VI: Proofs of Incompleteness Theorems

VI PROOFS OF INCOMPLETENESS THEOREMS 1. 2-undefinability of the set ZR. We prove in this section the important theorem stating that if 9 is a closed consistent set of matrices, then the set G A R is not 9-definable. This theorem has many interesting applications, e.g. it provides us with examples of non-recursive sets and sets which are not definable in (S). We shall also see that the incompletenesstheorems are easy corollaries to the theorem to be proved here. The method of proof is clearly connected with the well-known argument with the help of which Cantor showed that the set of all reals is not denumerable.

R i s a closed consistent set of m a t r i c e s , then i s not 9-definable.

T h e o r e m 1. If the set

GA 9

Proof. Assume that the set G A 9 (i.e. the set of s e n t e n c e s which are elements of 9) is &-definable, and consider the set Izn[X(I,D,, n ) non E G A 91. According to theorems 2 4 and V !) 2 5 this set is $?-definable and hence there exists a m a t r i x @ in which exactly one f r e e v a r i a b l e a1 occurs and which satisfies the formulas

IV

#(I, D,, n ) non E G A 9 3@(Dn)E 9, X(1, D,, n ) E G A 9 3 *O(D,) E R. Remembering that @(D,) is an abbreviation for X( 1, D,, @) and that this m a t r i x is in G, we can rewrite these formulas as follows: (1) (2)

X(1, D,, n) non E G A 9 3S(1, D,,@) E G A 9, X(1, D,, n ) E G A 9 3 d J ( 1 , Dn,@)E G A 9.

Put n = @. Prom (1) we obtain then X(1, DO,@)E G A R and hence (2) yields *X(l, DO,@)E G A R. This proves that the set 9 is inconsistent.

R-UNDEFINABILITY

89

OF THE SET S p

It follows that our assumption was wrong and that the set G A9 is not &definable. Theorem 1 is thus proved. Taking 9 = Sr or R = Z we obtain the following corollaries : Corollary 2. The set 23 A (5 and hence also the set definable in ( S )l.

Sr is not

Corollary 3. The set S A G and hence also the set Z itself is not recursive 2. Corollary 4. The complement of the set S is not recursively enumerable Proof. If -S were recursively enumerable, it would be recursive, since the set S is recursively enumerable according to theorem V 5 1 (cf. theorem V 5 2, p. 87). We give still a positive formulation of theorem 1 : Theorem 5. Let ft be a closed consistent class of matrices, @ a m a t r i x in which exactly one free variable occurs, 0 a recursive definition of the function S ( x ) = S(1, D,, x ) , and Y the m a t r i x S(1,u(+),@). Under these assumptions the integer m =S( vlY)

satisfies either the conditions

m E R and @(D,,,)non

(1)

E

R

or the conditions

m non E 9 and ..n@(D,Jnon E 9.

(2)

Proof.

From m

= S(

I-

wY) we obtain

an M

-W)

since u is a recursive definition of S . It follows (cf. Chapter 111, section 4, theorem 1, p. 45)

I-@(W ++@(a(D( a 8

This corollary has been first proved by Tarski [22], p. 370. This corollary is due to Rosser [19], theorem V C. Cf. Rosser [19], theorem V A.

90

PROOFS OF INCOMFLETENESS TKEOREMS

whence by the definition of Y t-@(Dm) ++YU(D(*W)

i.e.

/-@(D,) + am.

(3)

The alternation of (1) and (2) is therefore equivalent to the trivial statement (m E

a).(a m non E 9)v ( mnon 6 9).(

v\

v * m non E

9)

which is obviously true since the class R is consistent. It is easy to explain why theorem 1 implies the incompleteness of the system (S): We can write down with the help of symbols of (S) m a t r i c e s which from the intuitive point of view are definitions of various consistent and closed classes 8.For instance we can take an arbitrary recursive definition of the relation 9

E

$w&

=x

and put an existential quantifier on its front. We are then inclined to believe that we have “expressed” in (S) the arithmetical statement x E S. Theorem 1 shows however that the m a t r i x which we have thus constructed is not a recursive definition of the class S,and hence either there are integers in 5 for which we cannot prove in (S) that they satisfy the m a t r i x , or there are integers outside S for which we cannot prove in (S) that they do not satisfy the m a t r i x . In both cases we infer that there exist s e n t e n c e s which are intuitively true but unprovable in (S). Such s e n t e n c e s are certainly undecidable in (S) because provable s e n t e n c e s are intuitively true. We elaborate this idea in the next section.

2. Proofs of the incompleteness theorem. I n this section we shall give three different proofs of the incompleteness theorem stating that there exist s e n t e n c e s undecidable in (S). A methodological discussion of these proofs will be given in section 4. T h e o r e m 1. There exist sentences 0 such that neither 0 nor a0 are in S. Moreover a 0 with these properties can be found among

PROOFS OF THE INCOMPLETENESS THEOREM

91

sentences of the form (Avk)z(v&.)where ,Z is a recursive definition of the set of all integers. First proof *. The set G A 5 is different from C 5A S r since the former set is definable in (S) (cf. Chapter V, section 2, theorem 10, p. 82) and the latter non-definable in (S) (cf. section 1, theorem 1, p. 88). Since S A G is contained in Sr A G (cf. Chapter IV, section 5, theorem 6, p. 69), there exist s e n t e n c e s 0 which belong to S ~ GAbut not to % A G. Every such s e n t e n c e is undecidable since its n e g a t i o n a0 does not even belong to Z r (cf. Chapter IV, section 5 , theorem 7, p. 69). To find the form of the s e n t e n c e 0 we remark that the set (5 A S is recursively enumerable since

m

E

G A 5E = ( 3 9 )[(g E '$%xu,) (m = g*) - (m E G ) ] .

It follows t,hat the set G A S possesses a formal definition of the

form (Ev,)@(a,, vk) where @ is a formal definition of the relation % = Amg[(gE '$3,u,j.(m= g * ) . ( m E G ) ] (cf. Chapter V, section 2, corollary 8, p. 80). Using theorem 1 5 we find a 0 of the form (Av,) a@(D,, vk) such that (1)

1

(0E Zr)-(Ev,)@(D,, vk)non E Sr v 0 lion E Z r - a(Ev,)@(D,, vk)non

E Sr.

The exact value of n could easily be deduced from theorem 1 5 but we shall not need it in our proof. We maintain that the second term of the alternation (1) is false. Indeed, (Ev,)@(D,, vk)non E 2 r implies that (Ev,)@(D,, v,) is in Sr since the set 'D A %r is complete (Chapter IV, section 5 , theorem 9, p. 69). This means that there is a q such that @(D,, D,) E Sr;@ being a formal definition of the relation 8,the last formula proves that % ( @ , q ) ,i.e. that q represents a formal proof of 0. Hence there exists a formal proof of 0, and we obtain 0 E S and a fortiori 0 E Sr.We have thus proved that the second This proof has been communicated to me orally by Tarski. It is contained implicite in Tarski [22], pp. 370-374.

92

PROOFS OF INCOMPLETENESS THEOREMS

term of the alternation (1) leads to a contradiction. It follows that the first term of this alternation is true, i.e. that

0 €XT, (Evk)@(D,,vk)non

E

Sr.

Observing that 0 = ( A V ~-@(On, ) vk) we obtain from (2) -@(D,,0,) E Zr for every q. Since @ is a recursive definition of a relation, we must have either k@(Dn, D,) or -@(Dn, D,). S being a subset of Sr,we infer that

k -@(Dn, D,)

(4)

for q

=

1, 2,

. ..

Formula (4) shows that the m a t r i x Z(aJ = v1 @(I),, a,) is a recursive definition of the set of all integers. Further we infer from (3) that the s e n t e n c e 0 does not belong to S since otherwise there would exist a q representing a formal proof of 0, and hence we would obtain k@(De,D,) which would imply that (Ev,)@(D,, vk) belongs to S.Finally 0 does not belong t o S since according to ( 2 ) it is not even an element of SL The proof of theorem 1 is thus complete. Second proof 6. Let @ be a recursive definition of the relation

-

'8 = M ( 9 E P%,)+ = @*)I. Since (Evk)@(aI,vk) is not a recursive definition of the set there exists a t least one integer m such that (5)

1

-

(mE S)- (Ev,)@(D,, vk)non E S v (mnon E Z ) . (Ev,)@(D,, vk)non

E

Z, S.

The exact value of m is irrelevant for the proof, but we may remark for the sake of completeness that m = S(1, D(Y),P)where Y = (Av,) -n@(o(aJ, vk) and where CT is a recursive definition of the function X(1, D,,x) (cf. section 1, theorem 5, p. 89). We shall show that the first term of the alternation ( 5 ) is false. Indeed, if m were an element of S,there would be an integer q such that %{m,q) which proves that @ (Dm, I),) would belong to %, This is approximatively the original proof of Godel [9], pp. 187 -189.

PROOFS O F THE INCOMPLETENESS THEOREbl

93

and hence (Ev,)@(D,,,, v k ) would be an element of 2. This shows that the first term of ( 5 ) leads to a contradiction. It follows that (6)

m non

(7)

(Avk) -@(D,, vk)non

E

S, E

S.

For every q the s e n t e n c e * @ ( D m , D , )is an element of 5. Indeed, if *@(Om, D,)were not in S, then @(Dm, D,)would be in S since @ is a recursive definition of a relation. Hence we would obtain %(m,q ) which would imply that m E S. Since this contradicts (6) we infer that

*@(Dm, De)E X for

(8)

q

=

1, 2,

...

Formula (8) proves that the s e n t e n c e 0 = (Evk)@(Dm, vk) does not belong to S ;indeed if this sentence were in 5, the set S would be o-inconsistent. Formula (7) shows on the other hand that a 0 does not belong to S. Finally, it follows from (8) that the m a t r i x .Z(a,) = a,) is a recursive definition of the set of all integers. Since t- 0 ff (Avk)z(vk), we see that the s e n t e n c e 0 satisfies all the conditions which we required in theorem 1. Third proofe. Let us consider the same recursive relation % as in the second proof, and let 17 be a recursive definition of % and v a recursive definition of the function a 2. It follows from these definitions that

= (3g)%(m,s), %Z(m9 ) 1-I n(%, DLA %(m,9 ) 3 I- * n ( D m , Dg),

mES

(9) (10)

-

(11)

v(D(m))m D(* m).

(12)

Let v, be a b o u n d v a r i a b l e which does not occur in 17,and Put (13)

= 17(a,, a2)

8& (AVh)[vh

4 a2

* n(v(al), Vh)].

We maintain that P is a recursive definition of 3.Indeed, if

* This proof

is due to Rosser [19], theorem 11, p. 89.

-

94

PROOFS O F INCOMPLETENESS THEOREMS

k *n(D,, D,) according to (11) and since it follows that * P(D,, D,). Let us now assume that %(m,9 ) . Since 5 is a consistent class (cf. Chapter IV, section 5, theorem 8, p. 69) it follows that no i satisfies the condition %( * m, i) and hence we obtain by (10) and (11) 8 ( m , g ) , then

t- *IT-+ *P,

t- WD,, DC7h

(14)

k * n(D(* m ) , Di)

(15)

for i

=

1, 2, 3,

. . ..

Formulas (15) and (12) together with theorem I11 4 1 yield

t-

(16)

*n(v(D,), D i ) for i

=

1, 2, 3,

. . .,

and we obtain by the propositional calculus

1[a3 M D,v a3 M D,v . . . v a3 M

-+

VI

n(v(D,), a3)

which implies (according to theorem I11 7 7) that (17)

(AVh)[Vh -4 D,-+ *

w@,), Vdl.

The conjunction of (14) and (17) gives according to (13) Do). Hence P is a recursive definition of the relation 8. We now apply theorem 1 5 taking 9 = 5 and

t-P(D,, (18)

@ = (Evk)P(al,

vk)

where vk is a bound v a r i a b l e which does not occur in P. We obtain a s e n t e n c e 0 such that either (19)

0 E '5. and @(D,)non E S

or ( 20)

0 non

ES

and *@(D@)non

E

S.

Moreover, (cf. formula (3) in the proof of theorem 1 5) (21)

1* 0 +>@(D,).

From the two possibilities (19) and (20) the first leads to a contradiction. Indeed, from 0 E S it follows that 8(0, 9 )for some g whence, by the property of P which we have just established,

95

PROOFS OF T E E INCOMPLETENESS THEOREM

t-

P(D,, D,)and hence (cf. Chapter 111, section 3, theorem 1, p. 44) t- (Ev,)P(D,, vk) i.e. @(Dg) E 5. Since this contradicts the second part of (19), we see that (19) is impossible and therefore that (20) must hold. The same reasoning shows also that

k * P(D,, D,)

(22)

for g

=

. . ..

1, 2, 3,

We can now prove that O is an undecidable sentence. Since O non E S by (2O), it remains to show that VI O non E S. Assume that

*O€%

(23)

and let g be an integer such that %( * O , g ) . We obtain therefore

-I

(24)

P(D(

@),

D,).

By (21) and (23) we obtain t-@(D,) whence by (18) and theorem I11 7 11

I- (EVk){bk -4 D,&

w

.

7

Vk)l v [Vk M

D, P(D,,

Vk)l

v

CD,-4 Vk & W , Y

Vk)l>.

Using the law of distributivity of the existential quantifier over the alternation we obtain from the last formula

i

I-

(25)

(EVk)[Vk -4 D, & w e , V k ) l v (EVk)[Vk M D, 85 Vk)l v (EVk)[D, -4 Vk c% PP,, V A l .

w,,

Let us write this formula as r l v r 2 v r 3 . We shall show that it follows from (13) that

t- * r.. Indeed

1P(D,, and from (24) that we get k * [PfD,,

t-

aS2)& D, C a2 --f * n ( D (

0),D,)

n ( D ( * O), D,).Using the law of transposition D, -4a2]whence

a2) &

(Avk) * [We, vk) & D, 4 vk] i.c.

r,.

It is known from the propositional calculus that if

t-rlv r2v I',

96

PROOFS O F INCOMFLETE!NESS THEOREMS

and t- * r,, then t-rlv r2.Hence the last term in ( 2 5 ) can be omitted. Using theorems I11 4 1 and I11 7 7 we obtain now from (25) the formula From ( 2 2 ) we obtain however

t-

P(D,, D1)&

.A

P(D,,D,)& . . . & * P(D,, D,)

Hence the assumption ( 2 3 ) leads to the result that the class S is inconsistent. This assumption is therefore wrong and the s e n t e n c e 0 undecidable. The s e n t e n c e 0 has the form (Avk)Z(vk)where the matrix Z(al) = .A P(D,, al) is according to ( 2 2 ) a recursive definition of the set of all positive integers. The proof of theorem 1 is thus complete. We note still separately for later use the result established in the second proof. We divide this result into two parts one of which uses only the consistency and the other the o-consistency of S: Theorem 2. If the set S is consistent, @ is a recursive definition of the relation $3 (or more generally of any recursive relation 9 such that m E S = (3n)9(m,n)),oa recursive definition of the function X(x) = S(1,D,,x) and !P = (Avk) *@(o(al), vk), m= S(1,D(!P),!P), then

(AvJ *@(Dm, vk)non E S, *@(D,,D,)E X for q = 1, 2, 3 ,

(26) (27)

.. .

( A h ) *@(o(D(YY)), Vk) + (AVk) -@(D,, Vk) E S. Theorem 3. If the set S i s o-consistent, then the sentence (Avk)*@(Om, vk) i s undecidable. Note that the formula ( 2 8 ) is the consequence of the definition of m and the formula 't-Dm M o(D(Y))which results immediately from the definition of o. Note further that (28)

(29)

1

(AVk)

.A@5(a(D(yu)),

vk) =

D(!P),(AVk)

*@(5(al)9

vk))

= S(1,D(Y),!P) = m.

GENERALIZATIONS

97

3. Generalizations. The three proofs given in section 2 are evidently very akin to each other and use one and the same fundamental idea. However, each of them uses different properties of the set S and the relation %. It follows that the three proofs will give rise to three distinct theorems when we enumerate separately those assumptions concerning the set S which are actually needed to carry out the proof. In this way we obtain the following theorems : Theorem 1. N o definable closed subclass $ of SrA rJ32 is complete 7. To prove this theorem we proceed exactly as in the first part of the first proof in section 2. The form of the sentence undecidable with respect to 9 depends on the class 9, and more exactly, on the form of matrices which can be taken as formal definitions of 9. In the special case, however, when the class $3' is recursively enumerable, i.e. has the form I n [ ( ~ q ) D ( Q)] n , with recursive D, the proof given on pp. 91-92 applies without changes with % replaced by 0 and we obtain a s e n t e n c e undecidable with respect to R of the form (Avk)L'(vk) where 2 is a recursive definition of the set of all positive integers. Theorem 2. No recursively enumerable, closed and consistent class of m a t r i c e s is complete8. Proof. A recursively enumerable class has the form

w I 3 q ) w h 411 where 0 is a recursive relation. It is now sufficient to repeat the third proof of section 2 replacing everywhere 8t by D. The undecidable sentence which we obtain in this way has exactly the same form as the sentence obtained in theorem 2 1. Slightly less obvious is the generalization of the second proof. Theorem 3. No definable closed and w-consistent class R i s complete. Proof. We first remark that there exist definable setswhich

'

This general formulation of incompleteness theorems is due to Tarski 1221, p. 370 and [23], p. 109. This theorem is due t o Rosser [19], theorem 11, p. 89. 7

98

PROOFS OF I N C O M P L E T E N E S S T H E O R E M S

are not %-definable (e.g. the set $t!itself, cf. section 1, theorem 1, p. 88). Using theorem V 4 1 we infer that there exists a &definable relation 3 with, say, n arguments such that for some j n the function F = mini 8 is not &definable. Let @ be a &definition of 3, vh a b o u n d v a r i a b l e which does not occur in@and y = (,uvh)s(j,v,,,@). Since ~1 is not a &definition of F , there exist integers pl, p2, . . . , pn-l such that

<

(1)

Q1(aPl),

-

*

>

-

D(PW-1))

D(F(i4, . *

-

9

Pm-1)) non

E

9.

We maintain that F(pl, . . ., P,,-~) = 1 and that (2)

N % ( P ~ ,-

3

pj-1,

q, p f ,

*

7

pn-1) for q

1,2,

* * *

Indeed, if there were a smallest q for which ( 2 ) would be false, we would obtain F(pl, . . . , pn-J = q and the following relations would hold :

@ ( D ( P ~* )- ,* > D(pi-11, Dq, D ( P ~ *) *, * * D ( ~ n - 1 ) )E 9 ' 3 * @ ( q P l ) , * * , QPj-J, D,,D(P,L * - * > W n - 1 ) ) E R ( r < q ) which proves (cf. Chapter 111, section 7, theorem 12, p. 53) that q(D(pl),. . ., D(P,,-~))m D,E 9. Since this contradicts ( l ) , we obtain (2). Let us put

6 =@(D(P~),

* * * 9

and

D(pj--l),aj, D ( P ~ )* ,* * D ( ~ n - 1 ) ) 9

-

=

(Pvh)s(j,v h , 6 ) *

From ( 2 ) we obtain and hence (3)

* 3 ( D a ) G 9 for q

=

1, 2 , 3,

(Ev,)X(j, vh>$)non

E

...

R

since the set 9 is w-consistent. Using theorem I11 3 3, we obtain therefore * ;i; M 1 non E R.Since ( 1 ) can be written as M 1 non E 9 we infer that the s e n t e n c e m 1 is undecidable with respect to 9. Theorem 3 is thus proved.

DISCUSSION OB THE INCOMJ%E!FENESS TaEOREMS

99

Remark. It is easy to show that also the s e n t e n c e 0 = (Evh)S(j,vh, 3) satisfies the conditions 0 non E 2, * 0non E R. First formula results immediately from (3). If the second were false, we would obtain (Av,)S(j, vh, a@) E 9 which would imply w 1 E 9. (cf. Chapter 111, section 3, theorem 3, p. 45) that Since this contradicts our previous result, we infer that a0 non E R.

4. Discussion of the incompleteness theorems. I n this section we compare the three theorems obtained in section 3. First we show that theorems 3 2 and 3 3 are not comparable. To achieve this we must compare the hypotheses concerning 9 under which theorems 3 2 and 3 3 have been proved. We see that one hypothesis of theorem 3 2 (recursive enumerability of 9) is stronger than the corresponding hypothesis of theorem 3 3 (definability of ft). The other pair of hypotheses behaves conversely: The assumption of consistency made in theorem 3 2 is weaker than the assumption of w-consistency made in theorem 3 3. I n order to make this heuristic argument really convincing we have to prove the following two theorems: T h e o r e m 1. There exists a definable and w-consistent closed class Rl of m a t r i c e s which is not recursively enumerable. T h e o r e m 2. There exists a recursively enumerable and consistent closed class ftz of m a t r i c e s which i s not w-consistent. Proof of theorem 1. Let 9 be the set of all recursive definitions of the set of positive integers, i.e.

Let 52 consist of all sentences of the form (Avk)Z(vk)where 2 E % and vk is a b o u n d v a r i a b l e which does not occur in Z. Finally put ftl = Sg. The set 2 is contained in S r since for every Z in % the s e n t e n c e s S( 1, Dn, 2 )( n = 1, 2, . . .) are all provable and hence (Avk)Z(vk) €St: (cf. Chapter IV, section 5, theorem 12, p. 72). It follows that

100

PROOFS OF INCOMPLETENESS THEOREMS

el

the set is contained in Zx and hence w-consistent (Chapter IV, section 5 , theorem 11, p. 71). The definability of Q is an immediate consequence of the formula (1) and the equivalence

sz E Q = ( 3 Z ) ( 3 k ) ( ( Z E ~ ) . [ O C ( V k , Z) = 11[n= (AVk)S(l, Z)l)* vk,

It follows that also the class 9, is definable (cf. Chapter V, section 3, theorem 10, p. 82). If 9, were a recursively enumerable class, we could apply theorem 3 2 to it and would obtain a s e n t e n c e (Avk)Z(vk) undecidable with respect to 9, and such that Z E 9. This, however, is not possible since all sentences of this form are 91-provable. Hence the class 9, is not recursively enumerable and theorem 1 is proved. Proof of theorem 2. Let Z be a recursive definition of the set of all integers such that the s e n t e n c e (Avk)Z(vk)be undecidable (vk is here an arbitrary b o u n d v a r i a b l e which does not o c c u r in Z). Put @ = S(1,(pk)S(1,v,,

.A

Z),

.A

Z) = ( E v ~ ) Z(V~).

Further, let 2 be the unit class consisting of @ alone and put

Q,

=

SQ.

? is recursive, the class 9, is recursively enumerable Since the set i (cf. Chapter V, section 5, theorem 1, p. 86). eZis w-inconsistent. Indeed, the s e n t e n c e s Z(D,), Z(D,), . . ., Z(Dn), . . . are all provable and hence belong to 9,. If 9, were w-consistent, the s e n t e n c e (EvJ -A Z(v,) would not belong to 9, which would lead to a contradiction since this s e n t e n c e is identical with @. Finally is consistent, since the s e n t e n c e @ is undecidable (cf. Chapter 11, section 6 , theorem 1, p. 41). Theorem 2 is thus proved. We note still that 9, contains false Sentences, e.g. @. A similar discussion shows also that theorems 3 1 and 3 2 are not comparable. We have again in these two theorems pairs of

101

DISUOSSION OF THE INCOllIPLETENESS THEOREMS

conditions imposed on a closed class 9 of matrices. We see that one condition imposed on 9 in theorem 3 1 (9C S r ) is stronger than the corresponding condition of theorem 3 2 (consistency of ft), whereas the second pair of hypotheses behaves conversely. This heuristic argument is supported by the following theorems : Theorem 3. There exists a definable closed subclass of i s not recursively enumerable.

Sr which

Theorem 4. There exists a recursively enumerable closed and consistent class which i s not contained in Sr. Proof. The class satisfies the conditions of theorem 3 and the class g2 the conditions of theorem 4. It remains to compare theorems 3 1 and 3 3. We shall show that theorem 3 3 is stronger than theorem 3 1. First of all it is evident that theorem 3 3 is a t least as strong as theorem 3 1 since subsets of S r are always w-consistent. To show that the converse implication does not hold we shall prove the following theorem: Theorem 5. There exists a closed class 9 of m a t r i c e s which is o-consistent but not contained in Sr. Proof. Let us denote by {@} the class whose sole element is @. We shall call a s e n t e n c e sf, o-consistent if the set S,, is w consistent. The following equivalence results immediately from the definition of w-consistency: [@ i s w-consistent]

(h) [Oc(a,,!P)

=

( W W d V k , ~= )

2

=@

E

G*(!P){(!PESm).

= h = l].(n)[X(l,D*,!P)E S,,]

11 -+

( E V k ) [ W , Vk,

-w

non

E

+

%{,I>}.

Since n E = ( 3 g ) [ ( g E '$lcp)). (g, = n ) ]and the class p,,, is recursive (cf. Chapter V, section 3, theorem 9, p. 82) this equivalence proves that the set (Iof w-consistent s e n t e n ce s is definable. Since the set G A %r is not definable according to theorem 1 1, it follows that the sets (I and G ASr do not coincide, and since GA S r is a subset of (I, there exist s e n t e n c e s which are o-con-

102

PROOFS OF INCOMPLETENESS THEOREMS

sistent but not true. For every such s e n t e n c e @ the set o-consistent but not contained in Sr. Theorem 5 is thus proved.

is

Remark. Using theorem 1 5 one can easily exhibit explicitly an o-consistent and false sentence@. It can be shown that the Q, obtained in this way has the form (Ev,)(Av,)(Ev,)D(v,, v,, v,) where Q is a recursive definition of a ternary relation. It is worth while to note in connection with this result that every o-consistent s e n t e n c e of the form (Av,)(Ev,)Q(v,, v,) where Q is a recursive definition of a binary relation is automatically true s. We remark still that results reached in theorems 3 2 and 3 3 are in a sense best possible. It is namely known that there exist defhable, consistent, and complete classes of m a t r i c e s lo. Hence we can neither replace in theorem 3 2 the assumption of recuisive enumerability of rie by the weaker assumption of definability nor in theorem 3 3 the assumption of o-consistency of 8' by the weaker assumption of the ordinary consistency. Turing [25], p. 194. The existence of such classes of matrices has beenfht established by Lindenbaum. Cf. Tarski [20], theorem I. 56, p. 394. The proof that classes constructed by Lindenbaum are definable is contained in the paper Novak [15], p. 95-99. 9

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