Sets of Theorems with Short Proofs Author(s): Daniel Richardson Source: The Journal of Symbolic Logic, Vol. 39, No. 2 (Jun., 1974), pp. 235-242 Published by: Association for Symbolic Logic Stable URL: http://www.jstor.org/stable/2272636 . Accessed: 12/06/2014 21:41 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp . JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. . Association for Symbolic Logic is collaborating with JSTOR to digitize, preserve and extend access to The Journal of Symbolic Logic. http://www.jstor.org This content downloaded from 62.122.72.154 on Thu, 12 Jun 2014 21:41:24 PM All use subject to JSTOR Terms and Conditions
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Sets of Theorems with Short ProofsAuthor(s): Daniel RichardsonSource: The Journal of Symbolic Logic, Vol. 39, No. 2 (Jun., 1974), pp. 235-242Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/2272636 .
Accessed: 12/06/2014 21:41
Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp
.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].
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TiE JOURNAL OF SYmBOLIc LOGIC Volume 39, Number 2, June 1974
SETS OF THEOREMS WITH SHORT PROOFS
DANIEL RICHARDSON
R. Parikh has shown that in the predicate calculus without function symbols it is possible to decide whether or not a given formula A is provable in a Hilbert-style proof of k lines. He has also shown that for any formula A(xl, - - *, x") of arith- metic in which addition and multiplication are represented by three-place predi- cates, {(a,, - . ., an): A(al, - - *, a.) is provable from the axioms of arithmetic in k lines} is definable in (N, ', +, 0). See [2].
In ?I of this paper it is shown that if S is a definable set of n-tuples in (N. +, 0), then there is a formula A(xl, ..* *, xn) and a number k so that (a1,***, an) E S if and only if A(al, - - *, an) can be proved in a proof of complexity k from the axioms of arithmetic. The result does not depend on which formalization of arithmetic is used or which (reasonable) measure of proof complexity. In particular, this gives a converse to Parikh's result.
In ?II, a measure of proof complexity is given which is based on counting the quantifier steps in semantic tableaux. The idea behind this measure is that A is k provable if in the attempt to construct a counterexample one meets insurmountable difficulties in the definition of the appropriate Skolem functions over sets of cardinality < k. A method is given for deciding whether or not a sentence A in the full predicate calculus is k provable. The question for the full predicate calculus and Hilbert-style systems of proof remains open.
In ?III, it is.shown that there is no formula A(x, y) and number k so that, for all natural numbers a and b, a : b if and only if A(a, b) is provable in the predicate calculus in a proof of complexity k. In a similar way, addition cannot be repre- sented by proofs of bounded complexity in the predicate calculus. This implies that the result of ?I depends essentially on the presence of infinitely many induction axioms. So, for example, every true statement in the diagram of addition can be proved in ring theory but the proofs have unbounded complexity.
?1. Let Fbe one of the usual formulations of arithmetic. A relation R(xl, * , xn) is said to be uniformly representable in F if for some k and formula A(xl,* , xn), R(al, - - *, an) holds if and only if F F A(a1, - - *, an) and F F A(al, - - *, an) implies F FI- A(al, - - *, an). Here a is the numeral representing a and F Fk A means that A is provable in F in k steps.
Equality is uniformly representable. The representing formula is x = y. Vx(x = x) is a theorem. So if a = b, we can prove a = b in a uniformly bounded number of steps.
Received September 30, 1971; revised July 30, 1972. ? 1974, Association for Symbolic Logic
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Addition is uniformly representable in F. If t is a term, let t(a) be the term which is obtained by applying the successor operation a times to t. Suppose a + b = c.
a + 0 = a = [Vx(a + x = x(a) : a + x' = x(a + 1)) z Vx(a + x = x(a))]
is an induction axiom. We can prove a + 0 = a in a uniform way. We can also prove a + x' = (a + x)' uniformly. Thus we get Vx(a + x= x(a)-a + x' = x(a + 1)). This implies Vx(a + x = x(a)). By substitution, a + b = c.
The argument above can be carried out even if F represents addition and multi- plication by relation symbols.
The family of uniformly representable relations in F is closed under conjunction, disjunction and existential quantification. Thus, x # y and x + y # z are uni- formly representable. It follows that if R(xl, *-* *, xJ) is any relation over the natural numbers with R(xl, *, xJ) =Yi>3 , YmP(X11, x* , yl, *, yi) where P is a conjunction of disjunctions of formulae of the form a + b = c or a + b # c, then R(xl, - - *, xn) is uniformly representable in F. It is known that any definable relation in (N, ', +, 0) can be so expressed [3]. Thus
THEOREM 1. Let R(xl, - - *, xn) be any definable relation in (N' +, 0). There is a formula A(x1, * - *, xn) and number k so that R(al, *, an) holds if and only if F - A(a1, *-* *, an) and F F A(a,, -* - , an) implies F Fk A(ai, * * *, an).
?11. No matter which formal measure of proof complexity is adopted, there are intuitively hard proofs which when formalized have low proof complexity and also there are intuitively easy proofs which when formalized have high proof complexity. This is a consequence of the fact that, although the results of mathematical thinking are reflected in formal systems, the processes of discovery are not.
The most that can be expected is that proof complexity should correspond to some semantic idea such as the following: A is provable in a proof of complexity k if in the attempt to construct a counterexample to A we meet insurmountable diffi- culties at the kth stage. The best systematic method of constructing counterexamples is the method of semantic tableaux invented by E. Beth. A version of this method is described below.
Let L be some first order language with logical constants A, V, V, , 3. B
and with finitely many function symbols and relation symbols. The equality axioms for L are the sentences which say that equality is reflexive, symmetric and transitive andf(xl, - - *, xn) = f(y1, - - , YJ) whenever xl = Yi and. . . and xn = yn for each function symbol f; and R(x1, - - *, x%) D R(y1,.-., yn) whenever xl = y, and ... and Xn = Yn for each relation symbol R. Let E be the conjunction of the equality axioms for L.
Let (pi) be an infinite sequence of constants which do not occur in L. Let L(p,) be the smallest language which is an extension of L and has (pi) among its constants. Let T(pi) be the constant terms of L(p,), i.e., terms without variables. Let A be a sentence of L. A Beth tree based on A will be a tree (growing downwards) labeled with sentences of L(pi) which is generated by the process described below.
Let (ti) be an enumeration of T(pi). A Beth tree based on A will be constructed in stages. "Tk(A)" will denote the tree constructed at stage k. Tk(A) depends upon k, A, (tJ) and a finite number of free choices made between stage 0 and stage k. For
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each n, the tree constructed at stage n + 1, Tn+1(A), will be a finite extension of Tn(A). The extensions will be made by adding new labeled nodes at the bottom. " To branch b add (x, y) " will mean that we put two new nodes immediately below the bottom most node of b and label one new node x and the other new node y. We will say x E b if the sentence x is a label of a node in branch b.
We define To(A), the tree constructed at stage 0, to consist of a single node labeled (A) A (E) where E is the conjunction of the equality axioms. Suppose we have carried out the process to stage n and constructed Tn(A), which is finite. Tn+1(A) is constructed by applying the following operations to the branches b such that there is no x with x E b and - (x) E b.
(1) If (x) A (y) eb, extend b to b' so that xeb' and yeb'. (1)* If - ((x) v (y)) E b, extend b to b' so that - (x) E b' and -(y) E b'. (2) If -(a (x)) E b, extend b to b' so that x E b'. (3) If (x) v (y) E b but x 0 b and y 0 b, extend b by adding (x, y). (3)* If - ((X) A (y))eb but - (x) b and - (y) b, extend b by adding
( (x), (Y))- (4) If (x) v (y) e b but - (x) 0 b and y 0 b, extend b by adding (a (x), y). (5) If ((x) - (y)) e b, extend to b' so that x e b' and (y) E b'. (6) If VvB(v) E b but B(t,) 0 b for some < n + 1, extend b to b' so that b(t1) E b'. (6)* If (3vB(v)) e b, extend b to b' so that Vv (B(v)) E b'. (7) If 3vB(v) E b but for no term t is B(t) in b, extend b to b' so that B(p) E b'
where pi is a constant which has not previously appeared in the tree. (7)* If (VvB(v)) E b, extend b to b' so that 3v (B(v)) E b'. The operations can be applied in any order whatever. When they have all been
applied as many times as possible, i.e., when no further extension is possible, we have T + 1(A).
The constants which are introduced by operation (7) are called the e constants of the tree.
The process given above always terminates in a finite labeled tree Tn+1(A). This is because each kind of operation can only be applied once to a label of any node, and the new nodes which are produced have labels which are essentially simpler.
A branch b of a Beth tree is closed if there is an x so that x E b and (x) E b. A Beth tree is closed if every branch is closed. If there is no closed Beth tree based on a sentence A, there is a model for A. In fact such a model is specified by any infinite branch of Lim,-.. Tft(A) for any enumeration. If there is a closed Beth tree based on -(A) we will write FA. If some Tk(-(A)) is closed, we will write Fk A. A closed tree Tk(- (A)) is considered a proof of A of complexity k. If there is no closed Tk(A) we will say that A is k consistent. A is consistent in the ordinary sense if and only if A is k consistent for all k. k consistency defined in this way is supposed to correspond to a semantic idea: The Skolem functions which would specify a model with domain T(p1) can be defined so as to be successful for any subset of T(pi) of cardinality k. Another way of putting this would be: Pseudo-Skolem functions can be defined which cannot be distinguished from real Skolem functions except possibly by looking at more than k values. The existence of this semantic idea distinguishes this notion of k provability from that used by Parikh.
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We now want to show that k provability is decidable. The proof has the following form. Given A and k we can find a finite set F of terms in T(pi) so that if there is a closed Tk(- (A)), there is a closed Tk( (A)) with an enumeration whose first k terms are in F. Since there are only finitely many essentially different such Tk(- (A)) we can just try them all to test whether or not Fk A.
By the terms of a formula A we mean all terms which occur in A and all subterms of these terms.
Sub(s, t, A) will be the result of uniform substitution of s for t in A. Thus, for atomic formulae R(rl, - - *, rn),
where Sub(s, t, r) is defined as follows: Sub(s, t, r) = s if r = t. Sub(s, tf(r1, . - -, rn)) = f(Sub(s, t, r1), ... , Sub(s, t, rn)) if f(r1, ... , rn) 9 t. Sub(s, t, r) = r if r # t and the length of r is less than or equal to the length of t. Given a formula A and constant terms tl,, - - -, tnwe define ClA({tl, - - *, tn}) to be
the set of constant terms which can be obtained by substituting tl,, ,tn for variables in terms of A.
LEMMA 1. Suppose Tk( (A)) is a closed Beth tree with enumeration (t1, * , tk)
and e constants q1, ... , q,. Let t, be one of t1..., 4, tk so that
ti 0Cl(A) A (E)({tlg ***, tkgfl qJ .. 9q} { til)-
Let s be any constant term not containing ql, - - *, q,. Let Sub(s, t,, Tk(- (A)) be the tree obtained by replacing each label X by Sub(s, tt, X). Then Sub(s, t,, Tk(- (A)) is a closed Beth tree proving Sub(s, ti, A).
PROOF. It is claimed that the sequence of operations which generated Tk( - (A)) with enumeration (tl, - - -, tk) also generates Sub(s, ti, Tk(-(A)) with enumeration (Sub(s, ti, t1), ... , Sub(s, t,, tk)). We will only show that a step corresponding to operation (6) remains valid.
Suppose b is a branch of a tree which appeared at some stage in the construction of Tk(- (A)) and VvB(v) e b and b was extended to b' with B(t,) e b'. In the con- struction of Sub(s, ti, Tk(- (A)) the branch Sub(s, ti, b) appears and is extended to Sub(s, ti, b') with Sub(s, t,, B(t,)) e Sub(s, ti, b'). We have to show that this exten- sion can be obtained by operation (6), with Sub(s, ti, t,) as the substituted term, i.e., that Sub(Sub(s, t, t,), v, Sub(s, ti, B(v))) = Sub(s, ti, Sub(t,, v, B(v))), where Sub(s, v, A) means the result of substituting s for v in A wherever v is free in A. This will be true if, for every term r of B(v),
In the following, the left-hand side of this equation will be called L(r) and the right- hand side will be called R(r).
Suppose we have proved that for all r in B(v) with less than n function symbols, L(r) = R(r). Assume that r has exactly n function symbols. There are three cases.
Case 1. r does not have v in it. Then Sub(s, ti, r) does not have v in it. So L(r) = Sub(s, ti, r). Also Sub(t,, v, r) = r. So R(r) = Sub(s, ti, r). Thus L(r) = R(r).
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Case 2. r = v. Then Sub(s, tt, r) = v. So L(r) = Sub(s, tt, tj). Also, R(r) = Sub(s, tt, tf).
Case 3. r # v and r does have v in it. Then there is a term 'r in (A) A (E) so that r can be obtained by substituting terms in {tj, *, tk, qj, *, qu} for variables in 'r.
Assume r = f(rl,* , r,). r # tt since tj does not have v in it. Let
L(rm) = Sub(Sub(s, ti, t,), v, Sub(s, tt, r,)),
R(r,.) = Sub(s, tt, Sub(tj, v,, rm)).
By the induction hypothesis, L(rm) = R(rm) for -m = 1,.. ,p. But L(r) = f(L(rl),, -*-, L(r,)) and R(r) = f(R(rl), * , R(r,)) unless Sub(tj, v, r) = tt. Thus L(r) = R(r) unless Sub(tj, v, r) = t,.
Suppose Sub(tj, v, r) = tt. t1 # t, since r # v. r is obtained by substituting terms in {tl, * * *, tk, q1, ... I, qj} for variables in r. t, is not a subterm of r so tt is not needed to get r from r. It follows that
tte Cl(A) A ()({tl, *, tk, q1, ., q,} - {tt})
which contradicts an assumption of the lemma. The proof of Lemma 1 is now complete.
Suppose we adopt a G6del numbering of T(pf). Let 0 be the constant with smallest G6del number.
Define FO(q1, - - *, q,) to be the union of {0, q1, *j*, qg} and the set of constant terms of (A) A (E). Then, for each k, define Fk+, (qj,* , q,) to be the set of con- stant terms which can be obtained by substituting terms of Fk(ql1, , q- ) for variables in terms of (A) A (E).
LEMMA 2. Suppose one of (t1,* , tk) is not in Fk(q1, ... *, qn). Then there is a t, in (tl .,* tj) so that tt s FO(ql, -.., qn) and tt * Cl(A) (A({tlj * tk, q1, *, q* } -
{tt}). PROOF. The proof is by induction on k. Suppose k = 1. Then t1 0 F1(q*,..*, qn).
So t1j Fo(q1,*... qn) and t1 0 Cl(A) A (E)(l,--* q). Now assume that Lemma 2 is true for k < N. Assume k = N and we are given tj, - - *, to with tN having maximal length. If {t1, - - * tN 1} $ FN - (q, - - - , qn), we apply the induction hypothesis. On the other hand, suppose {t1,* *-, tN._} ' FN-l(ql, qn). Then tN
LEMMA 3. If there is a closed Beth tree Tk(- (A)) based on -(A) with e constants q,5* ...* * qws then there is one whose enumeration has its first k terms chosen from Fk(qlg
.. * , qn).
PROOF. Suppose this were false. Then among the counterexamples, pick a closed Beth tree Tk(' (A)) with enumeration (tj, - *-, tk) so that the sum of the G6del numbers of t1, - - *, tk is minimal. Since this is a counterexample, some tj 0 Fk(ql, - *, qn). By Lemma 2, there is a term tt so that
tt F0(q1, ... *, qn) and tt 0 Cl(A) A (5)({tl I * ** tk, q1 ., q*n} - {tt})-
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By Lemma 1, Sub(O, ti, Tk(- (A))) is a closed Beth tree proving Sub(O, t, A). However t, 0 FO(ql1 * * *, qn) and so Sub(O, ti, A) = A.
THEOREM 2. There is an effective procedure which, given a sentence A and a number k, decides whether or not Fk A.
PROOF. Suppose k and A are given. We have to decide whether or not one of the trees Tk(- (A)) is closed. We can write out finitely many tree schemes (Sc(t, ... , tk, ql, ** *, qn)) so that any Tk( (A)) can be obtained by substituting constants from T(pi) for t1, * , tk and constants from (pi) for q1, - * *, qn. We may as well assume that (ql, * ., q) = (Pl, * *, p.). Then, using Lemma 3, we can assume that t1, - ., tk are all chosen from Fk(pl,- p,.p) which is finite. So we need to try only finitely many possibilities.
In terms of the semantic idea, Fk(pl, - - *, pn) is a finite subset of T(pi) so that if the Skolem functions can be defined successfully for any subset S of Fk(pl, - - *, p,) of cardinality k, they can be extended in such a way as to be successful for any subset S of T(pi) of cardinality k.
?m. Let A be a set of sentences in a language L. We will say A Fk A if there exists a conjunction of sentences from A, D, so that Fk (D) v (A). Supposing L has a constant 0 and a successor function symbol ' so that we can represent the natural numbers, we will say that a set S of n-tuples of natural numbers is uni- formly representable from A if there is a formula A(xl, - - *, xn) and a number k so that (al,* *, an) E S if and only if A F A(al, * , an) and A F A(al, * , an) implies A Fk A(al,* , an). For example, the diophantine predicates are uniformly repre- sentable from the diagram of arithmetic.
We define the height of a sentence to be the largest number a such that for some variable v, v(a) is a term of the sentence. The height of a set of sentences is either infinity or, if it exists, the least upper bound of the heights of its elements.
THEOREM 3. Suppose A u {A} has height < H and A has no numeral, a, with a 2 H. Let (N, M) be any interval of natural numbers with M -N > (k + 2)H. Let x be a variable not in A. Suppose A Fk A. Then A Fk+ I1 Vx Sub(x, a, A) for some as [N, M].
PROOF. Let Tk(A, - (A)) be a closed Beth tree which proves A from A. Suppose the enumeration for this tree begins t1, -., t-k. Let the numerals, in order of size, among t1, ..., t. which represent numbers in (N, M) be a,, * - *, a1. Set a,,, = M and ao = Nand let a be defined as the least a. + 1 such that a, + 1 - at > H. This must exist because M -N > (k + 2)H. Now let pi be a constant which does not occur in Tk(A, -(A)). The claim is that if
~((D) = (A)) I D
Tk(A, -(A)) is I (A) I Tree
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As in the proof of Lemma 1 of ?II, we need only show that for any term r in the original proof and for any tj in the enumeration,
Sub(pi, a, Sub(tj, v, r)) = Sub(Sub(pi, a, t1), v, Sub(pi, a, r)).
As before, this is true unless there is a term r in the original proof with Sub(tj, v, r) = a and v occurring in r. But this is impossible since the height of A U {A} is less than H and since there are no numerals a, in the enumeration with a - at < H.
COROLLARY 1. x # y is not uniformly representable from any finite set of sentences.
PROOF. Suppose A were finite and a # b if and only if Ai Fk A(a, b). Pick H big enough to satisfy the hypothesis of Theorem 3. A F, A((k + 3)H, 2(k + 3)H). By Theorem 3, A Fk+2 A((k + 3)H, p(a)) where p is a constant not in L. The height of A u {A(x, p(a))} is still less than H. So by Theorem 3 again, A F A(q(b), p(a)) where q is a constant not in L. Then by substituting a for q and b for p, we get A F A(a + b, a + b). Thus inequality is not uniformly representable from A.
COROLLARY 2. x + y = z is not uniformly representable from any finite set of sentences.
Corollary 2 can be proved by the same method used for Corollary 1. COROLLARY 3. Let A be any finite set of sentences so that, for all a,
A IVx(x = 0 V x = O' V ...v x = a v 3y(x =y(a + 1)). Then, for any sentence VxA(x), A F VxA(x) if and only if there is a k so that, for all a, A Fk A(a).
PROOF. Evidently, if VxA(x) is provable from A, then there is a k so that, for all a, A Fk A(a). On the other hand, assume that, for all a, A Fk A(a). By Theorem 3, A Fk+l VxA(x(a + 1)) for some a. Thus A F VxA(x).
Parikh has proved a result similar to Corollary 3 for the full system of arithmetic in which addition and multiplication are represented by relation symbols. His proof does not apply to small fragments of arithmetic since it requires the system to prove theorems about its own proof predicate.
The following conjecture, which says that there are economic translations of Hilbert-style proofs into semantic tableaux, would imply that Corollaries 1, 2, and 3 remain true when k provability is interpreted as provability in a Hilbert-style system of k lines.
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CONJECTURE. Given a formula A(xl, - *, x,,) and number k, we can effectively find a number j so that for any natural numbers a.,, *, an, if A(al, - - *, a.) can be proved in the predicate calculus in a Hilbert-style proof of k lines, then Fj A(al,, * * an).
A proof of this conjecture might also supply a decision procedure for k prova- bility in Hilbert-style systems.
REFERENCES
[1] E. W. BETH, The foundations of mathematics, North-Holland, Amsterdam, 1965. [2] R. J. PARIKH, Some results on the length of proofs, Transactions of the American Mathe-
matical Society, vol. 177 (1973), pp. 29-36. [3] M. PRESBURGER, OIber die Vollstandigkeit eines gewissen Systems der Arithmetik ganzer
zahlen, Comptes Rendus du I Congres des Mathematiciens des Pays Slaves, Warsaw, 1929-1930, pp. 92-101, 395.
UNIVERSITY OF BRISTOL
BRISTOL, ENGLAND
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