finite models for inequations

Finite Models for InequationsAuthor(s): M. D. GladstoneSource: The Journal of Symbolic Logic, Vol. 31, No. 4 (Dec., 1966), pp. 581-592Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/2269694 .

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THE JOURNAL OF SYMBOLIC LOGIC

Volume 31, Number 4, Dec. 1966

FINITE MODELS FOR INEQUATIONS1

M. D. GLADSTONE

? 1. Introduction. By inequation is meant an expression of the form,

(a # b), where a, b are terms taken from any first-order formal language. An inequation is consistent iff it cannot be brought to the form,

(c X c),

by substitution. The main result (Theorem One) of this paper is that, for any finite set

of consistent inequations, there is a model with a finite domain, in which they are all valid (with # being interpreted as inequality).

This result, in a different form, is asserted by Herbrand on p. 34 of "Sur le problem fondamental de la logique mathematique". However, he offers no proof. In a footnote on p. 161 of Grundlagen der Mathematik, Band 2 (1939), Hilbert and Bernays remark that the result has a deceptive plau- sibility ("tauschenden Plausibilitat"), and quote K. Schutte as having shown that the proof is by no means obvious. I am not aware of any other relevant literature; so, to the best of my knowledge, the present proof is the first to be offered.

The above references treat the result in connection with a certain decision problem in the Predicate Calculus. We shall discuss this application of Theorem One in ? 7, after a proof occupying ?? 2-6.

In ? 8 we shall see that, in a special case, the model of Theorem One can be arranged to be of a particular simple form. The paper concludes (? 9) with a discussion of Theorem Two (the result of ? 8).

? 2. Notation. Abbreviations: "arg" for "argument", and "vbl" for "term variable" (this is the only type of formal variable allowed).

The statement that there is a substitution under which the term a generates the term b, is written

a -fb.

We extend this ->-relation to inequations, etc.

Received December 19, 1965. l This is part of the author's doctoral thesis, submitted at Bristol in 1964. I am

very much indebted to J. C. Shepherdson for supervising the work, and to the Depart- ment of Scientific and Industrial Research for a grant. An abstract of the paper was read at the 1964 Logic Colloquium (Bristol).

581



582 M. D. GLADSTONE

We distinguish between model and submodel. The former assigns values merely to the function signs in question, whereas the latter assigns values to the vbls as well. We denote the value assigned to a formal entity in a model or submodel, by printing that formal entity boldface; thus, "a" for the value of the term a, and "O" for the function assigned to the function sign ?b.

Gothic capitals, A, 3, ..., are reserved to denote strings of positive integers. These arise in the following notation.

Let a be any term, and i any positive integer. Then

a. i (or simply ai)

denotes that term-occurrence, if any, which constitutes the ith arg of the outermost function sign of a. If there is no such term-occurrence, ai is undefined. We extend this so that, for instance,

a.i. j

denotes that term-occurrence which constitutes the 1th arg of the outermost function sign of the term having occurrence ai. Similarly for aijk, etc.

Suppose a is a term and 9t is an integer-string such that at is defined (When 9t is void we take at to denote a itself). Let n length of At, by which we mean the number of integers in the string. Then the term-occurrence at is said to take place at depth n in a.

The rank of a term is the greatest depth of term-occurrence taking place in it.

We write term(a%)

for the term having occurrence A. To clarify the foregoing by an illustration, suppose

a = (x, 0,(x, 6)),

where b is a 2-arg function sign, 6 an individual constant and x a vbl. Then

term(a. 1) term(a.2. 1) = x.

Furthermore, a is of rank 2.

? 3. Reduction-of-variables sequences. In ?? 4-5 we shall set up two types of finite model, at least one of which is appropriate for each single, consistent inequation. In the present section, an algorithm (to be known as a "reduction-of-variables sequence", or "rov sequence" for short) is given for reducing any inequation to a form for which the appropriate model is obvious.

An inequation, a # b, is terminal iff it is of at least one of the following three types:



MODELS FOR INEQUATIONS 583

(I) For some integer-string X, both term(aX) and term(bX) are defined, each has at least one function sign (individual constants are regarded as function signs), and the outermost function signs of these two terms differ.

(II) For some e both term(aX) and term(bX) are defined, one of them being a vbl, say x, and the other a term properly containing x.

(III) a = b. By a reduction-of-variables (roy) sequence is meant a finite sequence of

inequations,

(al # bi), (a2 # b2),*As

subject to the conditions: (1) Exactly one member of the sequence is terminal, namely the last; (2) (at X be) generates its successor, if any, by a substitution which

consists in replacing some vbl x by a term c other than x, where, for some

S, one of term(aiY,), term(biX) is x, and the other is c. Note that the substitution described in condition (2) deprives (a? X be)

of one of its variables (c cannot contain x, for this would make (a? X be) terminal, type II). Therefore each inequation in the sequence has one distinct vbl less than its predecessor (hence "reduction-of-vbls").

LEMMA 1. Every inequation is first member of a rov sequence. PROOF. We shall first show that any non-terminal inequation, a X b,

has a successor as described in condition (2) (above). The proof is by induction upon k = min(rank of a, rank of b).

For k 0, at least one of a, b, say a, is a vbl; then the required successor is b X b. Now consider some k > 0 and assume the result for lesser ranks. We can write

a =4ai, .0..*, am), b + (bil ...,I bm),

for some positive integer m, function sign b, and terms ai, ..., bm. (a and b must have the same outermost function sign, otherwise type I applies). Since (a # b) is not terminal, none of (al X bl), ..., (am X bm) is of type I or II, and at least one of them, say (al X bl), is not of type III. That is, (al # bl) is not terminal. Hence, by the induction hypothesis, it generates a successor through a substitution as described in condition (2). It is easy to see that this same substitution generates a permissible successor of

(a # b), when applied to the latter. Such a train of successors cannot continue indefinitely, because of the

reduction in the number of distinct vbls. Eventually we must arrive at a terminal inequation, thus concluding a rov sequence headed by the given inequation.

The next lemma relates the consistency or otherwise of the first member of a rov sequence to the type of the last member.



584 M. D. GLADSTONE

LEMMA 2. Let (al X bl), ..., (am # bm) be a rov sequence; then (A) If (al X b1) is consistent, then (am # bm) is of type I or II; (B) If (al X b1) is inconsistent, then (am # bm) is of type III; furthermore, the class of inequations of the form (c # c) generated through substitution by (al # bi) coincides with the class of inequations of this form generated by (am # bm).

PROOF. Let Ai be the class of inequations of the form (c 0 c) generated through substitution by (at # be). Since (ai # bi) (aj+j X bi+i), we have

Ai contains Ai+j. (1)

We may suppose that (as 0 be) generates (ai+l g bi+1) by replacing the vbl x throughout by the term d, where

term(aiX) x,

term(bi3X) = d.

Let (ai # bi) generate (c X c) by the substitution S. Then S sends x and d into the same thing, namely term(c^). So, if before S we insert the substitution which consists in replacing x by d, this will have no effect upon the endproduct. Therefore, S sends (aj+j # bi+) into (c # c), and we have

Aj+j contains Ai. (2)

Combining results (1) and (2), Ai = Ai+1 and so

Al Am. (3)

Now, Am is obviously void for (am W bm) of type I or II. Results (A) and (B) follow at once.

? 4. Type II models. Throughout this section,

(a 1 b 1), . .., (an # b,,)

stands for a particular rov sequence with last member of type II. We shall show how to construct a certain kind (called type II) of finite model in which (al X bi) is valid.

A model is said to be of type II when it is determined in the way described below, by a set of function signs L, a specified member a of L, a positive integer m, and a non-void integer-string U.

The description of the model is made easier by the following notation. For any term a, and positive integer r,

air

is to denote the term obtained from a by replacing every term-occurrence at depth r by a certain individual constant 5. Now, let ft be a function sign other than ac, but having the same number of args as c (subj ect to




these conditions, it does not matter exactly how f is chosen). Then the domain D of the model has as its elements all terms of the form

a/m + 1,

where a is a vbl-free term built up from the signs of L augmented (if these are not already members of L) by fi, 6. D is obviously finite, for finite L. To complete our model, we must assign functions to the signs of L. Let 0 be an s-arg sign of L, and let A1, .., A, be elements of D. Then, apart from one exceptional case, we have

O(1 ... *, A) = )0(Al/m, ..., As/m).

The exceptional case occurs when the following two conditions are both satisfied: (i) 0 =c; (ii) term((oc(Al, ..., As)) .) is defined and has a as its outermost function

sign. We then put

O(/A1, .*A) s) =(/Al/m, ..., As/m).

This completes the description of the general type II model. We return to the given rov sequence to choose particular values for L,

o, mY, U. Since (an X b.) is of type II, there exists a vbl w, and integer- string A, such that one of term(a.%), term(b.9t), say the former, is w, and the other is a term properly containing w. We define U to be the integer- string satisfying

term(b.%3) = w.

For the remaining particular values, we take L = the set of all function signs appearing in the given rov sequence; a= the outermost function sign of term(b.9t); m = np + 1, where p = max(rank of a., rank of b.).

(The exact value of m does not matter, as long as m > nP.) In the course of the next three lemmas, we prove that (al X bl) is valid

in our model. In making statements (e.g. the next lemma) about the values assigned to terms, the phrase, "in any submodel", is to be understood.

LEMMA 3. (A) For s > t, a/s = b/s implies a/t = b/t; (B) Term((a/r) . X) = term(a. X)/(r - length of S), provided that both

sides of this equation are defined, and that r ? m + 1. PROOF OF (A). Obvious. PROOF OF (B). Consider the special case when

a = 0(b),



586 M. D. GLADSTONE

0 being a 1-arg function sign. Then

a =(b/m),

where b is 0 or fi, it does not matter which. We have

term((a/r). 1) = b/r - 1 = term(a. 1)/r - 1.

It is clear that, in a similar way, we can prove (B) for every other case when length of S e 1. The general result then follows easily by induction on the length of N.

LEMMA 4. For any i, r, where 1 C i < n and r < 2p, if

ai/r bi/r, then

a?ij/r - p bi+j/r - p.

PROOF. We may take it that (at E bi) generates (ai+l E bi+j) by the substitution which consists in replacing x by c, where term(aix) = x and

term(biX) = c. Assuming ai/r = bi/r, we have

term((ai/r) . N) = term((bi/r) . X). (1)

Let q = length of N. Then, by lemma 3(B),

term((aj/r) . N)= term(a1X)/r - q, (2) and

term((bi/r) . N) = term(bjX)/r - q. (3)

Combining results (1), (2) and (3),

term(ajX)/r - q term(biX)/r - q.

i.e. x/r- q c/r q. (4)

Since p = max(rank of a., rank of b.), clearly p > q. Hence, applying lemma 3(A) to result (4),

x/r-p c/r P. (5)

By consideration of lemma 3(B) and the model definition, we see that

ai+i can be obtained from ai by a succession of two-part operations like the following (i) For some s > 0, replace an occurrence of x/(m + 1 - s) by one of

c/(m + 1 -s); (ii) Replace certain occurrences of a. (or fi) having the above occurrence

of x/(m + 1 - s) within their scope, by f (or ac). By result (5), all changes introduced by step (i) are at depth > r - p + s > r - p. With regard to step (ii), action never, in fact, needs to be taken.




This is because, relative to a candidate a. (or fi), any change introduced by step (i) is at depth > r - p > p ? length of V. To sum up, in going from ai to ai+1 there are no changes at depth < r- p. That is,

a1/r - p = a?+1/r - p. (6) Similarly,

bi/r - P b1i+/r - P. (7)

Since the lefthand sides of (6) and (7) are clearly equal, we conclude that

ai+,/r - p = bi+l/r - p.

LEMMA 5. a, 0 b1. PROOF. Our model has been arranged so that the outermost function

signs of term(ant), term(b,^) are different. If the former is a, then the latter is fi; and if the former is not a., then the latter is. Hence, by lemma 3(B), an and bn have a function sign difference at depth = length of t ? p. Therefore

anlp + 1 # bn/P + 1.

It follows that al # bi. If the contrary were the case, then by (n - 1) applications of lemma 4 we would have the contradictory result, an/P + 2 = bn/p + 2.

Lemma 5 says in effect that (a, E b1) is valid in our model.

? 5. Type I models. Throughout this section,

(al X bi), . . , (an X bn)

stands for a rov sequence whose last member is of type I. Again we construct a finite model in which (al # b1) is valid, proceeding exactly as in the last section, except that there is no longer a function sign a needing special treatment.

The definition of type I model is as for type II with all mention of a. and V omitted. So the model is determined by just the two quantities L, M.

L, M, p are defined as before. LEMMA 6. a, 0 bi. PROOF. (an # bn) is of type I. Therefore, an, bn have a function sign

difference, obviously at depth < p, which is transmitted to their values on the model domain. Hence

an/p + 1 =# bn/P + 1.

This can be shown to imply al # bi by the same argument as used in lemma 5. This argument depends upon lemmas 3 and 4. These can be



588 M. D. GLADSTONE

proved for type I models in the same way as for type II, except that all references to oc are omitted.

Type I models are attractively simple, but are not always adequate. LEMMA 7. There exists a consistent inequation which is valid in no

type I model. PROOF. We offer as our counter-example the palpably consistent in-

equation, x O O(X),

where x is a vbl, and 0 a 1 -arg function sign. Let m be any positive integer, and L any set of function signs including 0. Consider the type I model determined by L, m in the usual way. In the submodel which puts

x = (0(. . * WMb) ... ))

(m + 1 appearances of 0 are to be understood) both sides of our inequation take the same value.

? 6. Combination of models. Putting together our results so far, we have

LEMMA 8. For any consistent inequation there is a finite model in which it is valid.

PROOF. Every inequation, (a g b), is first member of a rov sequence (lemma 1). If (a # b) is consistent, the last member is of type I or II (lemma 2). In either case there is a finite model in which (a # b) is valid (lemmas 5 and 6).

THEOREM 1. For any finite set of consistent inequations there is a finite model in which they are all valid.

PROOF. In view of lemma 8, it is sufficient to show that any two finite models may be combined in such a way as to preserve finiteness and validity (the result for n models follows from that for two, by simple induction).

Take any two finite models M1, M2. There is no harm in assuming that every function sign given a value in M1 is given one in M2 also, and vice- versa. The giving of values to hitherto unvalued function signs cannot destroy the validity of an inequation, as long as the domain of the model is not enlarged.

Then the required model M is simply the direct product of M1, M2. That is, its domain consists of all ordered pairs {A, K}, where A, K belong to the domains of M1, M2, respectively; and functions are combined in the usual way; for instance, a one-arg function sign 0 would be given the value,

0({A, K}) = {#1(A), 02(K)},

where O1, 02 are its values in M1, M2, respectively. Clearly M has the required properties.




? 7. Discussion of Theorem 1. The discussion covers two topics. 1. Mixed systems o/ equations and inequations. Our results fail for very simple systems of the above kind. For instance,

let S be the system consisting of the equation,

((x + Y) + Z) = (x + (y + z)), (1) and the inequation,

x 0 (x + Y), (2)

where + is a 2-arg function sign, and x, y, z are vbls. S is consistent (e.g. interpret + as addition and x, y, z as positive integers). Let D be the domain of any model in which both (1) and (2) are valid. In any submodel, let A. stand for the value of any term built up from + and exactly n occurrences of x (all such terms must have the same value, because of the associative property, (1)). Now, because of property (2), Al, A2, ..., must all be distinct, i.e. D is infinite.

2. A special case of the decision problem for the predicate calculus. Let C be a closed wff of the predicate calculus, consisting of a string of

quantifiers followed by a disjunction,

(Cl V ... V Cm),

where each of Cl, .. ., Cm is of one of the two forms,

P(ai, .., an),

-,P(al, .. .,an)

for some n-adic predicate letter P, and terms ai, ..., an. (= negation.) A decision procedure for C is given by Hilbert and Bernays on pp. 160-1 of Grundlagen der Mathematik, Band 2. Our Theorem 1 provides an alternative decision procedure which has hitherto been assumed, but not proved, to exist.

First we make some simplifying assumptions about C: (1) No universal quantifiers appear. Hilbert and Bernays show how to

eliminate these by a method based on Herbrand's Theorem, which involves bringing in new functions signs.

(2) Each predicate letter is monadic. Each n-adic predicate letter can be replaced by a new monadic letter, and new n-arg function sign.

(3) There is only one distinct predicate letter. Let the one predicate letter be P. Then C may be rewritten as

P(al) V ... P(ar) V ,P(bl) V . .. P(bs).

Our final assumption is (4) The variables appearing in al, ..., ar are distinct from those

appearing in bl, . . ., b,. This depends upon the fact that existential quanti-



590 M. D. GLADSTONE

fiers allow replacements like that of (Ex) (A (x) V B(x)) by

(Ex) (Ey) (A (x) V B(y)).

Hilbert and Bernays show that F C iff there exist integers i, j, and a term d, such that

both P(ai) -? P(d)

and -,P(b1) -+-,P(d);

i.e. iff (ai X bj) is inconsistent. This is easily determined by the reduction of-vbls procedure (lemma 2). But it is interesting that an alternative decision procedure is given by our Theorem 1, which tells us that, if not F C, then C is refutable in a finite model. It can be verified that this still holds even if the simplifying assumptions are lifted.

? 8. Complete sets of terms. First, two definitions. A set of terms S is complete iff every vbl-free term built up from the signs of S is generated through substitution by at least one member of S. Two sets A, B of terms are mutually consistent iff there exists no term which is generated both by a member of A and a member of B.

For the rest of this section, A, B stand for two particular, finite, mutually consistent sets of terms, whose union is complete. To avoid trivialities, we assume that the function signs of A, B include at least one individual constant, and at least one sign with ? 1 args.

If we make the further assumption (which does no harm) that A, B have no common vbl, then the mutual consistency condition becomes equivalent to saying that, for a E A and b E B, all the inequations (a # b) are consistent. What we are about to do is to show that for these inequations an unusually simple, finite model suffices.

Before commencing the proof, we make one more definition: let m=

greatest rank of term in A or B. LEMMA 9. Let c be any term in which,

(i) the only function signs are those of A, B, (ii) each vbl-occurrence is at depth m; then some member of A or B generates c through substitution.

PROOF. By induction upon k = no. of vbl-occurrences in c. For k= 0. the result is immediate from the completeness of A plus B. Now take k > 0. Choose some vbl-free term g of greater rank that c,

having no function signs other than those of A, B. Next choose one occurrence of some vbl, say x, in c, and replace it by g, thus obtaining a new term d.

By the induction hypothesis there exists a member a of either A or B, which generates d through a substitution that we shall call D. Because its rank is too large, g has no occurrence in c, and hence only one occurrence




in d. This means that part of the effect of D must be to replace some vbl y by a term which we shall write as +(g), where I There is exactly one occurrence of y in a,

and exactly one occurrence of g in +(g). (1)

Let C be a substitution identical with D, except that y is replaced by +(x) (this bears the obvious meaning) instead of +(g). Then C sends a into c. Result (1) tells us that there are no unwanted side-effects.

LEMMA 10. Let a, b be vbl-free terms generated by members of A, B, respectively. Then a, b have a function sign difference at depth ? m.

PROOF. Let Xi, X2, . . ., be distinct vbls. Work through a from left to right, replacing the first term-occurrence at depth m by xi, the second by X2, and so on. Call the resultant term a,. In a similar way obtain bj from b. Clearly I al-a, (1)

*b fb.

By lemma 9, there exists a term a2, belonging to A or B, such that a2 -? ai. From result (1), a2 -* a. Hence, by the mutual consistency property, a2 must belong to A. There must be a term b2, analogously related to b. I.e. I a2e A and a2 al, (2)

b2 EBand b2 bl.

Viewing result (2) in the light of the mutual consistency property, we conclude that al # bi. Because of their form, the only possible difference between al and bi is in function signs. Since they are both of rank _ m, any such difference must be at depth ? m. Result (1) tells us that this function sign difference is transmitted to a, b.

THEOREM 2. Let A, B be two finite, mutually consistent sets of terms, whose union is complete. Then there exists a finite model with the properties: (1) ae A and be B imply a k b; (2) If c, d are two vbl-free terms built up from the signs of A, B, then

c = d iff c, d have no function sign difference at depth ? greatest rank of term in A or B.

PROOF. Let A, B, m retain their definitions at the head of this section. Let L be the set of all function signs found in A, B. Then the required model is the type I model determined by L, m as described in ? 5.

Suppose a e A and b EB. In any submodel, let a* denote the vbl-free term generated by a through the substitution which replaces each vbl x by its value x. Similarly for b, b*. Clearly a = a*, b = b*. From lemma 10 and property (2) of the model (which is obvious), we have a* # b*. Hence a 7 b.



592 MODELS FOR INEQUATIONS

RIDER TO THEOREM 2. If the union of A, B is not complete, then property (2) no longer follows.

PROOF. As a counter-example, take A, B to consist, respectively, of the single terms ip(x, x), y(y, +(y)), where b is a one-arg sign, 'V is a 2-arg sign, and x, y are vbls. The argument is as in lemma 7.

? 9. Discussion of Theorem 2. One application of the theorem is briefly discussed.

Suppose we negate the wff C of ? 7, thus obtaining

-,P(al) & . .. -,P(a,) & P(bi) & ... P(bs),

which we proceed to add to the pure predicate calculus as an axiom. We might well wish to think of the resultant system as applying to precisely all those terms whose function signs are limited to a certain set L. Suppose the axiom C is so strong that it completely determines the truth or falsity of P(d) for all vbl-free terms d that concern us. Then, identifying the A, B of Theorem 2 with {ai, . . ., ar}, {bl, ..., bs}, respectively, we have a very simple model for C, which gives us a decision procedure for any closed wff of our system ("truth" is not the same as "derivability" here, unless one introduces some induction schemata appropriate to L).

One possible generalization is to n-truth-valued systems. This would involve proving Theorem 2 for n mutually consistent sets, which is easily done.

UNIVERSITY OF BRISTOL



finite models for inequations

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