Incompleteness of a Formal System for Infinitary Finite-Quantifier Formulas

Incompleteness of a Formal System for Infinitary Finite-Quantifier FormulasAuthor(s): John GregorySource: The Journal of Symbolic Logic, Vol. 36, No. 3 (Sep., 1971), pp. 445-455Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/2269953 .

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THE JOURNAL OF SYMBOLIC LoGic Volume 36, Number 3, Sept. 1971

INCOMPLETENESS OF A FORMAL SYSTEM FOR INFINITARY FINITE-QUANTIFIER FORMULAS

JOHN GREGORY1

In [1], various formal proof systems for infinitary formulas were defined.2 3 The formal proof system 13 .(Z; Q2) is the result of extending the basic predicate calculus 93 . by adding a collection Z of axiom schemes and a collection Q of rules of inference. Let Taut be the collection of all infinitary propositional tautologies, considered as axiom schemes. Let QI consist of all the quantificational rules of independent choices. We will show, in ?2 (see Theorem 2.1), that 3 ..(Taut; 0) is not complete for L., (i.e., infinitary finite-quantifier) sentences; that is, we will exhibit an L.. sentence + such that - is true in all models, but -o is not provable in W3. (Taut; 0). (The unprovability is shown by a weak forcing version of Boolean general models.) This answers a question of Karp in [1, 12.1(i)]. In ?4, we will show that our cb is "93.. complete for L.. sentences."

By Theorem 11.5.1 of [1], 93..(Taut; El,) is complete for L,,,, sentences. Since we will show that 93 .,(Taut; 0) is not, there is a rule of independent choice which is not a derived rule of inference in 93,3(Taut; 0). At the end of ?2, we will give a specific instance of a rule of choice not derivable in 93x,(Taut; 0). In ?3, we will show that each rule of independent or dependent choice is a derived rule of inference in the formal system 930.. This will be a corollary of the following fact: if 13 is a S1 notion of proof such that only valid formulas are provable, and if L.. formula 0 is provable in 93, then 0 is provable in 93...

We will assume the informal axioms of Zermelo-Fraenkel set theory with the axiom of choice.

In ??3, 4 we will require both knowledge of the (bottom level of the) hierarchy in Levy's [3] and the ability to relate this hierarchy to (as in [2]) proof systems and to satisfaction of L.. formulas by Boolean models. For these sections, we will also use the fact (mentioned in [1, p. 141], with different notation) that an La, formula is consistent in 93. iff it is satisfiable in some Boolean model.

?1. Notation. We assume the reader is acquainted with infinitary languages and systems of proof for them. Our notation differs from [1] and [2], partly in that conjunctions and quantifications are over sets instead of over sequences or functions, and also in that -, A, V are our primitives instead of ---, A, V. As in [1]

Received August 3, 1970. ILThe work on this paper was partially supported by NSF grant GP-1 1263. 2 We wish to thank Professor Carol R. Karp for helpful suggestions on the first version of

this paper. 3 Our notation differs inessentially from Karp's [1]. The subscript "6 a)" indicates that there

is no restriction on the cardinality of conjunctions or quantifications.

445



446 JOHN GREGORY

and [2], the syntactical notions are definable in Zermelo-Fraenkel set theory with the axiom of choice, and the basic ones are 1:1 in the hierarchy of Levy's [3].

A finitary (i.e., ordinary) first order language L determines the collection of its (finitary) relation symbols and constant symbols. The corresponding language L.. is definable as follows. We assume a sufficient supply of variables. A term is a constant or a variable.

DEFINITION. L.,, is the smallest class such that: (0) atomic sentences R(tl, * * *, t.) are in L.., for each n E w, terms t1, * ,,

and n-ary relation symbol R of L; (1) if k is in L. , then so is its negation n.b; (2) if T is a subset of L..0, then the conjunction AT is in L..x; and (3) if b is in L. ,, and if v is a set of variables, then the universal quantification

Vvb is in L.W., Abbreviations. The infinitary disjunction VT is-AtA{--, I e IF}. The finitary

conjunction b A 0 is A{+, 0}. The other finitary connectives implication -*, disjunction v, and equivalence -* can be defined from A and -_. The existential quantification 3v is -- v --. Given a list x0, xl, * * * of the elements of v, we may write Vxoxl *... instead of Vvb. Thus, V{x}b becomes Vxb. We will sometimes use the symbols "V." "I-," etc., in abbreviations of the English language; the context will determine the usage.

The letters a, P, y, 8 always denote ordinals. We write x -< y iff x has cardinality (strictly) less than y. The letters K, A denote cardinals such that K > A 2 w.

The language L.,\ is the collection of those L.4, formulas b such that v < A for every quantification VvO occurring in 0. The language LKA is the collection of those LCx formulas 0 such that T'< K for every conjunction AT occurring in si. Thus, LOC LI# c L c C L .coco Lao is (essentially) the same language as the original finitary language L. L.,. (L.,,) is called the language of all infinitary (countable) finite-quantifier formulas, given L.

As usual, given a formula O(xl, *. * * x,,), 0(Yi ,... * y,,) is defined to be the result of substituting Yi for xl, ... , y,, for xn in 0, with suitable modifications of the bounded variables if some yi is not free for xi in 0.

The relativization G(*) of formula 0 to 0(x) is defined recursively: atomic 0 = 0); (-_0)'*) is _m&*); (As)(*) is A{0(*) I 0 e 0}; and (Vv@)<*) is Vv((Apetby)) + 0w*).

Given a formula xb and functionf, define Sf(b) to be the result of replacing w by f(w) at each occurrence in 0 of a w e Dm(f). Given a formula b and a function f from a set of variables to terms, define SFf(b) to be the result of replacing x by f(x) at each free occurrence in b of an x e Dm(f).

DEFINITO. The basic predicate calculus 93.. with equality has the following axioms and rules of inference:

Basic propositional axiom schemes. 0 -4. (-9 -* 0). (A (fo b)

(AT) for each e T.



INCOMPLETENESS OF A FORMAL SYSTEM 447

Basic propositional rules of inference. Modus Ponens. Infer -q from premises 0 and 0 -0 Conjunction. Infer A'T from the set TF of premises.

Basic quantificational axiom schemes. Vv( -* )-* (0 -? Vve) if no variable in v is free in 0. (Vv0) -+ SFf(9) if 0 has no free occurrence of an x e v bound in 0 by f(X).

Basic quantificational rule of inference. Generalization. Infer VvO from 0. Equality axioms.

x = X. (A i\n(xi = ye)) -> (R(xo, * * *, x.) -+ R(yo, * , n))

for each n + I-ary relation symbol R and variables. Then the definition of a proof in 93.. is assumed known (given a finitary lan-

guage L). If we add a collection I of axiom schemes and a collection Q of rules of inference

to the above, then the resulting proof system is called 93,. ,.(.; Q). If F or ?2 is empty, then we omit mentioning it; e.g., 93.,o(l; 0) is X,00(z).

If we restrict the proof system 9.,. by requiring that every formula occurring in a proof must be in the language L,.A, then the resulting proof system is called 93.,A; similarly, we can define 93.x(E:; Q).

We assume known the meanings of "structure at satisfies f," "formula b is satisfiable (i.e., consistent)," and "formula & is valid." A sentence 0 is (semantically) complete for L* sentences iff, for every L* sentence 0, either 0 - or 0 o- is valid.

A proof system 93 is complete for L* sentences iff, for every L* sentence a, a is valid iff a is provable in 93. A formula 0 is called 93 consistent iff there is no proof of m 0 in 9. A sentence 0 is 93 complete for L* sentences iff both 0 is 9 consistent and, for every L* sentence 0, either 0 --> a or 0 -of. -,af is provable in 93.

The following is essentially Theorem 11.4.1 of [1]. COMPLETENESS THEOREM FOR L4,.. (KARP). 93,.,. is complete for L.,1,. sentences. A propositional formula 0 is one built up using only 0-ary relation symbols and

the propositional connectives -A, A. Let I be the set of 0-ary relation symbols occurring in 0. Consider any 2-valued function i e 2'. For every propositional a built up from the symbols in I, the truth-value i*(a) is defined by induction: i*(R) = i(R); i*( = 1 - i*(); and i*(AT) = n I{i*() I a e T}. The above propositional formula 0 is a tautology iff, for all such i, i*(0) = 1. An instance of a tautology 0 is a formula of the form SF(O) for some F mapping 0-ary relations to formulas of L. ., (0 need not be in the L. . of the specified L).

DmunqmON. Taut is the collection of all propositional tautologies, considered as axiom schemes. Thus, each instance SF(O) of a propositional tautology 0 is an axiom of 93..(Taut).

The rule of independent choices is: From

V et infer V Vveni, fel {el

provided that the vj, i e I, are pairwise disjoint and no variable of vj occurs free in any 0i such that j :A i.



448 JOHN GREGORY

The rule of dependent choices is: From

V 6d infer V 3wflVvflG6, R< aO<a

provided that the v0, fi < a, are pairwise disjoint, no variable of vf occurs free in any Oy such that y < f, and Wfl contains every free variable x of Od such that x E v6 for some 8 < 8.

We write QI, D, QID for the collections of rules of independent choices, dependent choices, of both kinds of choices, respectively.

A derived rule of inference in a proof system 93 is one such that the conclusion of the rule is provable in 93 whenever all the corresponding premises are provable in 93.

?2. Incompleteness of 930 .(Taut) for L . . sentences. 2.1. THEOREM. The predicate calculus 93 , cm(Taut) is incomplete for infinitary

finite-quantifier sentences. PROOF. We will find an infinitary finite-quantifier sentence # such that # has no

model, but --b has no proof in 93 . (Taut). We have arranged that S be a modifica- tion of a sentence a, used in Malitz's [4, pp. 170-173]. Let X be the first cardinal (210)+ cardinally greater (>) than the set of all subsets of the set w of natural numbers. Consider the language L which has the following relation symbols: equality symbol =; unary T, A; unary Qa for each a E R; binary <, R. The inter- pretations of L are structures <X, T, A, Qa, ?-, R>ac-..

Let the sentence b be the conjunction of the set 'D of the universal closures of the formulas occurring among (1) to (5) below. (1) " X is the disjoint which union of T and A":

T(x) v A(x) -,(T(x) A A(x)).

(2) " <T, < >is a full binary tree":

T(x) <-?3y(x < y V y < x)

(x < y A y < z)-? x < z X < z A y < z-*x < y V y < x [Vx3yzVw((x < wr" y < w V z < w) A y # z A -m3y < z A -iz <yYI((x))

[3yoy1Vz((yo < z V Yi < z) A yo # Yi A 'yo < Yi A 'Yl <Y yo)]f( VE c. * y..VZ(Z < x Z = yO V **. V Z = yn).

(3) "R c T x A and distinct elements b of A correspond to distinct maximal branches {t I tRb} of T":

xRy -> T(x) A A(y) x < y A yRz-*xRz xRz A yRz-+x < y v y < x A(x) -> 3y(yRx) yRx -+ 3z(y < z A zRx) A(x) A A(y) A Vz(zRx +-+ zRy) -+ x = y.




(4) " The sets Q,, are nonempty and pairwise disjoint subsets of A":

Qe(x) -* A(x) for a E M. 3xQa(X) for a e X, --(Qa(x) A Q8(x)) for distinct cc, e e

(5) "The sets Qa partition A, and, for each Qa and each element t of T. there is an element of Qa that corresponds to a maximal branch through t":

T(x) 3y(xRy A Qa(Y)) for a e X, A(x) V{Qa(x) I a e h}*

The formulas (5) will only be needed for ?4, where we show that # has a completeness property. Thus, A(Q c L,,,,) will also be inconsistent and 903 .(Taut) consistent.

Our q has no model. For suppose <X, T, A, Qa, <, R>ae% were a model of 0. By (3), A can be identified with a set of subsets of T. By (2), T is countable. Thus, A :? 20 < M. By (4), A ? M, a contradiction.

We must still show that -of is unprovable in X30*, .(Taut). This will be shown by a version of weak forcing in which large quantifiers are interpreted as in the general models of [1, 12.3].

Let T = Un ..2n + the set of all nonnull 2-valued finite sequences t. We consider T to be a set of formal constants. Find a sequence <Qa I a e X> of infinite pairwise disjoint sets of constants not in T. Let A = Ure~xQcc and X = T U A. For each set v of variables, let Sv be the set of all functions s: v -* X with finite range. We will consider the forcing language L' which allows just those sentences of the form SF3(0) where 0 e L.,, and s e Sv for some set v of variables.

Let the set C of forcing conditions consist of all 1-1 functions p: Dm(p) -> 20 such that finite Dm(p) c A. Order C by reverse inclusion: p < q iffp - q. C has the property that

2.2. Vpq3r(p $ q -*r p A Vr1 -n(r1 ? r A r1 < q)) (so C could be identified with a dense subordering of an algebraically unique complete Boolean algebra).

2.3. DEFINITION. The weak forcing relation. 2.3.1. Define the weak forcing relation IF for atomic sentences of L' by:

pIFc= c for ceX pIFIT(t) for teT plIFA(a) for aeA p IF Qa(a) for a e Qa pIFIt<tt for tc t1,tteT p IF tRa for t c p(a)

not p IF 0 for all other atomic 0 and conditions p. 2.3.2. Weak forcing is defined over the propositional operations by:

p IF I- iff Vq < p (notqlF 0) p IF A eA iff VO E Q(p IF 0).

2.3.3. Forcing over universal quantifications is defined by: p IF VvO iff Vs e Sv(p IF SF3(O)).



450 JOHN GREGORY

Just as with other concepts of weak forcing, we have the following lemmas. 2.4. CoNSISTENCY LEMMA. There is no condition p and sentence 0 such that both

pl-OandplIF -,0. 2.5. ExTENSION LEMMA. If p < q IF 6, then p IF 0. 2.6. LEMMA. For atomic sentences 0, p IF m -,0 iffp IF 0. 2.7. LEMMA. p1F-n -,-, iifpl1F0. If not pIF 0, then there is q <p such that

qII IF . This is proved by induction on formulas, using 2.3.2, 2.3.3, and the last lemma. Forcing over the abbreviations -A, V, and 3 is determined by:

p 1F 0 -X- iff Vq < p(q IF 0 implies q IF -i) p 1F Ve iff Vq < p3r < q3O E) 0(r IF 0) p IF 3v0 iff Vq < p3r < q3s E Sv(r IF SF8(0)).

If p(a) defined, then p IF tRa or p 1F -,tRa, depending on whether t c p(a) or t c4 p(a). Also, p F -,cRd for c 0 Tor d MA.

2.8. LEMMA. If 0 is afinite-quantifier L' sentence not containing R, then

p 1F 0 if <X, T, A, Qa ,>,Xe asatisfies0 (where constants are interpreted by themselves).

2.9. LEMMA. Vp(pI1IF). PROOF. This will follow from the fact that, for every 0 among (1) to (5) with

some set v of free variables, and for every s E Sv, p IF SF8(0). Lemma- 2.8 can be used for (1), (2), (4), and the last formula of (5). We write out

two of the other possible cases (all cases are straightforward). p 1F (A(c) A A(d) A Vz(zRc zRd) -# c = d): Suppose q IF A(c) A A(d) A

Vz(zRc " zRd), to show q 1F c = d. Then c, d E A. Assuming c + d, then there is r < q such that r(c) and r(d) are defined. Since r is 1-1, there is t c r(c) such that t c: r(d). Then r IF tRc, but not r IF tRd. This contradicts q 1F tRc -+ tRd. Thus, c = dandq1- c = d.

p IF T(c) -+ 3y(cRy A Qt(y)): Assume P, IF T(c). Then c E T. Assume q < pl, to find r < q and a E Qc, such that r IF cRa A Qa(a). There exists some a E Qa - Dm(q). There exists some f e 2 - Rg(q) such that c c f. Put r = q u {(a,f)}.

2.10. THEOREM. If L< c formula 0 is provable in 93 (Taut), and if v is the set of free variables of 0, then Vp(p 1F Vv0).

PROOF. This is shown by induction on proofs. There are five possible cases to consider at the induction step.

Case 1. 0 is one of the basic axioms. Case 2. 0 is inferred by modus ponens from some X and X 0 for which the

inductive hypothesis holds. Case 3. 0 is inferred from the rule of inference for conjunction from a set e

of formulas for which the induction hypothesis holds and 0 = Ad. Case 4. 0 is inferred from infinitary generalization applied to a formula & for

which the induction hypothesis holds and 0 = Vvo#. Verifications for Cases 1 through 4 are straightforward and do not depend on the

particular nature of the partial ordering C and 2.3.1, provided Lemma 2.6 holds.




The verification for forcing of the quantification axioms is analogous to the verification for the general models of [1, 12.3].

Case 5. 8 is an instance of an infinitary propositional tautology. For this case, we require a few definitions and lemmas.

Suppose g satisfies the following condition: 2.11. g is a permutation of X such that g [ T is an isomorphism of T and g maps

Q. onto Q, for every a E M. 2.12. DEFINITION. Let g+(f) = U{g(t) If D t e T} for each fe 20). Define

g*: C-t. C by g*(p) = {(g(a), gI(f)) I (a, f) e p}. 2.13. LEMMA. g* is an order isomorphism of C. 2.14. LEmmA. For every sentence 0 of the forcing language L', p IF 0 iffg*(p) 1F

SFg(O). PROOF. The proof is by induction on formulas. We consider the case of the

induction where 0 is Vvb and the induction hypothesis holds for all SF8(O) such that s e Sv. Note that So = {gs I s e Sv}. Thus,

p 1F 0 iff Vs E Sv(p IF SF,(O)) iff Vs e Sv(g*(p) IF SFg(SFs(b)) = SFgs(SFg(0))) iff Vs E Sv(g*(p) 1F SF,(SFg(#))) iff g*(p) IF SFg(G).

2.15. LEMMA. If q, r < p, X0 is afinite subset of X, and X0 n A c Dm(p), then there is a g such that g satisfies condition 2.11, q U g*(r) E C, and g r X0 is identity.

PROOF. Find integer n large enough so that, for all distinct go, fi e Rg(q) u Rg(r),fo r n # fi r n, and so that all t e X0 have domain less than n. For t e T, g(t) is defined by: g(t) and t have the same domain, and, for all m E Dm(t),

g(t)(m) # t(m) if m = n and t [ n is a segment of some r(a), a 0 Dm(p),

g(t)(m) = t(m) otherwise. Find permutation g [ A such that g r Dm(p) is identity, g maps every Qa onto

Q,, and g maps Dm(r) - Dm(p) into A - Dm(q). Then g satisfies condition 2.11 and g r X0 is identity. If a E Dm(p), then gI(r(a)) = r(a) = p(a) = q(a). Thus, q U g*(r) is a function. We show that it is 1-1. Consider any a E Dm(r) - Dm(p) and any b E Dm(q). Then g+(r(a)) r n # q(b) [ n if r(a) # q(b); and g+ (r(a))(n) # r(a)(n) = q(b)(n) if r(a) = q(b). For a e Dm(p) and a # b 1E Dm(q), g+(r(a)) -

q(a) # q(b). Therefore, q U g*(r) is 1-1. 2.16. DEFINITION. If p and L' sentence 0 are such that p(a) is defined whenever

a e A and a occurs in 0, then we say that p is large enough for 0. Note that, for each L' sentence. 0 and each p, there is q < p such that q is large enough for 0.

2.17. LEMMA. Ifp is large enough for L' sentence 0, then p 1F 0 or p IF -,O. PROOF. Suppose p is large enough for 0. Assume that neither p IF 0 nor p IF -- .

By Lemma 2.7, there are q, r < p such that q F1--, 0 and r IF 0. Let X0 be the set of constants occurring in 6. We can now apply Lemma 2.15 to get a g such thatg satisfies condition 2.11, q U g*(r) E C, and g r X0 is the identity. Then

qLg*(r) < qlIF--,O, q L g*(r) < g*(r) 1F SF/(O) = 0,

which contradicts the Consistency Lemma 2.4.



452 JOHN GREGORY

PROOF OF CASE 5. Suppose 0 is an instance SF(+) of a propositional tautology e built up from a set Dm(F) of propositional variables. Let v be the set of free variables of 0. We show p IF VvO. If not, then, for some s e Sv, not p IF SF(0). By Lemma 2.7, there is q < p such that q IF -,SF,(G). There is r < q such that r is large enough for SF,(0). Now SF&(0) = S1(#) for the composition f = S&F.. Define i: Dm(F) -* 2 by: i(w) = 1 iffr Wf1(w); i(w) = 0 iffr IF -nf(w). By Lemma 2.17, i is indeed totally defined. By induction on subformulas of b, for every sub- formula M of #: r IF Sf('q) if the truth value i*(j) = 1; r IF -,Sf(n) if i*(n) = 0. But i*(b) = 1, so r IF SF&(0). This contradicts r < q IF -,SF&(0).

This finishes the proof of Theorem 2.10. If -- were provable in Xpj(Taut), then by Theorem 2.10, 0 IF --. But Lemma

2.9 shows that 0 IF q, which contradicts Consistency Lemma 2.4. This finishes the proof of Theorem 2.1.

By Theorem 2.1 and [1, 11.5.1], we know that some rule of independent choices for L.., formulas is not a derived rule of inference in q3a,(Taut). We now give a specific instance where the rule fails.

Assign a variable xt to each t e T in a 1-1 manner. For each n E W, let v. = {xt I Dm(t) = n + 1}, and let m(n) be the finite cardinal of vn. Let v = Unevi. For each n e co, let T^(x) be the formula

T(x) A 3Yo YnVz(Aicjsn(Yi < y,) A (z S X4-4Z = yo V * V Z = Yr)) (informally, "x is of height n in T"). Let 0,(vn) be the formula

- (O A A{Tn(X) A Tn(y) A x # y I distinct x, y e vj}) (informally, "if i, then v. is not a set of distinct elements of height n in the tree T"). Our instance of failure of the rule of independent choice is given by:

V n co(vn) is provable in 93 ,,,(Taut), but nVc-.v~n0n(Vn) is not provable in 93. .(Taut).

The unprovability of the last formula follows from 2.10, 2.9, 2.4, and 0 IF m 1Vnsec4VO{Rn(Vnl), We now informally show that Vnec,,,O(vn) is provable in T3..(Taut). Assume

m Vnv o~n(Vn),

Then i, Tn(x), and x # xl for all n e co and distinct x, xl e v.. It can be shown from formulas (2) of the definition of # that Vy(T(y) - V n, Tn(y)) and

Vyo *Ym(n)((At m(n) Tn(Y1)) -+ Vs < y a men) = yj) Thus, Vy((y) -e VXVx = y).

In the following, let 00(x, y) denote xRy, and let #1(x, y) denote .-xRy. For each a e K, there is some y such that Q,(y). By a tautology,

V A #f(x)(x, y). fe2' xet'

This shows that AV3Y(Qa(Y) A A \f(x)(x, Y).

By a tautology applied to this (noting that K > 2v),

VV 3Y(Qe(Y) A A\ #f(X)(X, Y)) A\ 3Y(Qa(Y) A A\ #1(X,(X y)) fe~~~~~vG XEI3l~ Iz e




Then there are a < # e X, y, z, anddfe 2V such that

Qx(y) A A ~f(x)(X, y) A Q8(Z) A A 0f(X)(X9 z). XGV XGV

Then A(y), A(z), and Vx(xRy ?- xRz). Thus, y = z. But not Q,(y) A Q8(y), a contradiction. Thus, Ve,,6AG(v0) is provable in V3.,,(Taut).

Remark. In the proof of 2.1, our use of weak forcing could have been re- placed by the use of the pre-Cohen Boolean general models. Boolean general models arise from the general models of [1, 12.3] by allowing the values of formulas to lie in any Boolean algebra instead of requiring them to lie in the usual two-valued Boolean algebra. In our case, the Boolean value of an L.. sentence 0 is {p I p IF 9}. The Boolean algebra is the set of all these values, ordered by inclusion. This Boolean algebra is not K-distributive for infinitive cardinals K. but (by Case 5 of the proof of 2.10, the main point of that proof being 2.17), each expression of the distributive law by a sentence of the forcing language has Boolean value unity.

?3. The rules of choice in IP.. In this section and the next, we assume familiarity with Levy's [3] hierarchy En, I% of formulas .of the formal set theory ZFC (for- malized Zermelo-Fraenkel set theory with the axiom of choice).

Let V be the class of all sets. A S1 (H 1) relation is any class R such that, for some 11 (11l) formula at of ZFC, R = {(a,, - - *, aJ) I <V, e> satisfies s1(al, * ? *, a,)}. If a proof system 93 is such that the unary relation " 0 is provable in V " is El (HI1)

in the argument 0, then we will say that V3 is a El (H1i) notion of proof. 13a, co cog and 93<, -2nID) are 1 notions of proof. (The verification of this is nearly as in (2]. The main change is that conjunctions and quantifications operate over sets now, instead of over sequences; the rule of dependent choices is especially affected by this.)

301. LEMMA. Xis a l1 notion of proof. This is because 0 is unprovable in 93.,. iff either 8 is not an L.. formula or -1 8

is satisfiable in some Boolean model. But "L., formula -- 0 is satisfiable in some Boolean model" is a El relation of 0.

In the next lemma, we write ZFC F V iff d is provable in the theory ZFC. 3.2. LEMMA. Let d(x) and YR(x) be S1 and H1 formulas, respectively, in the one

free variable x. Let <<(x, y) be a El formula. If ZFC F Vx3yq(x, y) and ZFC, F Vxy(6(x, y) A y ? cO A d(x) - (x)),

then ZFC F Vx(d(x) - (x)). This can be proved by using Theorem 36 of Levy's [3]. 3.3. THEOREM. Suppose 93 is a El notion ofproof such that only vaildformulas are

provable. If L ,, formula G is provable in X, then 0 is provable in 93s . PROOF. Let a correspond to "L.. sentence 0 is provable in 93;" let .4 corres-

pond to " 0 is provable in 93 Let W correspond to "all conjunctions AT occurring in 0 are over sets T -c." x, y correspond to 0, a, respectively.

Assume L.. formula 8 is provable in T3, to show that it is provable in 13X<, By Lemma 3.1 and the last paragraph, we can assume 0 is in L,,.. But 0 is valid. By Karp's completeness theorem for L,01W, 0 is provable in 93,0



454 JOHN GREGORY

3.4. THEOREM. The rules of independent and dependent choices are derivable rules in Wa3cocC.

PROOF. Suppose 6 and b are the premise and conclusion, respectively, of one of these rules. Suppose 0 is provable in 93.,, to show that f is provable in 93.,,. Then 0 is also provable in X0 0(KID). Applying the rule in 93 .(QIm), 6 is provable there. X0 0(KID) is a T1 notion of proof. By 3.3, 6 is provable in S3..,

Remarks. The referee points out that the above implies the known fact (from [2]) that Taut is not El. Indeed, l3.,,(Taut; K2) is complete for L.. sentences (by 11.5.1 of [1]). If Taut were El, then by 3.3, 93.,(Qj) would also be complete for Le o sentences. By 3.4, 93, would be also. This contradicts 2.1, so Taut cannot be F1.

The addition to a proof system 3,43, (Z*; Q) of a El collection s of valid L.. schemes yields no new theorems since, by 3.3, each instance of an element b of s is already provable in 9 ,3Ow (provided one uses a reasonable definition of "instance" so that every Loo instance 0 of such an infinitary finite-quantifier formula b is provable in 93,, when ,b is provable in A3 ). Therefore, by 2.1, there is no S1 collection Z of valid L w . schemes such that 93, (Taut u F) is complete for Le w sentences. To some extent, this justifies the use in [1] of a El collection QI of L., rules of inference (as opposed to a El collection I of valid L.. schemes) to get a complete proof system for L.. sentences.

Theorem 36 of [3] can also be used to show, e.g., that the inequality of [1, 7.1.1(iv)] (for singular cardinal y) is strict whenever 26 < 2y for all P < y. (The converse is also true.) More generally, assume infinite cardinals y, 8 < 'y. If D is a set of pairs (a, of) such that U(0.D)eDP < y6, then, in the Boolean algebraic termino- logy of [5], there is (for every cardinal K > y6) a K-complete Boolean algebra which is not (8, y)-distributive but which is (a, fl)-distributive for each (a, P) E D.

?4. S13< completeness of a sentence with no model. From Lemmas 3.1, 3.2, and from Malitz's proof of the completeness of his sentence a,, we now prove

4.1. THEOREM. Our b is 93, complete for L, sentences. PROOF. Given ordinal a > w, define Ac to be that L.. sentence defined by the

result of replacing X by a in the definition of sb. Let La denote the corresponding language (thus, Q0 is a relation symbol of Da iff e a). Then Ox is i, and LK is L. We show that, if # is some 0a, and if 0 is an Lax,, sentence, then either b -? 0 or

0 m,0 is provable in 93,oco. We can assume that the relation symbols have been chosen so that the relation

"for some a > co, ,b is S., and 0 is an Lao<, sentence" is El in argument (0, 0); let this relation correspond to Qf of Lemma 3.2. By Lemma 3.1, the relation "either

- 0 or I -* --0 is provable in T.,4," is I 1 in argument (,, 0); let it correspond to .Y of Lemma 3.2. Let W correspond to " all conjunctions of 0, b are over sets :?P." x, y correspond to arguments (/, 0), P, respectively. Applying Lemma 3.2, we need only show that either #a -* 0 or ,- 0 is provable in S when a is countable and 0 is an L,,1, sentence. Consider such an a and 0. With changes in notation, it was shown in Malitz's [4] that 0ba is semantically complete forL,,,. (Our a is Malitz's Al; our Qu is Malitz's {a I aEd} where d = # E a. Our {Qo I P E a} corresponds to {B1 I 0 e i e w} in [4, Lemma 3].) Thus, either bAd -+ 0 or , -m -,0 is valid. By




Karp's completeness theorem for L one of these sentences is provable in 93.., 4.2. Remarks. A similar proof shows that A(Q n L@,0) is 3nd complete for

Lx.. There is also a (longer) more elementary proof of 4.1. 4.3. Remarks. 0 is not 93,.(Taut) complete for L.. sentences. For let 0 be

V{xnInew-} V A V (xm=xn). fliteucco meco neu

Then 01 F 0, so 0 - 9 is unprovable in 93w .(Taut). But q -. 0 is also unprovable in 93X, .(Taut); this can be shown by the methods of ?2. (The main modifications of ?2 are as follows. Fix some a* e A. Let D = {do I n E w} for the constant functions dn: (n + 1) -> {0}. Letf* be the constant mapf*: W -c {O}. Letp* be {(a*,f*)} Replace Sv by {s I s: v -+ X, and Rg(s) - D is finite}. Replace C by {p I p e C and p < p*}. In condition 2.11, add the clause "g r D is identity and g(a*) = a-." Replace "0 IF" by "p* IF.")

However, it is possible to show that j A 0 is 93 . .(Taut) complete for L. l, sentences and that there is no inconsistent Lww sentence which is 93w, complete for Lww0 sentences.

REFERENCES

[1] C. R. KARP, Languages with expressions of infinite length, North-Holland, Amsterdam, 1964.

(21 - , Nonaxiomatizability results for infinitary systems, this JOURNAL, vol. 32 (1967), pp. 367-384.

(3] A. LEVy, A hierarchy of formulas in set theory, Memoirs of the American Mathematical Society, no. 57 (1965).

[4] J. MALrrZ, The Hanf number for complete L,,10, sentences, The syntax and semantics of infinitary languages, Springer-Verlag, Berlin-Heidelberg-New York, 1968, pp. 166-181.

[5] R. SiKORSK[, Boolean algebras, 2nd ed., Academic Press, New York, 1964.

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Incompleteness of a Formal System for Infinitary Finite-Quantifier Formulas