6, which can use an A-permission to act. Suppose y enters A at stage s. An R,-strategy a: is able to use the permission if y < p&x)1 for some x. An R,,i-strategy /3 is able to use the permission if p$(x)l for some x and some k E {c, d, w}, and y is less than the smallest element in this marker’s use block. To initialize a function or functional means to declare it to be undefined on all arguments. In addition, when a p-marker is initialized, its requests are cancelled. To initialize a strategy means to cancel its witnesses and cycles and to initialize its functions and functionals. If p is an R,,i-strategy, /3-(k) G (Y, and a: initializes all 5 such that LY<,_c, then all P-cycles in any state (k’) such that (k) CL(k)) are
Infima in the d.r.e.
initialized define
also.
A satisfied
a parameter
strategy
large means
which
to define
245
degrees
is initialized it bigger
becomes
than
the construction. At each a-expansionary stage s + 1, for &-strategy markers and note that (Y must also follow the marker
unsatisfied.
any number (Y, we rules
yet used in
(re)define
given
To
above.
the We
describe the process for r only since it is similar for E. First, if there is no a-expansionary state t < s + 1 such that r has not been initialized at any stage at, then,
for each x < I(a, s) in order,
the smallest
define
y(x, s + 1) large.
such stage. For each x < E(cu, s) determine
cu-expansionary
Otherwise
whether
fix t to be
there is a smallest
stage u 2 f such that y(x, v)J and
B, r y(x, v) = B, r y (x, V)
and
C, r x = C, r x.
If so redefine y(x, s + 1) = y(x, v). Finally, for each x < l(cu, s) in order such that y(x) has not yet been redefined, define the marker large. Suppose (Y and s + 1 are as in the previous paragraph and let t + 1 be the a-expansionary stage immediately preceding stage s + 1. Further suppose some /3 with r(p) = (Ywas permitted to put a number x in the use block for pi(y) into B at a stage ~3 t + 1, where p;(y) was defined at stage s’ + 1 =Zt + 1. If pi(y) has not been initialized since stage s’ + 1 and (i) k = c and there is some z
Construction Stage 0. Initialize
all nodes.
State s + 1. Define S,,, by induction until either /&+,I = s or some strategy ends the induction. Then check whether some strategy wants to act on A-permission before ending the stage. Let S,,, 1 II = (II. We define (Y+ = &+, r (n + 1) as follows. If (Y is satisfied then set (Y+ = cy-(0). Otherwise define (Y+, if possible, using the appropriate case below and execute the actions mentioned, else end the definition of hs+ 1. Case 1: (Yis an R,-strategy. Pick the first subcase below which applies. (1.1) Stage s + 1 is not a-expansionary. Then a+ = a-( 1). (1.2) a is ready to execute the disagreement strategy for some x. Choose the smallest such x and define pm(x) =x. Set S,Y+,= (Y. Initialize r,, &, and all 5‘> a: (1.3) Some highest priority /? has requested that LY= r(p) take control of the markers. Let (Y carry out the second step of this process and then redefine (Y’S markers. Cancel p’s request. Set LY+= a-(m).
246
D. Kaddah
(1.4)
Otherwise.
Redefine
Case 2: 1y is an R,,i-strategy. (2.1)
For some
(necessarily
(Y’Smarkers Then
and set (Y+ = (u”(oo).
pick the first subcase
unique)
x, the x-cycle
below which applies. has progressed
to Step
(w) since the last a-stage (i.e., p”(x) has been defined by r(a) in subcase since the last a-stage). Set Ss+, = a. Initialize all 5 > (Yand a-cycles >(w). (2.2)
Some x-cycle
is ready
to progress
Let the cycle progress for the largest 6 s+l = cy, Initialize all 5; 3 a-(d). (2.3)
to Step (2) (i.e.,
such x (i.e.,
define
@(x)J p”(x)
(1.3)
< o,(,)(x)). = qd(x)).
Set
For some
(necessarily unique) x, the x-cycle has progressed to Step (i.e., q”(x) has been defined by r(a) in subcase (1.3) since the last a-stage.) Then set a+ = (u-(d).
(d) since the last (Y stage.
(2.4) Some x-cycle is ready to progress to Step (e) (i.e., I&,“(x)J < rF(&x)). Let the cycle progress for the largest such x (i.e., define p’(x) = r/~‘(x)). Set S,,, = cy, and initialize all c 2 a-(c). (2.5) For some (necessarily unique) x, the x-cycle is ready to progress to Step (c). Let the cycle progress (i.e., let a define r/~‘(x)), and set (Y+ = a-(c). (2.6) Otherwise. If there is no cycle in Step (l), then pick a new large witness x and say the x-cycle is now in Step (1). Let LY+= a-( 1). Case 3: a is an unsatisfied NC-strategy. Initialize all c> a: (thus in effect restraining B r (s + 1)). Strategy a: is now satisjied. Finally, if A,+1 -A,=0, orifyEA,+i-A,but no a~&+, isabletouse this permission, then initialize all c such that as+, <,_ Z; and end the stage. Otherwise fix the highest priority node or cycle which is able to use the permission. Let a denote this node and, if appropriate, let the associated cycle have witness x. Choose the first case below which applies. (i) (Y is an R,-strategy. Then there is some least x such that y < p(x)j,. Remove p(x) from B and initialize all c > (Y. (Note that we always have p(x) =x.) Request that all strategies above a: immediately reset their markers. Strategy (Yis now satisfied. (ii) The x-cycle is in Step (w) (i.e., p”(x) can use the permission). Put x into X,, and the largest fresh number in p”(x)‘s use block into B. Request that r(a) take control of the markers. Initialize all f > (Y. Cancel all of LY’Scycles and declare cy satisfied. (iii) The x-cycle is in Step (2) (i.e., p*(x) can use the permission). Put the largest fresh number in pd(x)‘s use block into B. Request that t(a) take control of the markers. Initialize all c > (Y and cu-cycles ~(2). Put the (iv) The x-cycle is in Step (c^) (i.e., p’( x ) can use the permission). largest fresh number in p’(x)‘s use block into B. Request that t(a) take control of the markers. Initialize all f 3 a-(c) and &-cycles a(?). This ends the construction. Verification Assume A is nonrecursive. We define the true path 6 by induction. Let 6 r 0 = 0. Suppose we have defined 6 1 n = a. Then let 6(n) be the least k so that
Infima in the d.r.e.
a-(k)
E 13, for infinitely
path to be a marker Lemma
degrees
many s, if such k exists.
on the path
247
We define
a marker on the true
6.
4.2. Each node a E 6 is initialized only finitely often and initializes nodes
below it on 6 only finitely often. If a = 6 r n then 6(n) is defined. Thus 16 1= 03. If (Y is an N,-strategy,
then a is satisfied.
Proof. We prove the lemma
by induction,
for all LY’c CY.By the inductive all t a s, a is not initialized
hypotheses
so assume
LY5 6 and the lemma
is true
we can can fix a stage s so that,
by any a’ c a at stage t and 6, # (Y. Thus
for
any node
which initializes (Y after stage s must be to the left of a and must have been waiting for A-permission at stage s, and there are only finitely many of these. Once such a node receives permission it will act at most one more time via permission, so we may fix a stage s’ > s so that (Y is not initialized after state s’. The lemma is now clear if a: is an N,-strategy. Let (Y be an &-strategy. If permitting case (i) is executed by (Y after stage s’ then LYis satisfied forever and S(n) = 0, so assume this does not happen. If (Y executes subcase (1.2) infinitely often then we claim A is recursive: To determine whether n E A, find a stage t b-s’ and an x such that p,(x)J > n, and then n E a e n E A,. Thus LYinitializes nodes below it at most finitely often and 6(n) exists. Now let a be an &-strategy. If a executes permitting case (ii) at any stage t >s’ then it is satisfied forever and 6(n) = 0, so assume this does not happen. Strategy a cannot execute subcases (2.1) or (2.2) (and thus permitting case (iii)) infinitely often, since then A would be recursive. To see this, use the same argument as in the previous paragraph but also require that n is less than the p-marker’s use block. So assume permitting cases (ii) and (iii) and subcases (2.1) and (2.2) are only executed finitely often for (Y. Now if LYexecutes subcase (2.3) infinitely often then 6(n) = d and the lemma is proved, so assume this does not happen. Finally, many permissions
if (Y executes are received
subcase (2.4) infinitely often, then for pc and A is recursive as before.
only finitely Thus in any
case 6(n) is defined. Node (Ymay initialize nodes below a-(c) infinitely subcase (2.4), but then s(n) G d. Thus the lemma is true for a. Cl
often via
Lemma 4.3. Suppose V(X) is a VW-marker on the true path, or V(X) = ye(x) o,(x) where a-(m) c 6. Then Y(X) is moved finitely often and 3”s (Y(X, s)?)
+
or
3s Vt 2s (Y(X, t)J,).
Proof. We drop the subscript LYin the following. Note that Y can only be initialized by nodes or cycles to its left. Using Lemma 4.2 and the fact that Y is a marker on the true path, we may fix a stage s so that Y is never again initialized and, if Y is y or CJ, then
D. Kaddah
248
Also we may fix a stage t > s so that Y(X, t)i, Now let 5 be some N,-strategy Again
by Lemma
since otherwise the lemma is clear. so that (Yc c c 6 and 5 has not acted by stage t.
4.2, we may fix a stage n > t so that
stage r~ and is satisfied
forever.
Then
no element
5; acts for the final time at
or leave
B after
stage V. This is because c initializes all nodes and cycles > < and if any node or cycle
or undefined
after stage u if B changes
on some element
less than the
marker, and this will not happen. If Y(X, u)? then the marker can only be redefined if it is returned to some previous position, and this will only happen if some element less than the marker’s last defined position leaves of these previous positions are
B. Note that all will
never
be
0
Let (Y be an &-strategy with infinite outcome on the true path. Note that (Y redefines yIy and a, at each a-expansionary stage if necessary. From Lemma 4.3 we know that these markers reach a limit for each x. The rules for redefining markers
automatically lim y(-,
s
Next we prove Lemma
guarantee
s) Gr B G3C,
similar
and
lim a(-,
s
s) +. B 63 0,.
facts for q-markers.
4.4. Let /3 be an R,,i-strategy
with outcome k E {c, d} on the true path.
Then lim #‘(-,
s
s) =+ B
and
3”~ 3s Vt > s @(x, t)J.
Proof. We only prove the lemma for the case k = c, since the other case is similar. By Lemma 4.2 we may fix a least stage s such that 1~,= q’p is not initialized at any stage t 2 s. Fix X. First we prove that lim, I/J(X, s) + B. Note that V(X) is a marker on the true path so it satisfies the conditions in Lemma 4.3. Now B can determine whether I/J(X) is undefined at all stages t s s by searching for any stage t s s such that V(Y, t)l f or some y > X. If v(x) has not yet been defined by stage t, then q(x) will never be defined, so assume there is some least stage t s s such that V(x, 94. Note that no node or cycle
lnjima
q-markers
to their
positions
in the d.r.e.
degrees
249
at stage s’ and declares
defined at stages 3s ‘. Thus, if ZJ< n’ are p-( c)-stages
undefined
all markers
first
u’>J=
~1,
such that t s u, then
(r/G,
~J)L and (B,,
1 144~ ~1) = (B, 1 VG
(r/G,
v)? and r/G,
u’)l)
~1)) +
Vk
VG
and 3
3~ < 21(Y E B, - B,,),
and so lim, r/~(x, s) + B. We also note limit using a B-oracle is uniform in n. Finally
we show that there
that
are infinitely
this procedure
many
+-markers
for determining
the
which are eventually
always defined. Suppose this is false and fix a stage s after which the finitely many permanently defined markers are fixed. Now let x be the largest number so that the marker I/J(X) is permanently defined. Let < be any N,-strategy on the true path such that P-(c) E 5; and 5 1s . not yet satisfied at stage s, and let t’ > s be the stage at which c acts for the last time. We know that stage t’ exists by Lemma 4.2. Then there must be some y >x such that 1+9(y, t’)L since 5 is below the c-outcome of /3. As in the proof of Lemma 4.3 we see that rj~(y, t’) is fixed at its present position because no node or cycle of higher priority than 5 will ever act and I$’initializes all nodes and cycles of lower priority. 0 Lemma
4.5.
Let
LX be an R,-strategy
on the true path.
If a’s hypotheses
are
satisfied then a-( 1) $6. Proof. We drop the subscripts on C and D in the following. Assume (Y’S hypotheses are satisfied and suppose, by way of contradiction, that cu-( 1) c 6. Note that, if x E C, then there is some stage se(x) such that K(B;x,
sc(x))l
and
Vt 2s (B, 1 &(B;x,
se(x))
= B 1 &(B;x,
We similarly define sD(x) for x E D. Fix the largest x so that at infinitely a-stages s we have I( a; s) = x. Finally fix a smallest stage s such that 1. (Yis not initialized at any stage t 2 s,
se(x))). many
2. for all y 6x, if y E C, then y E C, and s > s”(y), and similarly for D, and 3. for all y =Gx and all t 3 s, if y 4 C then y 4 C,, and similarly for D. Let g be an N-strategy on the true path so that a-( 1) G < and < has not acted by stage s. Fix an CY-( 1)-stage t > s so that c acts for the final time at stage t. Since l(c~, t) SX, there must be some y GX such that y E (U,JB) - Ct) or y E (V&(B) - 0,). Now, as in the proof of Lemma 4.3, we have B, 1 t = B 1 t. Thus in the first case we have y E U, and in the second we have y E V,. In either case a’s hypotheses are false, a contradiction. Finally, if X(LY)J then, by Lemma 4.4, (Y will always eventually be able to define its markers without violating the marker rules. Thus LY-( 1) 4 6. 0 Lemma 4.6. Let p c 6 be an Reli-strategy /3-(O) # 6, then R,,i is satisfied.
and suppose
R,‘s hypotheses
are true.
If
250
D. Kaddah
Proof.
Note that for this lemma we do not need to assume that r(p) is the final &-strategy on 6. First suppose p-( 1) c 6. Then R,,i is clearly satisfied unless p has some permanently uncancelled witness x so that @(B;x)J = 0, but the x-cycle is never able to progress to Step (c). If qi is a primary marker then this cannot happen, so assume n(p)J. By Lemma 4.4, there are infinitely many permanently defined qZurJ-markers on 6 so the x-cycle will eventually be able to progress. Thus R,,i is satisfied since its hypothesis is not. Next suppose /3-(c) c 6. (The argument for P-(d) c 6 is similar.) Fix a stage s so that I&; is never again initialized. To determine whether IZE C,, use a B-oracle to find a P-stage t >s and an x > n so that ~Cp(x,t)l = lim, q@). Lemma 4.4 ensures that this is possible. Using the fact that R,‘s hypotheses are true, it is clear that, if IZE C,,,, then B can determine whether n E C,. On the other hand, if 12$ C,,, then it 4 C,, since otherwise we would define pi(y) for some y ax and initialize $$. Thus C, + B and R,,i is satisfied. 0 Lemma m-(m)
4.7. Let cz be the final R,-strategy c 6 and p c S is an R,,i-strategy
on 6. Then R, is satisjied.
Zf in addition
with t(P) = a; then R,,i is satisfied.
Proof.
Using Lemma 4.5, the first part of this lemma is clear if (u-( 1) c 6. Using Lemma 4.2 and the definition of the disagreement strategy, it is also clear if cum(O) c 6, because if a executes this strategy, then its disagreement will not be injured unless a is also initialized. So assume (u-(~) c 6 and fix a least stage s such that, for all stages t 2 s, we have 6, +,_ a-(m) and 1~ is not initialized at stage t. We drop most strategy subscripts in the following. Suppose that p acts to put x into X, where r(p) = a, but do not necessarily assume that /3 c 6. We now examine in some detail the sequence of events leading up to this action. Note that x’s cycle cannot be initialized at any time during this sequence since otherwise x would be cancelled and p would never enumerate x into X. Also recall that a strategy always defines a marker position as large as possible. Let s(O) >s be the stage at which p picks witness x, and s(l) the stage at which it defines v’(x) > B(B;x, s(l)). Also let s(2) be the stage at which /3 defines p’(x) = V(x, s(2)) < Y(X, s(2)). Th en, because the x-cycle has not been initialized by stage s(2), we have B,(l) r B(B; x, s(l)) = R(z) r o(B; x, 4)). At stages s(2) all nodes and cycles >fi-(c) are initialized. Let s(3) be the stage at which p”(x) receives A-permission to put some element n from its use block into B. Then, because of the initialization at stage s(2) and the fact that the x-cycle remains uncancelled at stage s(3), we have (K(Z) r Y(X, 42)))
U {n> = K(X) 1 vk
s(2)).
Note that this implies B,(i) 1 o(B; x, s(l)) = K(3) 1 o(B; x, s(l)).
lnjima in the d.r.e.
251
degrees
are again initialized and p requests At stage s(3) all nodes and cycles ~/3-(c) that a take control of the markers. At the next a-expansionary stage s(4), if (Y does not act via the disagreement strategy, then it moves all markers on the path between greater
cu”(oo) and /? which shared than
a position
with v’(x,
s(2))
to a new position
the new verifying
use on y(x). Also LYassigns r@‘(x) to this same position and redefines y(x, s(4)) even larger. Note that any new cycles which are started at stages >s(4) by strategies p’ such that (Yc j3’ c p will pick witnesses larger than this new position. We claim that if y >x and y E CSc2) then executed
the disagreement (R,,,,
1 Ye> 43))
strategy, u W(x>>
y E CSc4). Otherwise
(Y would
have
unless #K(4)
r Yk
(9)
m).
But if equation (9) is true, then some node or cycle
then
p(x) G y(x, s(6)), Also note that p(x)
o(x, s(6)),
and
R,+) 1 e(B; x, s(l))
= Kclj
1 @B;x,
4)).
is defined
larger than the essential uses for y(x) and a(x) at we will say that y is critical. We claim that stage s(5). If x Gy G p(x) R.&Y) = R(Y) f or all critical y. Note that if this claim is true then T(x) and Z(x) are correct because witnesses are always chosen large, so that y(x) and a(x) have not yet been defined at stage s(O) and will always be defined to be larger than if the claim is true and /3 c 6 then requirement R,,iis s(0). Furthermore, satisfied, because if /3 ever acts on a witness x and is never initialized afterwards, then Rsc6) r p(x) = B 1 p(x) and thus O,(B; x) = 0 while x E X. We assume in the rest of the proof that B 1 x = Bsc6) 1 x, because otherwise it is clear that T(x) and Z(x) are correct. To prove the claim we first note that at stage s(6) all nodes and cycles >p are initialized so that no such node or cycle will ever again change Bsc6) r s(6). Also, because the x-cycle is not cancelled at any stage as(6), we know that if any node of cycle CL p changes B(y) at at stage >s(6), then either y
D. Kaddah
252
after
this initialization.
when
/3’ defines
critical.
p;,(z),
On the other
Thus y E B for each critical
Finally
we show
then there
that
hand,
and thus y
if P’-(k)
contradicting
G j3, then
c 6 and
t)i and w >s(6)
t > s(6)
is some
z so that
t&(z)J
= p(x).
is a P’-(k)-stage,
for each w in this marker’s
use block. Note that this will finish the proof of the claim chooses the largest r&-marker when defining a &-marker. there
that y is
y E BsC6). E /3’-(k)
is some z such that t&z,
stage s(6)
J3 is initialized
our assumption
because /3’ always First note that at
At the next
a-expansionary
stage s(7) >s(6), strategy a, moves I&(Z) and all other markers along /3 below (u”(w) which share a position with p(x) together to a new large position. Now at each p’-( k)-stage as(7) we will have q$(z) 2 $$,(z, s(7)) unless I&(Z) has been declared undefined, so suppose this happens at stage u 2 s(7). If Q!&(Z) is declared undefined because all /3’-cycles in Step (k) are initialized then the result is clear, so assume that the marker is declared undefined because enumerates a number y into B, where y is in the use some node or cycle ~/3’-(k) block for &(z’) < q;.(z). Note that as long as I&(x)~, it is the smallest &.-marker which is >s(6). Thus the node or cycle which enumerates y must be -CUP and we must have r&(z’)
0
This establishes
Theorem
4.1.
0
Next we show that not all r.e. degrees are the infima of two d.r.e. degrees. In the following theorem we will build a low, r.e. degree which is nonbranching in D2. The construction and proof for this theorem are similar to those of Theorem 4.1, and we will use some of the same notation and strategies. Theorem
4.8. There is a low, r.e. degree b which is nonbranching
in the d.r.e.
degrees. Proof. We will build an r.e. set B whose degree satisfies the theorem. Below we indicate the similarities to the construction for Theorem 4.1 and discuss in detail the differences. We will use N-requirements to force B to be of low, degree. These requirements are essential to the proof and they are of a different form than those in Theorem 4.1. Suppose a, has been assigned to requirement N,. Roughly, we will say that an a-stage is expansionary if 1We(B)1 is larger than ever before. Then requirement N, is as follows: N,:
3” a-expansionary
stages
e
1W,(B)1 = 00.
Injima
in the d.r.e.
253
degrees
At each a-expansionary stage there will be some new x in W,(B) with verifying use cpe(B;x). Strategy a tries to keep x in the set by defining a marker qIol(-lc) larger than the verifying use and restraining B. These markers will never be enumerated
into B, but otherwise
markers
in Theorem
markers
associated
4.1.
they function
There
with different
will
strategies
in the end, we will be able to prove
very much
be no
actual
like the R,,j-strategy
a-restraint
will be coordinated
that there
are infinitely
stages if and only if there are infinitely many permanently An N-strategy a has the two outcomes 00< 1. Suppose
B.
Instead
as before
on
so that,
many
a-expansionary
defined a-markers. that each N-strategy
on
the true path satisfies its requirement. Then we claim that B is low,. To see this note that there will be a unique strategy on the true path associated with W,(B). Thus we will have FinE ef {e: Wt
is finite} =+ 0”.
Now FinB is 2:-complete, so B” c1 FinB and thus B” ST@“. We next turn to the nonbranching requirements. Let (Y work on requirement R,. In Theorem 4.1, the strength of the N-requirements allowed us to prove that if (Yc 6 and a?s hypotheses were true, then a-( 1) # 6. The N- strategies in this proof do not guarantee the same result and so we weaken R,‘s hypotheses: R,:
C, E U,(B) 3X,,
and
0, c V,(B)
+
&, ,I$ (Xc is r.e. and X, = T,(B 63 C,), ,Ze(B 43 D,)).
The requirement’s hypotheses are used to define l(a; s), m(cu, s), and (Yexpansionary stages much as before, although we delay the formal definitions until after the marker rules are given. Care must be taken to ensure that the functionals r = r, and 2 = Z, are well-defined. The marker y(x) is moved whenever some y < x enters C = C,. It is also moved if some y s y(x) enters B and no substrategy of a: has an uncancelled request that the marker not be moved. We will be able to show that if (Y’S hypotheses are true then there are infinitely many such uncancelled requests, so that y(x) reaches a limit. The rules for moving a(x) are analogous. Strategy a may still need to redefine these markers to earlier positions when numbers leave C or D. Since a cannot remove numbers from B, it has no disagreement strategy. Thus the permitting function pa is no longer needed and a has only the two outcomes m < 1. Below the infinitary outcome of an R,-strategy a: we again have subrequirements: R,,i:
X, = O;(B)
3
C,+
B
or
D,+B.
Let /I be an R,,;-substrategy of R,-strategy (Y. The basic cycle from Theorem 4.1 must be altered in several ways. We eliminate the permitting functions pk and the permitting steps (2) and (2). Previously, if the x-cycle executed a permitting step, say (e), then either the cycle advanced (if a new verifying use was found) or (Y
254
took
D. Kaddah
over
and
construction disagreements.
tried
to create
the x-cycle
a disagreement
may act and achieve
However,
(if some neither
each time this happens
result
y
left
C).
since a: cannot
some y
must
leave
In this create C, and
so we will be able to prove that no marker is enumerated into B infinitely often. Here the secondary function of the P-markers is to drive numbers out of C or D, enabling
us to prove
We now describe
C+
B or D +
B if p fails to diagonalize.
the basic cycle for p. Note
states and will never return Step (d).
to a previous
(1) Pick a new large witness say x has been realized.
state,
that the cycle can be in one of 4 although
it may repeat
x. Wait for O@(B, x)s = 0. When
Step (c) or
this happens
we
(c) Define q”(x) > e,(x). (Note that if C 1 x changes then we can move y(x) to a new position >$J’(x).) Wait for some y
then y(x) and a(x) will never need to be returned both y(x) and a(x) have been redefined by a uses on y and z.) We say /3 is satisfied. (We now unless B r e,(x) changes.) Theorem 4.1 the outcomes of the cycle are 0
to a previous position to be larger than the have R,,i permanently < d < 1. The priority
ordering is extended to cycles as before. The tree T is defined as in the previous proof, except that now R,- and N,-strategies both have the two outcomes 00< 1. We also use the same definition of a marker on path or node a, with the addition that, if /3 is an NC-strategy and p-(w) E (Y, then $J~ is a marker on (Y. The essential use for markers r/~,“,qd, y, and o is defined as before. We note that, if p is an N,-strategy, then r/~~(x) will only be defined if n E W,(B), in which case the essential use for VP(x) is defined to be the verifying use on x. For R,,i-strategies p, the function r(p) is defined as before. For any strategy (Y we define n(a) to be the longest R,,i- or N,-strategy /3 c cy which a guesses to have infinite outcome, if such exists. We eliminate the use blocks from Theorem 4.1 because elements can no
lnfima in the d. r. e. degrees
255
longer be extracted from B. This makes the marker rules considerably simpler. The first marker rule is still enforced, namely if 3t(p)l, then P-markers must share positions with n(p)-markers on the path /3. To ensure no marker is moved infinitely often we enforce two additional marker rules: If I/31= IZ, then p must avoid the first IZmarkers associated with n(p) on the path /3, and if /3 is initialized at stage s or if some p-marker is moved at stage S, then all P-markers first defined after stage s must avoid the first s markers associated with J@) on the path p. For a working on R, we define the length of agreement I(&, s) to be the largest x so that for every y
Y e 4,
+ Y e K,(B),
and a is allowed to define y(y) and a(y) larger than their essential uses without violating the marker rules. We define m(a, s) and cu-expansionary stages as before. Let s + 1 be an a-expansionary stage. The strategy (re)defines its markers for each x < l((~, s) as follows. First, if a has been initialized since the most recent a-expansionary stage then, for each x in order, define y(x, s + 1) large. Otherwise fix t to be the smallest a-expansionary stage such that LXhas not been initialized at any stage at. For each x determine whether there is a smallest a-expansionary stage v 2 t such that y(x, v)J and B,, /y(x,v)=B,
ry(x,~)
and
C,, rx=C,
rx.
If so redefine y(x, s + 1) = y(x, v). Now let s’ + 1
256
D. Kaddah
advances. However, in this construction the procedure becomes more complicated if the x-cycle does not advance because the markers associated with each 5 2 /3-(k) must be corrected. In particular we must ensure that if r(E) c a and (Y moves t&y), then at the end of the procedure we still have r/.$(y) G yrYrts,(y).If n(a)l this is accomplished by linking the new position of the &markers to the position of Y(W). There is the danger that Y(W) may become undefined before a is able to complete the procedure, but we will show that (Y must have been initalized in that case. If ~(a)? then a simply moves the E-markers to a position which is not too large, for example the position s + 1. We give the details of the procedure in the following two paragraphs. Whenever some marker Y(Z) is moved, all Y(W) are declared to be undefined for all w > z. Also, if (Yis initialized before it has completed all steps, then the rest of the procedure is cancelled. To execute the first step, a immediately corrects all markers 1+9~.(y)on the path a which share a position with r&(x) by moving them together to some new large position. We note that t(j3’) will redefine yscBrJ(y) large at the next @‘)-expansionary stage if r&(y) is moved. Also note that if n( (Y)Tthen there are no such markers qa(y), and thus this step is not needed. Strategy cr must wait for its next expansionary stage t + 1 > s + 1 to execute the second step. We give the details in outline form in order to make the cases clearer. First assume k = c. Form the set S of all VP.-markers on the path p, where (Yc /3’ G 0 and the marker shares a position with r/$(x). Case (I): C, r x = C, 1 x. In this case the x-cycle is ready to progress to Step (d) and we may ignore all nodes 5 1 p-(c) . M ove all of the markers in the set S to some new large position and define r/$(x) to be this number. Note that this position will be 2 the position used in the first step. Case (II): otherwise. In this case the x-cycle is not ready to advance and thus we must correct all markers for E 2 /3-(c). Form the set S’ of all such vE-markers which share a position with r#i(x). Subcase (i): PDF. Move all markers in S U S’ to position s + 1. Note that, for any 5 and z such that I&(Z) ES’, the marker y&z) will already have been moved by r(E) to a position 2s + 1. Subcase (ii): otherwise. Move all markers in S U S’ to the current position of Y(W). We will show in the proof that, for any g and z such that r&(z) ES’, we must have ‘y&z) > Y(W) else /3 would have been initialized.
The procedure is analogous for k = d. In this case, (I) may be executed twice for the x-cycle: once after the cycle enumerates I&X) into B in Step (d), and again after the cycle enumerates r/&x) into B in Step (0). When (I) is executed for the first time for the x-cycle, we say that the cycle is ready to execute Step (0). When (I) is executed for the second time for the x-cycle, j3 requests that for all y GX, y,(y) and o,(x) not be moved, and we say that p is satisfied.
Infima in the d.r.e.
257
degrees
Construction Stage 0. Initialize
all nodes.
until either 16,+11 = s or some strategy ends Stage s + 1. Define &+i by induction the induction. Let 6,+i 10 = 0. If S,,, 1 IZ = a, then define S,,, 1 (n + 1) = a+ as follows. If a is satisfied then set ay+ = a-(O). Otherwise define my+, if possible, using the appropriate
case below
definition. Case 1: (Yis un R,-strategy.
and execute
the actions
Pick the first subcase
mentioned,
else end the
below which applies.
(1.1) Stage s + 1 is not a-expansionary. Then a(+ = (w-( 1). (1.2) Some highest priority 6 has requested that (Y= r(p) take control of the markers. Let o carry out the second step of this process and then redefine a’s markers. Cancel p’s request. Set cy+ = LY-( m). (1.3) Otherwise. Redefine a’s markers and set (Y+ = a-(m). Case 2: IXis an R,,i-strategy. Pick the first subcase below which applies. (2.1) For some (necessarily unique) x, the x-cycle is ready to execute Step (0). Let the cycle act (i.e., enumerate x into XZca, and q”(x) into B, and request that t(m) take control of the markers). Strategy (Y is not satisfied. Set S,,, = a. Initialize all 5‘> (Yand cancel all a-cycles. (2.2) Some x-cycle is ready to execute Step (d) (i.e., @(x)l< r+,(x)). Let the cycle act for the largest such x (i.e., enumerate r/~“(x) into B, and request that r(a) take control of the markers). Set S,,, = (Yand initialize all 5 2 a-(c). (2.3) For some (necessarily unique) x, the x-cycle is ready to progress to Step (d) (i.e., $J~(x) has been defined by r(a) in subcase (1.2) since the most recent LYstage). Set a+ = a-(d). (2.4) Some x-cycle is ready to execute Step (c) (i.e., IJJ’(X)J < JJ~(~,(x)). Let the cycle act for the largest such x (i.e., enumerate q’(x) into B, and request that r(a) take control of the markers). Set S,,, = (Yand initialize all 5‘2 a-( 1). (2.5) For some (necessarily unique) x, the x-cycle is ready to progress to Step (c). Let the cycle progress (i.e., let (Ydefine q’“(x)). Set LY+= a-(c). (2.6) Otherwise. If there is no cycle in Step (l), then pick a new large witness x and say the x-cycle is now in Step (1). Set &+ = a-( 1). Case 3: (Yis an NC-strategy. Pick the first subcase below which applies. (3.1) Stage s + 1 is a-expansionary. Let (Y define the new marker and set Ly+= a_(m). (3.2) Otherwise. Set my+= a^( 1). Finally, initialize all 5‘ such that asfl < L 5‘ and end the stage. This ends the construction. Verification We define the true path 6 by induction. Let 6 10 = 0. Suppose we have defined 6 112 = (Y. Then let 6(n) be the least k so that a-(k) G 6, for infinitely many S, if such k exists. We define a marker on the true path to be a marker on the path 6.
258
Lemma
D. Kaddah
4.9. Zf LY= 6 1 n then 6(n)
is defined and (Y initializes each /3 2 a-(n)
at
most finitely often. Thus 161 = w and each node on 6 is initialized at most finitely often. Proof. We prove the lemma by induction, so assume a: E 6 and the lemma is true for each a’ c LY.By the inductive hypotheses we can can fix a stage s so that, for all t 2 s, a is not initialized at stage t. If LYis an N,- or &-strategy then the lemma is clear. Suppose o! is an &-strategy. If a executes subcase (2.1) at any stage t > s then it is satisfied forever and 6(n) = 0, so assume this does not happen. Suppose (Y executes subcase (2.2) infinitely often. Then we claim a also executes subcase (2.3) infinitely often. If the claim is false then there are only finitely many x so that @(x) is ever defined, and thus a must execute subcase (2.2) infinitely often for some x. But, since a! never advances to subcase (2.1), some y
(we say such a
Fix a P-marker r+!~ on 6. Note that, by the procedure for taking control of markers, q(x) is only moved when the current position of v(x) is enumerated in /3, and if some y < I&(X)is enumerated in B then q(x) is undefined from then on. We first claim that any &-strategy I; will only enumerate finitely many numbers into B for any fixed cycle. To see this, note that c enumerates a marker ~~(2) into B only if I#~(z) = $$(z) < y&z) or r&(z) = r&z) < o,o,(z). In either case the z-cycle advances unless some z’
Injima in the d.r.e.
degrees
259
on q(x), and from above we know that /I’ will only enumerate V(X) into B finitely often for the z-cycle. Thus we may fix a stage s so that no p’ 2 p enumerates V(X) into B at any stage 2s. Finally, for /3’ c /I, if p’ enumerates q(x) into B after stage s, then the cycle for /I’ does not advance else /3 would be initialized. Thus r(p’) moves all markers which share ~(x)‘s position together to some new position. By the inductive hypotheses, v(x) will only be moved finitely often. Now we show that (i) implies (ii). If (ii) is false then, by (i), we may fix a least stage s so that the finitely many permanently defined q-markers have finished moving. Suppose p is an &-strategy and qj = v’ (the other two cases are similar). By assumption /3-(c) c 6 and at the next /I-( c)-stage after s some new 1+9(y)will be defined. Note that all 0’ such that P-(c) c/I’ place their markers on r/$-markers. By definition of stage s, no such p’ will enumerate a marker position
4.11. Every
N, is satisjied.
Proof.
Fix 6 c 6 so that /3 works on N,. Suppose W,(B) is an infinite set. Then there are infinitely many P-stages at which IW,(B)I is larger than ever before. At infinitely many of these stages /3 will be able to define a new marker, where we use Lemma 4.10 to prove this if n(P)j,. Thus ,6-(m) c 6. Now assume /I-( co) c 6. Again by Lemma 4.10, there are infinitely many permanently defined P-markers. Note that when We is first defined we have q&) ’ $e(B; .x). Th us if r+!+r(x)is a permanently defined marker then x E W,(B) and so IW,(B)I = 03. q Lemma 4.12. Suppose R,‘s hypotheses are satisfied. such that p-(O) # 6. Then R,,i is satisjied.
Let /3 c 6 be an R,,i-strategy
Proof. Let /I’ be the immediate successor of j? on 6. Fix a stage s so that for all t 3 s, /3 is not initialized at stage t and 6, #,_ B’. Let C = C,. Suppose /I’ = /I-( 1). If R,,i is not satisfied then there must be some realized
witness x so that /3 is never able able to progress to Step (c). But by Lemma 4.10 this cannot happen. Now suppose /-I’ = /l-(c). (Th e case where /3 has outcome d is similar.) We claim C cT B. By Lemma 4.10 there are infinitely many permanently defined t/$-markers. From the method for moving these markers it is clear that
260
D. Kaddah
lim, I#‘;(-, s) + B. Then
to determine
whether
n E C, use a B-oracle
to find an
x > n and a stage t so that r,!$j(x, t)J and
Note that if n E C,, then p can determine
whether
n E C because
R,‘s hypotheses
are satisfied. On the other hand, if n E C - C,, then some y-cycle, where y ax, would advance to Step (d), say at stage u > t, and we would have 6,
proving
R,-markers, similarly
the next lemma
we automatically
we remark
have lim, y(-,
that,
from the method
for moving
B @ C, if this limit exists,
s) +
and
for u and B G3D.
Lemma 4.13. Let LXbe the final R,-strategy on 6. Zf (Y+ = a-( 1) c 6 then R, is satisfied. Zf R,‘s hypotheses are satisfied, then a+ = (u-(m) c 6 and (Y’S markers are moved finitely often. Proof. Note that if R,‘s hypotheses are satisfied then a will always eventually be able to define markers by Lemma 4.10. Thus if (Y+ = a-( 1)) then R, is satisfied since its hypotheses are not. Now assume that R,‘s hypotheses are satisfied, show that y(x) and a(x) are only moved finitely every stage t 2 s, cx is not initialized, C,,, 1 x = C,,, 1 x There
are infinitely @(B;y)J
many =O
and indices
o,,,
1 x = Q,,
and so (Y+ = cr-( m) . Fix x. We often. Fix a stage s so that, at
1x.
i so that for all y,
with use 8i(B;y)
~0.
Fix such an index i >s and an R,,i-strategy #I c 6 so that r(p) = LYand p has not yet acted at stage s. Then each p-witness will be >x. Now /3 will want to act on each of its witnesses y and, by Lemma 4.10, the y-cycle will advance to Step (c). Thus /3-( 1) # 6. By Lemma 4.12 and our assumptions on R,, /3-(c) # 6 and /T(d) $6, and so we may fix a stage t > s so that j3 acts for the final time to put witness y >x into B. At the next a-expansionary stage >t, all p’-markers on the path /?, where a-(m) c p’, are moved together to a new large n(m)-marker position m. Also, /3 requests that the markers y(z) and a(z) not be moved for all z < y, so in particular m > y(x), a(x). Note that this request will not be cancelled since otherwise j? would be initialized. By the definition of stage t and the fact that each strategy chooses the largest possible marker position when it acts, we know that no p’ G 6 will act again using a position
in the d.r.e.
Infima
261
degrees
Lemma 4.14. Let a, be the final R,-strategy on 6 and suppose the requirement’s hypotheses are true. Then R, is satisfied. Let 0 be an R,,i-strategy such that t(P) = cx and /3-(O) c 6. Then R,.i is satisfied. Proof. By Lemma We must prove
4.13 we know
a-(m)
that these functionals
c 6 and r, and 2, are total functionals.
are correct.
Fix a smallest
stage s so that a
is not initialized at any stage t 2 s. We drop most subscripts in the following. Let /3 be any R,,i-strategy with r(p) = (Y, but do not yet assume that /3 c 6. Let x be a witness and
a(x)
for /3. We first prove
are moved
when
necessary.
a useful
fact, namely
Suppose
I/&X)
that the markers
is enumerated
into
Y(x) B at
stage s’ and the x-cycle is not initialized. Then we must show that, at stage s’, no a-substrategy p’ has an uncancelled request that the a-markers y(x) and a(x) not be moved. Suppose otherwise and note that p’ must have made this request at a stage t when it executed subcase (2.1) for some y-cycle, where y > x. From the definition of the tree and the fact that y > x we must have /3-( 1) =Gp’. Now if r&(x) is enumerated into B by p or by some 5 2 /3-(d), then p’ would be initialized and its request cancelled. The only other possiblity is that r&(x) is enumerated into B by some Z+cycle in Step (k), where c-(k) G p. Suppose this is true and let r&(z) be the marker which shares a position with r&x). We claim that in this case /I?’ was initialized at a stage IJ such that t < u s s’. This is clear if c-(k) cL/3’, so assume g-(k) c p’. Then at stage t there was some t&z’) which shared a position with +$(y). By the construction we have z < z’. Then r&z’, s’)f because otherwise E would have chosen this marker for its action at stage s ’ . But r+!$(z;, s’)? implies that /3’ has been initialized at some stage u with t < v 6s’. Thus the m-markers will be moved whenever I&(X) is enumerated into B. Now suppose /3 picks witness x at stage s(O) > s and enumerates x into X, at stage s(6). We examine in detail the sequence of events leading up to this action. Fix a stage s(l) >s(O) at which p defines v”(x), so that $“(x, s(l))
’ oi(R; x, s(l)).
Then we have
B~(I, I ei(B; XT s(l)) = Bs(6)I ei(B; x, s(l)) since otherwise some higher priority node or cycle must x-cycle would have been cancelled. Now fix stages s(4) > s(3) >s(2) > s(1) so that /3 executes final time at stage s(2), (Y defines vd(x, s(3)) at the next s(3), and p executes subcase (2.2) for the final time at stage rI%
s(2)) < Y(X) s(2))
and
have
acted
and
the
subcase (2.4) for the a-expansionary stage s(4). Then
qd(x, s(4)) < a@, s(4)).
From the rules for taking control of the markers, Cs(2j r x = Csc3) r x. Also (Y redefines I/J’(x, s(3)) < ym(x, s(3)) an d makes sure that each marker is larger than
D. Kaddah
262
both of their verifying uses. We then have r,!~“(x,s(4)) < y&, s(4)) because, as was noted at the beginning of this proof, y(x) is moved whenever I/J’(X) is moved. At the next a-expansionary stage s(5) >s(4), a will redefine vd(x, s(5)) smaller than both y&, s(5)) and a&, s(5)) and will also make sure that each marker is larger than all three of their verifying uses. Strategy /3 is now ready to execute subcase (2.1) for the x-cycle, and at the next P-stage s(6) it does so by enumerating x into X, and Wd(x, s(6)) into B, and requesting that CYtake control of the markers. At the following o-expansionary stage s(7), a: executes the second step of this procedure. Arguments similar to the above show that, for all t as(6) Y&G 9, 4~
0 ’ v%,
s(6)),
and
B, r &(B;-G s(l)) = Bsc6) 1 f%(B;x, s(l))
unless some y < IJJ~(X,s(6)) is enumerated into B. We claim that if y is enumerated into B after stage s(6), then either y