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Chapter 6 Additional First Order Results
CHAPTER 6
ADDITIONAL FIRST ORDER RESULTS
0 1. Compactness We call an infinite set S of signed formulas realizable if there is a model (9, W ,C, 9')and a re9 such that for any formula X
T X E S *X E @( T ) and T t X , F X E S X E @ ( r ) and r y X . There is a similar concept for sets of unsigned formulas U. We say Uis satisfiable if there is a model (9, 92, C, 9') and a I ' E g such that for any formula X X E U + X X E @ ( ~and ) rkX.
Lemma 1.1: Let U be a set of unsigned formulas and define a set S of signed formulas to be { TX I XEU}.Then (1). U is satisfiable if and only if S is realizable (2). U is consistent if and only if S is consistent. Pro08 Part (1) is obvious. To show part (2), suppose U is not consistent. Then some finite subset {ul, ..., us}is not consistent, so from it we can deduce any formula. Let A be an atomic formula having no predicate symbols or parameters in common with {ul, ..., u,}. Then kl(U1 A * * * A U n ) 3 ~ .
72
ADDlTIONAL FIRST ORDER RESULTS
~~1.661
Hence there is a closed tableau for
so there is a closed tableau for
By the way we have chosen A , there must be a closed tableau for { T(u, A A u,)} and hence for { Tu,, ...,Tu,}. Thus S is not consistent. The converse is trivial.
-..
Because we have this lemma, we will only discuss realizability and consistency of sets of signed formulas. Lemma 1.2: Let S be a set of signed formulas. If S is realizable, S is consistent. Pro08 If S is not consistent, some finite subset Q is not consistent. That is, there is a closed tableau Vl, V, ..., V, in which gl is {Q}. If Q were realizable, by theorem 5.2.7 every gi would be, but a closed configuration is not realizable. Lemma 1.3: Let S be afinite set of signed formulas. If S is consistent, S is realizable. Pro08 Let S be { TX,, ..., TX,, FYI,..., FY,}. S is consistent if and only if (F(X1 A
**.
A
x,) 13 (Y1 V * * * V Y,)}
is consistent. If this is consistent, (XIA AX,)^ (& v v Y,) is a . non-theorem, so by the completeness theorem, there is a model (9, 2,k, 9)and a r E 3 such that X j € @ ( r ) ,$ ~ @ (and r)
rpc(xlA...Ax,)D(Y~ V - v Y,). But then for some r* r*tx,A - A x,, r*)cY, v . - v Y,, so r*realizes S. This method does not work if S is infinite, but the lemma remains true, at least for sets with no parameters. The result can be extended to sets with some parameters, but we will not do so.
m.651
73
COMPACTNESS
Lemma 1.4: Let S be an infinite set of signed formulas with no parameters. If S is consistent, S is realizable. Proofl The proof can be based on either of the two tableau completeness proofs. If we use the first proof, that of ch. 5 0 5, change step 0 to: “Sis consistent. Extend it to a Hintikka element with respect to PI. Call the result r1”.Continue the proof as written. The lemma is then obvious. If we use the proof of ch. 5 0 6 the result is even easier. S is consistent, so by lemma 5.6.4, we can extend S to a set I‘ which is good with respect to P,.The result follows immediately. Theorem 1.5: If S is any set of signed formulas with no parameters, S is consistent if and only if S is realizable. Corollary 1.6: If every finite subset of S is realizable, so is S. Corollary 1.7: If U is any set of unsigned formulas with no parameters, U is consistent if and only if U is satisfiable.
Remark 1.8: The last corollary could have been established directly by adapting the completeness proof of ch. 5 Q 10. Definition 1.9: For a set of formulas U, by X E u.
r b U we mean I‘ C X for all
Corollary 1.10 (strong completeness): Let U be any set of unsigned formulas with no parameters. Then UtIX if and only if in any modelu { F X } is inconsistent,
I
Corollary 1.11: (cut elimination, Gentzen’s Hauptsatz): If S is a set of signed formulas with no constants and {S, T X } and {S, F X } are inconsistent, so is {S}.
Remark 1.12: This may be extended to sets S with some parameters. To be precise, to any set S which leaves unused a countable collection of parameters. It follows that in the completeness proof of ch. 5 Q 6 a set A maximal consistent with respect to P actually contains TX or FX for each X with parameters from P.
74
ADDITIONAL FTRST ORDER RESULTS
CH. 682
$2. Concerning the excluded middle law If S is a set of unsigned formulas, by S t , X and S k , X we mean classical and intuitionistic derivability respectively. Let X(a,, ..., a,) be a formula having exactly the parameters al, ..., a,. By the closure of X we mean the formula (VxiJ
(Vxi.) X(xil, ' " 9 xi,)
(where x i j does not occur in X ( a l , ..., a,,)). Let M be the collection of the closures of all formulas of the form X v -X. We wish to show:
Theorem 2.1: If X has no parameters,
t C Xe M F I X . We first show:
Lemma 2.2: Let (9,92,k, 9')be a model, r E 9 , and suppose Y E M rk-Y. Then r can be included in a complete .%chain V such that %" is a truth set (see ch. 4 Q 6). Pro08 Enumerate all formulas beginning with a universal quantifier:
XI,x2,x,,....
Let T o=r. Having defined r,,,consider X , + l . If X,+ $ @ (r:)for any T:, let I'n+l=I',,. Otherwise there is some r,*such that X,+,E@(~,*). Say X,,, is ( V x ) X ( x ) . We have two cases: (1). If r,*b-(Vx)X(x), let
r,+,=r:.(2). If r,*pC (Vx) X ( x ) , there is a r,**and an a ~ P ( r , *such *) that r,** y X(a). Let r,+,be this r,**. Let the %chain V be {r0,rl,r2,...I. Since YEM=-Tb Yand T = T 0 ,
V is a complete %chain by the definition of M, and so V' is an almosttruth set. Thus we have only one more fact to show:
Y (ct) E V' for every parameter CI of
Q'
+ (Vx) Y (x) E V' .
Suppose (Vx) Y (x, al, ..., a,,) $ V' (where al, ..., a, are all the parameters of Y ) . If some ai is not a parameter of W, we are done. So suppose each at occurs in V'. Then for some r,,€V, all ai@(I',,) and r. j! (Vx) Y(x, al, ..., an). But by the construction of Q, there is a rm(m2n) such that r, pC Y(b,al, ..., a,) for some b~P(r,,,).But
r t (vxl) ...(vx,)
(VX) [Y (x, x ~..,., x,) v
Y (x, x ~..., , x,,)]
~1-1.683
and r9Wm, so
75
SKOLEM-LOWENHEM
r,t
Y ( b , cxl ,... a,) v
N
Y ( b , 011, ..., a n ) ,
thus rmk-Y(b, a,, ..., a,). -Y(b, a1,..., cx,)~%‘, so Y(b,a,..., cxn)$V’ for a parameter b of V’. Now to prove the theorem itself: If M t , X then for some finite subset {ml, ..., m,} of M
t,(ml A A mn)3 X . By theorem 4.8.2 (and the completeness theorems) kc(ml A * . * A m,)
3
x.
But kcml A .--Am,, hence tCX. Conversely, if MY ,X,let S be the set of signed formulas
I
{ F X ) u {TY Y E M } . Since MY ,X, S is consistent. Then by the results of the last section, S is realizable. Thus there is a model (’3, W ,k, 9’) and a r E B such that Y E M TC ~ Y, X e @ ( T ) and ry X. But X has no parameters, so X v X E M .Thus k X v X , so C X. Now by lemma 2.2 there is a truth set containing -X. Hence Y .X. N
-
-
0 3. Skolem-Ldwenheim By the domain of a model (9, 9, k, 9’) we mean ursgY(r). So far we have only considered models in which the domain was at most countable. Suppose now we have an uncountable number of parameters and we change the definitions of formula, model and validity accordingly, but not the definition of proof.
Theorem 3.1: X is valid in all models if and only if X is valid in all models with countable domains. Pro08 One half is trivial. Suppose there is a model (9, W , k, 9’) with an uncountable domain in which X is not valid. The correctness proof of ch. 5 # 2 or 9 is still applicable. Thus Xis not provable. Since Xis not provable, if we reduce the collection of parameters to a countable number (including those of X), X still will not be provable. Then any of the completeness proofs will furnish a counter-model for X with a countable domain.
76
c1i.694
ADDITIONAL FIRST ORDER RESULTS
This method may be combined with that of Q 1 to show Theorem 3.2: If S is any countable set of signed formulas with no parameters, S is consistent if and only if S is realizable in a model with a countable domain. Theorem 3.3: If U is any countable set of unsigned formulas with no parameters, U is consistent if and only if U is satisfiable in a model with a countable domain.
Remark 3.4: In part 11, we will be using models with domains of arbitrarily high cardinality.
5 4.
Kleene tableaus
The system of this section is based on the intuitionistic system G3 of [lo]. The modifications are due to Smullyan. The resulting system is like that of Beth except that sets of signed formulas never contain more than one F-signed formula. Explicitely, everything is as it was in ch. 2 Q 1 and ch. 5 Q 1 except that the reduction rules are replaced by the following, where S is a set of signed formulas with at most one F-signed formula.
KT v
S,TXvY S, T X ]S, T Y .
KFV
ST,FXVY ST,
FX
ST,F X v Y KTA
S,TXAY S, T X , T Y
KTKT
3
KT 3 KT V
S,T-X ST, FF S,TX3Y ST,F X I s, T Y S, T ( h )X ( X ) S, TX(a> S, T ( V X )X (x) S, T X ( a )
where in KT3 and W V the parameter a does not occur in S or X ( X ) .
c~.6§4
77
KLEENE TABLEAUS
There are several ways of showing this is actually a proof system for intuitionistic logic. We choose to show it is directly equivalent to the Beth tableau system, that is, we give a proof translation procedure. We leave it to the reader to show the almost obvious fact that anything provable by Kleene tableaus is provable by Beth tableaus. To show the converse, we need
Lemma 4.1: If a Beth tableau for { TX,, ..., TX,, FYI,..., FY,) closes, then there is a closed Kleene tableau for
{TX,, ..., TX,, F ( Y , v ..*v Y,)}. Proofi The proof is by induction on the length of the closed Beth tableau. If the tableau is of length 1, the result is obvious. Now suppose we know the result for all closed Beth tableaus of length less than n, and a closed tableau for the set in question is of length n. We have several cases depending on the first step of the tableau. If the first step is an application of rule F A , the Beth tableau begins
{{% FXi, .-.,FXn, FY A z}}, {{ST,FXI, * - * , FXn, F Y } {ST, F X I , * * * , FXn, F Z } } 2
3
and proceeds to closure. Now by the induction hypothesis there are closed Kleene tableaus for {&, F(X1v v Xn v Y ) } and {ST,F ( X , v v Xn v Z ) } . We have two possibilities: (1). If Y is not “used” in the first tableau, or if Z is not “used” in the second tableau, a Kleene tableau beginning
---
{{ST, F(X1 {{ST, F ( X 1
x,
v *’* v v (Y v *..v Xn))}
A
Z))}}
9
9
must close. (2). If both Y and Z are “used”, a Kleene tableau beginning {{snF(Xi v * * . v Xn v ( Y {{ST,
F(‘Y
A
A
z))}],
Z)>>
9
{{ST, F Y I , {ST,FZ}} Y
must close. The other cases are similar and are left to the reader.
78
ADDITIONAL FIRST ORDER RESULTS
CH.685
Thus the two tableau systems are equivalent. Now we verify a remark made at the end of ch. 5 0 10. Lemma 4.2: (Godel, McKinsey and Tarski): t,X v Y iff tlX or tl Y. Proofi Immediate from the Kleene tableau formulation. Lemma 4.3: (Rasiowa and Sikorski): If tl(3x) X ( x , al, ..., a,) where a, ,..., a, are all the parameters of X , then klX(b, a,, ..., a,) where b is one of the ai.If X has no parameters, b is arbitrary and kl(Vx) X(x). Pro08 A Kleene tableau proof of (3x) X ( x , a,, ..., a,) begins
{ { W x ) X(X, a,, a,)}}, {{FX(b, a,, --.a,)}}, a*.,
Y
and proceeds to closure. If b is some ai,we are done. If not, we actually have a proof, except for a different first line, of (VX)
8 5.
X ( X , 01,. * * Y a,).
Craig interpolation lemma
Theorem 5.1: If FIX= Y and X and Y have a predicate symbol in common, then there is a formula 2 involving only predicates and parameters common to X and Y such that k , X > Z and kIZ 3 Y; if X and Y have no common predicates, either tl X or t, Y.
-
The classical version of this theorem was first proved by Craig, hence the name. The intuitionistic version is due to Schiitte [17]. Essentially the same proof was given for a natural deduction system by Prawitz [15]. We give basically the same proof in the Kleene tableau system. For another proof in this system see [ll]. We find it convenient to temporarily introduce two symbols t and f into our collection of logical symbols, letting them be atomic formulas, and letting them combine according to the following rules.
x v t=
tv
x =t,
Xvf=f vx=x, X A t= t A X=X, x A f = f AX=f, -t=f, -f=t,
CH.645
CRAIG INTERPOLATION LEMMA
79
x 3 t = f 3 x = t, t3X=X X3f=-X, (3x)t = (Vx)t = t ,
( 3 x ) f = (Vx)f = f . By a block we mean a finite set of signed formulas containing at most one F-signed formula. When we call a block inconsistent, we mean there is a closed Kleene tableau for it. By an initialpart of a block we mean any subset of the T-signed formulas. We make the convention that if S is the finite set of unsigned formulas {XI,..., X,,)then TS is the set {TX,, ..., TX,,}. We further make the convention that for a set S of formulas, S, and S, represent subsets such that S, nS, =8 and S, u S, =S. By [S] we mean the set of predicates and parameters of formulas of S, together with t and$ Now we define an interpolation formula X for the block (TS,P Y ) (where S is a set of unsigned formulas and Y is a formula) with respect to the initial part TS,, which we denote by { TS, F Y } / {TS,}, as follows ( X may be t or f,but we assume t and f are not part of S or Y ) : Xis an ( T S , FY}I(TS,} if (1). [XI E CSll" CSZY y1, (2). { TS,, F X } is inconsistent, (3). (TX,TS,, FY} is inconsistent (we have temporarily added to the closure rules: closure of a set of signed formulas if it contains Tf or Ft). Lemma 5.2: An inconsistent block has an interpolation formula with respect to every initial part. Proofi We show this by induction on the length of the closed tableau for the block. If this is of length 1, the block must be of the form
{ TS,TX,FX}. We have two cases: Case (1). The initial part is ( T S , , T X } . Then X is an interpolation formula. Case (2). The initial part is { T S , ) . Then ( TS,, TX,FX} is inconsistent and t is an interpolation formula. Now suppose we have an inconsistent block, and the result is known for all inconsistent blocks with shorter closed tableaus. We have several cases depending on the k s t reduction rule used.
80
ADDITIONAL FIRST ORDER RESULTS
c~.6§5
K T v : The block is { T S , TX v Y, FZ}, and { T S , TX, F Z } and { T S , TY, F Z } are both inconsistent. Case (1). The initial part is { T S , , TX v Y } . Then by the induction hypothesis there are formulas U, and U, such that
U,is an {TS, T X , FZ}/{TS,,T X } , U2is an { T S , T Y , F Z } / { T S , , T Y } . Then U, v U, is an { TS, TX v Y, F Z } / {TS,, TX v Y } . Case (2). The initial part is { T S , } . Again, by hypothesis, there are U,, U2such that U,is an { TS, T X , F Z } / { T S , } , U,is an {TS, T Y , F Z ) / { T S , } . Then U, A U2is an {TS, TX v Y, F Z } / { TS,}. KF v : The block is { T S , FX v Y } , and { T S , F X ) or { T S , FY} is inconsistent. Suppose the first. Let the initial part be { T S , } . By hypothesis there is a U such that
U is an {TS, F X ) / { T S , ) . Then U is an { TS, FX v Y } / {T S , } . K T A : The block is { T S , TX A Y, F Z } , and { T S , TX, TY, F Z } is inconsistent. Case (1). The initial part is { TS,, TX A Y } . By hypothesis there is a U such that U is an {TS,T X , T Y , F Z } / { T S , , T X , T Y ) . Then U is an { T S , TX A Y, F Z } / {TS,, TX A Y ) . Case (2). The initial part is { T S , } . By hypothesis there is a U such that U is an { TS, T X , T Y , F Z } / { T S , } . Then U is an { T S , TX A Y, F Z } / {T S , } . KF A : The block is { T S , FX A Y } , and { T S , FX) and ( T S , FY} are both inconsistent. Suppose the initial part is { T S , } . By hypothesis there are U,, U2such that U, is an { TS, FX}/{T S , } , U, is an { T S , F Y } / { T S , } . Then U,A U, is an { T S , FX
A
Y}/(TS,}.
c~.685
CRAIQ INTERPOLATION LEMMA
81
KF- : The block is { TS, F - X ) , and { TS, T X } is inconsistent. Suppose the initial part is { TS,}. By hypothesis there is a U such that U is an ( T S , T X } / {T S , } . Then U is an {TS, F - X } / { T S , } . KT- : The block is {TS, T - X , FY), and { TS, F X } is inconsistent. Case (1). The initial part is { TS,}. By hypothesis there is a U such that
-
U is an { TS,FX}/{ TS,}.
Then U is an { TS, T X , FY}/{TS,}. Case (2). The initial part is {TS,, T - X ) . By hypothesis there is a U such that U is an { T S , F X } / { T S 2 } . We claim that U is an { T S , T X , F Y ) / {T S , } .
-
-
First we verify its predicates and parameters are correct. By hypothesis [UJc[S2]n[Sl,X], so immediately [-U]c[S,,-X]n[Sz, Y ] . We have the following two blocks are inconsistent:
{TSz,F U } {TS,, T U , F X ) . 3
It follows that the following two blocks are also inconsistent:
{TS,, T {TS2, T
- u, N
X,F U}, FY) N
3
and we are done. K F 3 : The block is { TS, F X 2 Y } , and { TS, T X , F Y } is inconsistent. Suppose the initial part is {TS,}.By hypothesis there is a U such that
u is an ( T S , TX,F Y ) / { T S , } . Then U is an { TS, FX =I Y ) / {TS,}. K T 2 : The block is { TS, TX 3 Y, F Z ) , and { TS, F X ) and { TS, TY, F Z } are both inconsistent. Case (1). The initial part is {TS,). By hypothesis there are U,, U2 such that U, is an { T S , FX}/{T S , } , U2 is an {TS, TY, F Z } / { T S , ) ,
82
ADDITIONAL FIRST ORDER RESULTS
c~.6$5
Then U,A U2 is an { TS, TX 3 Y, F Z } / {TS,}. Case (2). The initial part is {TS,, T X = Y}. By hypothesis there are U,, U2 such that U , is an { T S , F X } / { T S 2 ) , U2 is an { T S , T Y , F Z } / { T S , , T Y } . We claim U,= U2 is an { TS, TX 3 Y, F Z } / {TS,, TX 2 Y}. By hypothesis cull [S21 n [ S l y XI CU2l CS,, y] n [S2, 21 so [ul= U2]E [SlyX = Y] n [S2, Z] . 9
We have that the following four blocks are inconsistent: (1). w 2 , (2). {W,,TS,, (3). { TSI, TYY W), (4). { m2,TS2, and we must show the following two blocks are inconsistent:
fw, w, m,
The first follows from (2) and (3), and the second from (I) and (4). KF3 : The block is { TSyP(3x) X ( x ) } , and { TS, FX(a)}is inconsistent. Suppose the initial part is { TS,}. By hypothesis there is a U such that U is an { T S , FX (a)}/{ T S , } . Then [U]E[Sl]n[S2, X(a)]. Case (1). a# [U]. Then U is an { TS, F(3x) X ( x ) } / {TS1} Case (2). UE [ U ] , U E [SJ Again U is an { TS, F(3x) X ( x ) } / {TS,} Case (3). U E [U], a$ [S2]. Then ( 3 x ) U(z) is an {TSY F (3.1 x ( x ) } / { T S J* K T 3 : The block is { TS, T(3x)X(x), FZ}, and { TS, TX(a), FZ} is inconsistent, where a# [S, X ( x ) , Z ] .
m.685
CRAIG INTERPOLATION LEMMA
83
Case (1). The initial part is { TS,, T(3x) X ( x ) } . By hypothesis there is a U such that U is an {TS,T X ( a ) , F Z } / { T S , , T X ( a ) } . Then U is an { TS, T ( 3 x ) X ( x ) , F Z } / {TS,, T(3x) X ( x ) ) . Case (2). The initial part is { TS,). By hypothesis there is a U such that
u is an { TS,T X (a), F Z } / {TS1}. Then U is an { TS, T ( 3 x ) X ( x ) , F Z } / {TS,}. KFV: The block is { TS, F(Vx) X ( x ) ) , and { TS, FX(a)} is inconsistent, where &[S, X ( x ) ] . Suppose the initial part is {TS,}. By hypothesis there is a U such that U is an {TS, F X ( a ) ) / ( T S , ) . Then U is an { TS, F(Vx) X ( x ) } / {TS,}. KTV: The block is { TS, T ( V x ) X ( x ) , F Z } , and { TS, TX(a), F Z } is inconsistent. Case ( 1 ) . The initial part is { TS,, T ( V x ) X ( x ) } . By hypothesis there is a U such that U is an {TS,T X ( a ) , F Z } / { T S , , T X ( a ) } . Case (la). a # [ U ] . Then U is an { T S , T ( V x )X ( X ) , F Z ) / { T S , , T ( V x )X ( X ) } *
Case (lb). a E [ U ] , aEISl, X ( x ) ] . Again U is an {TS, T ( V x ) X ( x ) , F Z } / { T S , , T ( V x ) X ( x ) } . Case (Ic). a ~ [ v ] a#[S,, , X(x)]. Then (Vx) U(:) is an { TS, T ( V x ) X ( x ) , F Z } / {TS,, T ( V x ) X ( x ) } . Case (2). The initial part is { TS,}. By hypothesis there is a U such that
u is an { T S , T X ( U ) ,FZ)/{TSl}. Case (2a). a # [ U ] . Then U is an { TS, T ( V x ) X ( x ) , F Z } / { TS,). Case (2b). a€ [ U ] , a€ [Sz, X ( x ) , Z ] . Again U is an { TS, T ( V x ) X ( x ) , F Z } / {TS,}. Case (2c). a € [ U ] , a$ [Sz, X ( x ) , Z ] .
84
ADDITIONAL FIRST ORDER RESULTS
c~.656
Then (3x) U ( l )is an { TS, T ( V x ) X ( x ) , F Z } / {TS,). Now to prove the original theorem 5.1 : Suppose t I X 3 Y. Then { TX, F Y ) is inconsistent. By the lemma, there is a U such that U is an { TX,F Y } / {T X } . We have three cases: (1). U = t . Then since { Tt, F Y } is inconsistent, t, Y . (2). u=f. Then since (TX, F f ) is inconsistent, { F - X } is also inconsistent cf is not in X).Thus t-,-X. (3). U # t , u z f . Then U is a formula not involving t orf, all the parameters and predicates of U are in X and Y, and since { T X , FU} and (TU, F Y } are both inconsistent, tIX=, U and kI U 2 Y.
5 6.
Models with constant B function
In part I1 we will be concerned with finding countermodels for formulas with no universal quantifiers, and we will confine ourselves to models with a constant B function. To justify this restriction, we show in this section Theorem 6.1: If X is a formula with no universal quantifiers and y ,X, then there is a counter-model (’3, 9, k,P) for Xin which B is a constant function. Definition 6.2: For this section only, let a,, u2, a3,... be an enumeration of all parameters. We call a set r of signed formulas a Hintikka element if r is a Hintikka element with respect to some initial segment of a,, u2, a3,... (see ch. 5 Q 4). Lemma 6.3: If S is a finite, consistent set of signed formulas with no universal quantifiers, S can be extended to a Jinite Hintikka element. Proof: Suppose S is the set ( X , , X,, ..., Xn} where each X i is a signed formula. We d e h e the two sequences {pk},{&) as follows: Let
P, = 0, Qo = x,, ..., X”. Suppose we have defined Pkand Qkwhere Pk = Y,, ..., q , Q k = w,,..., w,,
85
MODELS WITH CONSTANT 9 FLINGTION
CH.686
and Pk u Qk (considered as a set) is consistent. To define Pk+, and Qk+l we have several cases depending on W, : Case atomic: If W, is a signed atomic formula, let pk+l
= Y,,
. * a ,
Y,, WI
Qk+l
= W2, ..*,Ws.
Case T v : I f W, is TX v Y, either TX or TY is consistent with Pk u Qk, say TX. Let Pk+i
= Y1, ...,
x, TX V
= w2, ...,
Y,
w,,T X .
Case F v : If W, is FX v Y then FX, FY is consistent with Pku Qk. Let Pk+l
= Y1, ..-,Y,, FX
V
Y,
Qk+l
= W2,
..., W,,F X , F Y .
CasesTA, F A , T - , T I aresimilar. Case T3 : If W,is T(3x) X(x), let a be the first in the sequence a,, a2,... not occurring in Pk or Q k . Then TX(a) is consistent with Pk u Qk. Let Case F 3 : If W, is F(3x) X(x), let {a,,, ..., sit} be the set of parameters occurring in P k u Q k such that no FX(a{,) occurs in PkuQk. Then {FX(ai,),...,FX(a,,)} is consistent with PkuQk. Let Pk+l
=pk>
Qk+l
=
W ~ , * *w,,FX(ai,),...,FX(ai,),F(3x)X(x), .Y
After finitely many steps there will be no T-signed formulas left in the Q-sequence because each rule T v , T A , T-, T 3, T3 reduces degree, and no rule F v , F A , F 3 introduces new T-signed formulas. When no T-signed formulas are left in the Q-sequence, no new parameters can be introduced since rule T3 no longer applies. After finitely many more steps we must reach an unusable Q-sequence. The corresponding P v Q-sequence is finite, consistent, and clearly a Hintikka element.
Remark 6.4: The above proof also shows the following which we will need later: Let R be a finite Hintikka element. Suppose we add (consistently) a finite set of F-signed formulas to R and extend the result to a finite Hintikka element S by the above method. Then
R, = s*.
86
ADDITIONAL FIRST ORDER RESULTS
c~.656
Since R s S , certainly R T s S T .That STGRT also holds follows by an inspection of the above proof; no new T-signed formulas will be added. Now we turn to the proof of the theorem itself. We have no universal quantifiers to consider, so we may use the definition of associated sets in ch. 2 9 4. Suppose X is a formula with no universal quantifiers, and y ,X. Then { F X } is consistent. Extend it to a finite Hintikka element S:. Let T,,,.., T,,be the associated sets of S:. Extend each to a finite Hintikka element, Sy, ..., S: respectively. Thus we have
s:, sy, ...)s:. For each parameter Q of some S: and each formula of the form P(3x) X ( x ) in St, adjoin FX(u) to S8 and extend the result to a Hintikka element S;. Do the same for S:, .... S,",producing S:, ..., S: respectively. Thus we have now
s;, s:, ..., s; .
Let T,,+iy...,T, be the associated sets of SAYS ; , . . . ,S:.Extend each to a Hintikka element, S:+,, ..., Sz respectively. Thus we have now
st, s:, ...)s,1,s,,,0 1, ..., sl),. For each parameter u used so far, and for each formula of the form F(3x) X ( x ) in S;, adjoin FX(u) to St and extend the result to a finite Hintikka element St. Do the same for each. Thus we have now
s;, s:, ...) s,",s;+ 1, ..., s;
.
Again take the associated sets, and extend to finite Hintikka elements, producing now 2 1 0 s;,~:,...r~n,Sn+l,...,S~,Srn+l ,...,s~.
Continue in this manner. Let 4)
W
s o = U S:, s l =U s:, k=O
k=O
By the remark above, for each n, S,T=S;T=S;T=.*..
Thus if S: has as an associated set S;, S,,, ES,.
etc.
c~.6$6
MODELS WITH CONSTANT
9 FUNCTION
87
It now follows that {So,S,, ..,} is a Hintikka collection. For example, suppose F- YES,..Let k be the least integer such that F- YES:. By the above construction, there is some set S," such that S: is an associated set of S: and TYES:. But then Sjk,sS,?, so by the above SjTcS,,and TYES,.The other properties are shown similarly. Moreover, 9(S,,)=9(Sm)for all m and a, as is easily seen. (Recall that 9's) is the collection of all parameters used in S.) Now as in ch. 5 0 3 there is a model for this Hintikka collection, and this model will have a constant B map, so the theorem is shown.
ADDITIONAL FIRST ORDER RESULTS
0 1. Compactness We call an infinite set S of signed formulas realizable if there is a model (9, W ,C, 9')and a re9 such that for any formula X
T X E S *X E @( T ) and T t X , F X E S X E @ ( r ) and r y X . There is a similar concept for sets of unsigned formulas U. We say Uis satisfiable if there is a model (9, 92, C, 9') and a I ' E g such that for any formula X X E U + X X E @ ( ~and ) rkX.
Lemma 1.1: Let U be a set of unsigned formulas and define a set S of signed formulas to be { TX I XEU}.Then (1). U is satisfiable if and only if S is realizable (2). U is consistent if and only if S is consistent. Pro08 Part (1) is obvious. To show part (2), suppose U is not consistent. Then some finite subset {ul, ..., us}is not consistent, so from it we can deduce any formula. Let A be an atomic formula having no predicate symbols or parameters in common with {ul, ..., u,}. Then kl(U1 A * * * A U n ) 3 ~ .
72
ADDlTIONAL FIRST ORDER RESULTS
~~1.661
Hence there is a closed tableau for
so there is a closed tableau for
By the way we have chosen A , there must be a closed tableau for { T(u, A A u,)} and hence for { Tu,, ...,Tu,}. Thus S is not consistent. The converse is trivial.
-..
Because we have this lemma, we will only discuss realizability and consistency of sets of signed formulas. Lemma 1.2: Let S be a set of signed formulas. If S is realizable, S is consistent. Pro08 If S is not consistent, some finite subset Q is not consistent. That is, there is a closed tableau Vl, V, ..., V, in which gl is {Q}. If Q were realizable, by theorem 5.2.7 every gi would be, but a closed configuration is not realizable. Lemma 1.3: Let S be afinite set of signed formulas. If S is consistent, S is realizable. Pro08 Let S be { TX,, ..., TX,, FYI,..., FY,}. S is consistent if and only if (F(X1 A
**.
A
x,) 13 (Y1 V * * * V Y,)}
is consistent. If this is consistent, (XIA AX,)^ (& v v Y,) is a . non-theorem, so by the completeness theorem, there is a model (9, 2,k, 9)and a r E 3 such that X j € @ ( r ) ,$ ~ @ (and r)
rpc(xlA...Ax,)D(Y~ V - v Y,). But then for some r* r*tx,A - A x,, r*)cY, v . - v Y,, so r*realizes S. This method does not work if S is infinite, but the lemma remains true, at least for sets with no parameters. The result can be extended to sets with some parameters, but we will not do so.
m.651
73
COMPACTNESS
Lemma 1.4: Let S be an infinite set of signed formulas with no parameters. If S is consistent, S is realizable. Proofl The proof can be based on either of the two tableau completeness proofs. If we use the first proof, that of ch. 5 0 5, change step 0 to: “Sis consistent. Extend it to a Hintikka element with respect to PI. Call the result r1”.Continue the proof as written. The lemma is then obvious. If we use the proof of ch. 5 0 6 the result is even easier. S is consistent, so by lemma 5.6.4, we can extend S to a set I‘ which is good with respect to P,.The result follows immediately. Theorem 1.5: If S is any set of signed formulas with no parameters, S is consistent if and only if S is realizable. Corollary 1.6: If every finite subset of S is realizable, so is S. Corollary 1.7: If U is any set of unsigned formulas with no parameters, U is consistent if and only if U is satisfiable.
Remark 1.8: The last corollary could have been established directly by adapting the completeness proof of ch. 5 Q 10. Definition 1.9: For a set of formulas U, by X E u.
r b U we mean I‘ C X for all
Corollary 1.10 (strong completeness): Let U be any set of unsigned formulas with no parameters. Then UtIX if and only if in any model
I
Corollary 1.11: (cut elimination, Gentzen’s Hauptsatz): If S is a set of signed formulas with no constants and {S, T X } and {S, F X } are inconsistent, so is {S}.
Remark 1.12: This may be extended to sets S with some parameters. To be precise, to any set S which leaves unused a countable collection of parameters. It follows that in the completeness proof of ch. 5 Q 6 a set A maximal consistent with respect to P actually contains TX or FX for each X with parameters from P.
74
ADDITIONAL FTRST ORDER RESULTS
CH. 682
$2. Concerning the excluded middle law If S is a set of unsigned formulas, by S t , X and S k , X we mean classical and intuitionistic derivability respectively. Let X(a,, ..., a,) be a formula having exactly the parameters al, ..., a,. By the closure of X we mean the formula (VxiJ
(Vxi.) X(xil, ' " 9 xi,)
(where x i j does not occur in X ( a l , ..., a,,)). Let M be the collection of the closures of all formulas of the form X v -X. We wish to show:
Theorem 2.1: If X has no parameters,
t C Xe M F I X . We first show:
Lemma 2.2: Let (9,92,k, 9')be a model, r E 9 , and suppose Y E M rk-Y. Then r can be included in a complete .%chain V such that %" is a truth set (see ch. 4 Q 6). Pro08 Enumerate all formulas beginning with a universal quantifier:
XI,x2,x,,....
Let T o=r. Having defined r,,,consider X , + l . If X,+ $ @ (r:)for any T:, let I'n+l=I',,. Otherwise there is some r,*such that X,+,E@(~,*). Say X,,, is ( V x ) X ( x ) . We have two cases: (1). If r,*b-(Vx)X(x), let
r,+,=r:.(2). If r,*pC (Vx) X ( x ) , there is a r,**and an a ~ P ( r , *such *) that r,** y X(a). Let r,+,be this r,**. Let the %chain V be {r0,rl,r2,...I. Since YEM=-Tb Yand T = T 0 ,
V is a complete %chain by the definition of M, and so V' is an almosttruth set. Thus we have only one more fact to show:
Y (ct) E V' for every parameter CI of
Q'
+ (Vx) Y (x) E V' .
Suppose (Vx) Y (x, al, ..., a,,) $ V' (where al, ..., a, are all the parameters of Y ) . If some ai is not a parameter of W, we are done. So suppose each at occurs in V'. Then for some r,,€V, all ai@(I',,) and r. j! (Vx) Y(x, al, ..., an). But by the construction of Q, there is a rm(m2n) such that r, pC Y(b,al, ..., a,) for some b~P(r,,,).But
r t (vxl) ...(vx,)
(VX) [Y (x, x ~..,., x,) v
Y (x, x ~..., , x,,)]
~1-1.683
and r9Wm, so
75
SKOLEM-LOWENHEM
r,t
Y ( b , cxl ,... a,) v
N
Y ( b , 011, ..., a n ) ,
thus rmk-Y(b, a,, ..., a,). -Y(b, a1,..., cx,)~%‘, so Y(b,a,..., cxn)$V’ for a parameter b of V’. Now to prove the theorem itself: If M t , X then for some finite subset {ml, ..., m,} of M
t,(ml A A mn)3 X . By theorem 4.8.2 (and the completeness theorems) kc(ml A * . * A m,)
3
x.
But kcml A .--Am,, hence tCX. Conversely, if MY ,X,let S be the set of signed formulas
I
{ F X ) u {TY Y E M } . Since MY ,X, S is consistent. Then by the results of the last section, S is realizable. Thus there is a model (’3, W ,k, 9’) and a r E B such that Y E M TC ~ Y, X e @ ( T ) and ry X. But X has no parameters, so X v X E M .Thus k X v X , so C X. Now by lemma 2.2 there is a truth set containing -X. Hence Y .X. N
-
-
0 3. Skolem-Ldwenheim By the domain of a model (9, 9, k, 9’) we mean ursgY(r). So far we have only considered models in which the domain was at most countable. Suppose now we have an uncountable number of parameters and we change the definitions of formula, model and validity accordingly, but not the definition of proof.
Theorem 3.1: X is valid in all models if and only if X is valid in all models with countable domains. Pro08 One half is trivial. Suppose there is a model (9, W , k, 9’) with an uncountable domain in which X is not valid. The correctness proof of ch. 5 # 2 or 9 is still applicable. Thus Xis not provable. Since Xis not provable, if we reduce the collection of parameters to a countable number (including those of X), X still will not be provable. Then any of the completeness proofs will furnish a counter-model for X with a countable domain.
76
c1i.694
ADDITIONAL FIRST ORDER RESULTS
This method may be combined with that of Q 1 to show Theorem 3.2: If S is any countable set of signed formulas with no parameters, S is consistent if and only if S is realizable in a model with a countable domain. Theorem 3.3: If U is any countable set of unsigned formulas with no parameters, U is consistent if and only if U is satisfiable in a model with a countable domain.
Remark 3.4: In part 11, we will be using models with domains of arbitrarily high cardinality.
5 4.
Kleene tableaus
The system of this section is based on the intuitionistic system G3 of [lo]. The modifications are due to Smullyan. The resulting system is like that of Beth except that sets of signed formulas never contain more than one F-signed formula. Explicitely, everything is as it was in ch. 2 Q 1 and ch. 5 Q 1 except that the reduction rules are replaced by the following, where S is a set of signed formulas with at most one F-signed formula.
KT v
S,TXvY S, T X ]S, T Y .
KFV
ST,FXVY ST,
FX
ST,F X v Y KTA
S,TXAY S, T X , T Y
KTKT
3
KT 3 KT V
S,T-X ST, FF S,TX3Y ST,F X I s, T Y S, T ( h )X ( X ) S, TX(a> S, T ( V X )X (x) S, T X ( a )
where in KT3 and W V the parameter a does not occur in S or X ( X ) .
c~.6§4
77
KLEENE TABLEAUS
There are several ways of showing this is actually a proof system for intuitionistic logic. We choose to show it is directly equivalent to the Beth tableau system, that is, we give a proof translation procedure. We leave it to the reader to show the almost obvious fact that anything provable by Kleene tableaus is provable by Beth tableaus. To show the converse, we need
Lemma 4.1: If a Beth tableau for { TX,, ..., TX,, FYI,..., FY,) closes, then there is a closed Kleene tableau for
{TX,, ..., TX,, F ( Y , v ..*v Y,)}. Proofi The proof is by induction on the length of the closed Beth tableau. If the tableau is of length 1, the result is obvious. Now suppose we know the result for all closed Beth tableaus of length less than n, and a closed tableau for the set in question is of length n. We have several cases depending on the first step of the tableau. If the first step is an application of rule F A , the Beth tableau begins
{{% FXi, .-.,FXn, FY A z}}, {{ST,FXI, * - * , FXn, F Y } {ST, F X I , * * * , FXn, F Z } } 2
3
and proceeds to closure. Now by the induction hypothesis there are closed Kleene tableaus for {&, F(X1v v Xn v Y ) } and {ST,F ( X , v v Xn v Z ) } . We have two possibilities: (1). If Y is not “used” in the first tableau, or if Z is not “used” in the second tableau, a Kleene tableau beginning
---
{{ST, F(X1 {{ST, F ( X 1
x,
v *’* v v (Y v *..v Xn))}
A
Z))}}
9
9
must close. (2). If both Y and Z are “used”, a Kleene tableau beginning {{snF(Xi v * * . v Xn v ( Y {{ST,
F(‘Y
A
A
z))}],
Z)>>
9
{{ST, F Y I , {ST,FZ}} Y
must close. The other cases are similar and are left to the reader.
78
ADDITIONAL FIRST ORDER RESULTS
CH.685
Thus the two tableau systems are equivalent. Now we verify a remark made at the end of ch. 5 0 10. Lemma 4.2: (Godel, McKinsey and Tarski): t,X v Y iff tlX or tl Y. Proofi Immediate from the Kleene tableau formulation. Lemma 4.3: (Rasiowa and Sikorski): If tl(3x) X ( x , al, ..., a,) where a, ,..., a, are all the parameters of X , then klX(b, a,, ..., a,) where b is one of the ai.If X has no parameters, b is arbitrary and kl(Vx) X(x). Pro08 A Kleene tableau proof of (3x) X ( x , a,, ..., a,) begins
{ { W x ) X(X, a,, a,)}}, {{FX(b, a,, --.a,)}}, a*.,
Y
and proceeds to closure. If b is some ai,we are done. If not, we actually have a proof, except for a different first line, of (VX)
8 5.
X ( X , 01,. * * Y a,).
Craig interpolation lemma
Theorem 5.1: If FIX= Y and X and Y have a predicate symbol in common, then there is a formula 2 involving only predicates and parameters common to X and Y such that k , X > Z and kIZ 3 Y; if X and Y have no common predicates, either tl X or t, Y.
-
The classical version of this theorem was first proved by Craig, hence the name. The intuitionistic version is due to Schiitte [17]. Essentially the same proof was given for a natural deduction system by Prawitz [15]. We give basically the same proof in the Kleene tableau system. For another proof in this system see [ll]. We find it convenient to temporarily introduce two symbols t and f into our collection of logical symbols, letting them be atomic formulas, and letting them combine according to the following rules.
x v t=
tv
x =t,
Xvf=f vx=x, X A t= t A X=X, x A f = f AX=f, -t=f, -f=t,
CH.645
CRAIG INTERPOLATION LEMMA
79
x 3 t = f 3 x = t, t3X=X X3f=-X, (3x)t = (Vx)t = t ,
( 3 x ) f = (Vx)f = f . By a block we mean a finite set of signed formulas containing at most one F-signed formula. When we call a block inconsistent, we mean there is a closed Kleene tableau for it. By an initialpart of a block we mean any subset of the T-signed formulas. We make the convention that if S is the finite set of unsigned formulas {XI,..., X,,)then TS is the set {TX,, ..., TX,,}. We further make the convention that for a set S of formulas, S, and S, represent subsets such that S, nS, =8 and S, u S, =S. By [S] we mean the set of predicates and parameters of formulas of S, together with t and$ Now we define an interpolation formula X for the block (TS,P Y ) (where S is a set of unsigned formulas and Y is a formula) with respect to the initial part TS,, which we denote by { TS, F Y } / {TS,}, as follows ( X may be t or f,but we assume t and f are not part of S or Y ) : Xis an ( T S , FY}I(TS,} if (1). [XI E CSll" CSZY y1, (2). { TS,, F X } is inconsistent, (3). (TX,TS,, FY} is inconsistent (we have temporarily added to the closure rules: closure of a set of signed formulas if it contains Tf or Ft). Lemma 5.2: An inconsistent block has an interpolation formula with respect to every initial part. Proofi We show this by induction on the length of the closed tableau for the block. If this is of length 1, the block must be of the form
{ TS,TX,FX}. We have two cases: Case (1). The initial part is ( T S , , T X } . Then X is an interpolation formula. Case (2). The initial part is { T S , ) . Then ( TS,, TX,FX} is inconsistent and t is an interpolation formula. Now suppose we have an inconsistent block, and the result is known for all inconsistent blocks with shorter closed tableaus. We have several cases depending on the k s t reduction rule used.
80
ADDITIONAL FIRST ORDER RESULTS
c~.6§5
K T v : The block is { T S , TX v Y, FZ}, and { T S , TX, F Z } and { T S , TY, F Z } are both inconsistent. Case (1). The initial part is { T S , , TX v Y } . Then by the induction hypothesis there are formulas U, and U, such that
U,is an {TS, T X , FZ}/{TS,,T X } , U2is an { T S , T Y , F Z } / { T S , , T Y } . Then U, v U, is an { TS, TX v Y, F Z } / {TS,, TX v Y } . Case (2). The initial part is { T S , } . Again, by hypothesis, there are U,, U2such that U,is an { TS, T X , F Z } / { T S , } , U,is an {TS, T Y , F Z ) / { T S , } . Then U, A U2is an {TS, TX v Y, F Z } / { TS,}. KF v : The block is { T S , FX v Y } , and { T S , F X ) or { T S , FY} is inconsistent. Suppose the first. Let the initial part be { T S , } . By hypothesis there is a U such that
U is an {TS, F X ) / { T S , ) . Then U is an { TS, FX v Y } / {T S , } . K T A : The block is { T S , TX A Y, F Z } , and { T S , TX, TY, F Z } is inconsistent. Case (1). The initial part is { TS,, TX A Y } . By hypothesis there is a U such that U is an {TS,T X , T Y , F Z } / { T S , , T X , T Y ) . Then U is an { T S , TX A Y, F Z } / {TS,, TX A Y ) . Case (2). The initial part is { T S , } . By hypothesis there is a U such that U is an { TS, T X , T Y , F Z } / { T S , } . Then U is an { T S , TX A Y, F Z } / {T S , } . KF A : The block is { T S , FX A Y } , and { T S , FX) and ( T S , FY} are both inconsistent. Suppose the initial part is { T S , } . By hypothesis there are U,, U2such that U, is an { TS, FX}/{T S , } , U, is an { T S , F Y } / { T S , } . Then U,A U, is an { T S , FX
A
Y}/(TS,}.
c~.685
CRAIQ INTERPOLATION LEMMA
81
KF- : The block is { TS, F - X ) , and { TS, T X } is inconsistent. Suppose the initial part is { TS,}. By hypothesis there is a U such that U is an ( T S , T X } / {T S , } . Then U is an {TS, F - X } / { T S , } . KT- : The block is {TS, T - X , FY), and { TS, F X } is inconsistent. Case (1). The initial part is { TS,}. By hypothesis there is a U such that
-
U is an { TS,FX}/{ TS,}.
Then U is an { TS, T X , FY}/{TS,}. Case (2). The initial part is {TS,, T - X ) . By hypothesis there is a U such that U is an { T S , F X } / { T S 2 } . We claim that U is an { T S , T X , F Y ) / {T S , } .
-
-
First we verify its predicates and parameters are correct. By hypothesis [UJc[S2]n[Sl,X], so immediately [-U]c[S,,-X]n[Sz, Y ] . We have the following two blocks are inconsistent:
{TSz,F U } {TS,, T U , F X ) . 3
It follows that the following two blocks are also inconsistent:
{TS,, T {TS2, T
- u, N
X,F U}, FY) N
3
and we are done. K F 3 : The block is { TS, F X 2 Y } , and { TS, T X , F Y } is inconsistent. Suppose the initial part is {TS,}.By hypothesis there is a U such that
u is an ( T S , TX,F Y ) / { T S , } . Then U is an { TS, FX =I Y ) / {TS,}. K T 2 : The block is { TS, TX 3 Y, F Z ) , and { TS, F X ) and { TS, TY, F Z } are both inconsistent. Case (1). The initial part is {TS,). By hypothesis there are U,, U2 such that U, is an { T S , FX}/{T S , } , U2 is an {TS, TY, F Z } / { T S , ) ,
82
ADDITIONAL FIRST ORDER RESULTS
c~.6$5
Then U,A U2 is an { TS, TX 3 Y, F Z } / {TS,}. Case (2). The initial part is {TS,, T X = Y}. By hypothesis there are U,, U2 such that U , is an { T S , F X } / { T S 2 ) , U2 is an { T S , T Y , F Z } / { T S , , T Y } . We claim U,= U2 is an { TS, TX 3 Y, F Z } / {TS,, TX 2 Y}. By hypothesis cull [S21 n [ S l y XI CU2l CS,, y] n [S2, 21 so [ul= U2]E [SlyX = Y] n [S2, Z] . 9
We have that the following four blocks are inconsistent: (1). w 2 , (2). {W,,TS,, (3). { TSI, TYY W), (4). { m2,TS2, and we must show the following two blocks are inconsistent:
fw, w, m,
The first follows from (2) and (3), and the second from (I) and (4). KF3 : The block is { TSyP(3x) X ( x ) } , and { TS, FX(a)}is inconsistent. Suppose the initial part is { TS,}. By hypothesis there is a U such that U is an { T S , FX (a)}/{ T S , } . Then [U]E[Sl]n[S2, X(a)]. Case (1). a# [U]. Then U is an { TS, F(3x) X ( x ) } / {TS1} Case (2). UE [ U ] , U E [SJ Again U is an { TS, F(3x) X ( x ) } / {TS,} Case (3). U E [U], a$ [S2]. Then ( 3 x ) U(z) is an {TSY F (3.1 x ( x ) } / { T S J* K T 3 : The block is { TS, T(3x)X(x), FZ}, and { TS, TX(a), FZ} is inconsistent, where a# [S, X ( x ) , Z ] .
m.685
CRAIG INTERPOLATION LEMMA
83
Case (1). The initial part is { TS,, T(3x) X ( x ) } . By hypothesis there is a U such that U is an {TS,T X ( a ) , F Z } / { T S , , T X ( a ) } . Then U is an { TS, T ( 3 x ) X ( x ) , F Z } / {TS,, T(3x) X ( x ) ) . Case (2). The initial part is { TS,). By hypothesis there is a U such that
u is an { TS,T X (a), F Z } / {TS1}. Then U is an { TS, T ( 3 x ) X ( x ) , F Z } / {TS,}. KFV: The block is { TS, F(Vx) X ( x ) ) , and { TS, FX(a)} is inconsistent, where &[S, X ( x ) ] . Suppose the initial part is {TS,}. By hypothesis there is a U such that U is an {TS, F X ( a ) ) / ( T S , ) . Then U is an { TS, F(Vx) X ( x ) } / {TS,}. KTV: The block is { TS, T ( V x ) X ( x ) , F Z } , and { TS, TX(a), F Z } is inconsistent. Case ( 1 ) . The initial part is { TS,, T ( V x ) X ( x ) } . By hypothesis there is a U such that U is an {TS,T X ( a ) , F Z } / { T S , , T X ( a ) } . Case (la). a # [ U ] . Then U is an { T S , T ( V x )X ( X ) , F Z ) / { T S , , T ( V x )X ( X ) } *
Case (lb). a E [ U ] , aEISl, X ( x ) ] . Again U is an {TS, T ( V x ) X ( x ) , F Z } / { T S , , T ( V x ) X ( x ) } . Case (Ic). a ~ [ v ] a#[S,, , X(x)]. Then (Vx) U(:) is an { TS, T ( V x ) X ( x ) , F Z } / {TS,, T ( V x ) X ( x ) } . Case (2). The initial part is { TS,}. By hypothesis there is a U such that
u is an { T S , T X ( U ) ,FZ)/{TSl}. Case (2a). a # [ U ] . Then U is an { TS, T ( V x ) X ( x ) , F Z } / { TS,). Case (2b). a€ [ U ] , a€ [Sz, X ( x ) , Z ] . Again U is an { TS, T ( V x ) X ( x ) , F Z } / {TS,}. Case (2c). a € [ U ] , a$ [Sz, X ( x ) , Z ] .
84
ADDITIONAL FIRST ORDER RESULTS
c~.656
Then (3x) U ( l )is an { TS, T ( V x ) X ( x ) , F Z } / {TS,). Now to prove the original theorem 5.1 : Suppose t I X 3 Y. Then { TX, F Y ) is inconsistent. By the lemma, there is a U such that U is an { TX,F Y } / {T X } . We have three cases: (1). U = t . Then since { Tt, F Y } is inconsistent, t, Y . (2). u=f. Then since (TX, F f ) is inconsistent, { F - X } is also inconsistent cf is not in X).Thus t-,-X. (3). U # t , u z f . Then U is a formula not involving t orf, all the parameters and predicates of U are in X and Y, and since { T X , FU} and (TU, F Y } are both inconsistent, tIX=, U and kI U 2 Y.
5 6.
Models with constant B function
In part I1 we will be concerned with finding countermodels for formulas with no universal quantifiers, and we will confine ourselves to models with a constant B function. To justify this restriction, we show in this section Theorem 6.1: If X is a formula with no universal quantifiers and y ,X, then there is a counter-model (’3, 9, k,P) for Xin which B is a constant function. Definition 6.2: For this section only, let a,, u2, a3,... be an enumeration of all parameters. We call a set r of signed formulas a Hintikka element if r is a Hintikka element with respect to some initial segment of a,, u2, a3,... (see ch. 5 Q 4). Lemma 6.3: If S is a finite, consistent set of signed formulas with no universal quantifiers, S can be extended to a Jinite Hintikka element. Proof: Suppose S is the set ( X , , X,, ..., Xn} where each X i is a signed formula. We d e h e the two sequences {pk},{&) as follows: Let
P, = 0, Qo = x,, ..., X”. Suppose we have defined Pkand Qkwhere Pk = Y,, ..., q , Q k = w,,..., w,,
85
MODELS WITH CONSTANT 9 FLINGTION
CH.686
and Pk u Qk (considered as a set) is consistent. To define Pk+, and Qk+l we have several cases depending on W, : Case atomic: If W, is a signed atomic formula, let pk+l
= Y,,
. * a ,
Y,, WI
Qk+l
= W2, ..*,Ws.
Case T v : I f W, is TX v Y, either TX or TY is consistent with Pk u Qk, say TX. Let Pk+i
= Y1, ...,
x, TX V
= w2, ...,
Y,
w,,T X .
Case F v : If W, is FX v Y then FX, FY is consistent with Pku Qk. Let Pk+l
= Y1, ..-,Y,, FX
V
Y,
Qk+l
= W2,
..., W,,F X , F Y .
CasesTA, F A , T - , T I aresimilar. Case T3 : If W,is T(3x) X(x), let a be the first in the sequence a,, a2,... not occurring in Pk or Q k . Then TX(a) is consistent with Pk u Qk. Let Case F 3 : If W, is F(3x) X(x), let {a,,, ..., sit} be the set of parameters occurring in P k u Q k such that no FX(a{,) occurs in PkuQk. Then {FX(ai,),...,FX(a,,)} is consistent with PkuQk. Let Pk+l
=pk>
Qk+l
=
W ~ , * *w,,FX(ai,),...,FX(ai,),F(3x)X(x), .Y
After finitely many steps there will be no T-signed formulas left in the Q-sequence because each rule T v , T A , T-, T 3, T3 reduces degree, and no rule F v , F A , F 3 introduces new T-signed formulas. When no T-signed formulas are left in the Q-sequence, no new parameters can be introduced since rule T3 no longer applies. After finitely many more steps we must reach an unusable Q-sequence. The corresponding P v Q-sequence is finite, consistent, and clearly a Hintikka element.
Remark 6.4: The above proof also shows the following which we will need later: Let R be a finite Hintikka element. Suppose we add (consistently) a finite set of F-signed formulas to R and extend the result to a finite Hintikka element S by the above method. Then
R, = s*.
86
ADDITIONAL FIRST ORDER RESULTS
c~.656
Since R s S , certainly R T s S T .That STGRT also holds follows by an inspection of the above proof; no new T-signed formulas will be added. Now we turn to the proof of the theorem itself. We have no universal quantifiers to consider, so we may use the definition of associated sets in ch. 2 9 4. Suppose X is a formula with no universal quantifiers, and y ,X. Then { F X } is consistent. Extend it to a finite Hintikka element S:. Let T,,,.., T,,be the associated sets of S:. Extend each to a finite Hintikka element, Sy, ..., S: respectively. Thus we have
s:, sy, ...)s:. For each parameter Q of some S: and each formula of the form P(3x) X ( x ) in St, adjoin FX(u) to S8 and extend the result to a Hintikka element S;. Do the same for S:, .... S,",producing S:, ..., S: respectively. Thus we have now
s;, s:, ..., s; .
Let T,,+iy...,T, be the associated sets of SAYS ; , . . . ,S:.Extend each to a Hintikka element, S:+,, ..., Sz respectively. Thus we have now
st, s:, ...)s,1,s,,,0 1, ..., sl),. For each parameter u used so far, and for each formula of the form F(3x) X ( x ) in S;, adjoin FX(u) to St and extend the result to a finite Hintikka element St. Do the same for each. Thus we have now
s;, s:, ...) s,",s;+ 1, ..., s;
.
Again take the associated sets, and extend to finite Hintikka elements, producing now 2 1 0 s;,~:,...r~n,Sn+l,...,S~,Srn+l ,...,s~.
Continue in this manner. Let 4)
W
s o = U S:, s l =U s:, k=O
k=O
By the remark above, for each n, S,T=S;T=S;T=.*..
Thus if S: has as an associated set S;, S,,, ES,.
etc.
c~.6$6
MODELS WITH CONSTANT
9 FUNCTION
87
It now follows that {So,S,, ..,} is a Hintikka collection. For example, suppose F- YES,..Let k be the least integer such that F- YES:. By the above construction, there is some set S," such that S: is an associated set of S: and TYES:. But then Sjk,sS,?, so by the above SjTcS,,and TYES,.The other properties are shown similarly. Moreover, 9(S,,)=9(Sm)for all m and a, as is easily seen. (Recall that 9's) is the collection of all parameters used in S.) Now as in ch. 5 0 3 there is a model for this Hintikka collection, and this model will have a constant B map, so the theorem is shown.