Logic and CS / Gallier. Logic for Computer Science_ Foundations of Automatic Theorem Proving.pdf / book03all

is valid.

(a) If some of the predicate symbols Pi occurs in A(P1, ..., Pk, Q), by Craig’s theorem (theorem 6.5.1) applied to (1), there is a formula C containing

only predicate symbols, constant symbols, and free variables occurring in both

A(P1, ..., Pk, Q) Q(x1, ..., xn) and A(P1, ..., Pk, Q ) Q (x1, ..., xn), and such that the following are valid:

Since A(P1, ..., Pk, Q) Q(x1, ..., xn) does not contain Q and A(P1, ..., Pk, Q )

Q (x1, ..., xn) does not contain Q, the formula C only contains predicate symbols among P1, ..., Pk and free variables among x1, ..., xn.

By substituting Q for Q in (3), we also have the valid formula

The validity of (5) and (6) implies that

A(P1, ..., Pk, Q) (C ≡ Q(x1, ..., xn))

is valid, which in turns implies that C defines Q explicitly from P1, ..., Pk.

(b) If none of P1, ..., Pk occurs in A(P1, ..., Pk, Q), then by Craig’s theorem (theorem 6.5.1), part (ii), either

Using propositional logic, either

A(P1, ..., Pk, Q) ¬Q(x1, ..., xn)

is valid, or

A(P1, ..., Pk, Q) Q(x1, ..., xn) is valid, which implies part (b) of the theorem.

6.6.4 Beth’s Definability Theorem, With Equality

We now consider Beth’s definability theorem for a first-order language with equality. This time, we can define either a predicate symbol, or a function symbol, or a constant.

Theorem 6.6.2 (Beth’s definability theorem, with equality) Let L be a first-order language with equality.

(a)Let A(P1, ..., Pk, Q) be a closed formula possibly containing equal-

ity and containing predicate. symbols among the distinct predicate symbols P1, ..., Pk, Q (di erent from =), where Q has rank n > 0. Assume that Q is defined implicitly from P1, ..., Pk by the sentence A(P1, ..., Pk, Q). Then there is a formula D(P1, ..., Pk) defining Q explicitly from P1, ..., Pk.

(b)Let A(P1, ..., Pk, f ) be a closed formula possibly containing equal-

ity, and containing predicate symbols among the distinct predicate symbols

.

P1, ..., Pk (di erent from =), and containing the function or constant sym-

bol f of rank n ≥ 0. Assume that f is defined implicitly from P1, ..., Pk by the sentence A(P1, ..., Pk, f ), which means that the following formula is valid, where f is a new copy of f :

Then there is a formula D(P1, ..., Pk) whose set of free variables is among x1, ..., xn and not containing f (or f ) defining f explicitly, in the sense that the following formula is valid:

.

A(P1, ..., Pk, f ) x1... xn y((f (x1, ..., xn) = y) ≡ D(P1, ..., Pk)).

Proof : The proof of (a) is similar to that of theorem 6.6.1(a), but using theorem 6.5.2, which yields D(P1, ..., Pk) in all cases.

The last application of Craig’s interpolation theorem presented in the next section is Robinson’s joint consistency theorem.

PROBLEMS

6.6.1.Show that in definition 6.6.2, the definability condition can be relaxed to

A(P1, ..., Pk, Q) A(P1, ..., Pk, Q )

x1... xn(Q(x1, ..., xn) Q (x1, ..., xn)).

6.6.2.Give the details of the proof that explicit definability implies implicit definability.

6.7 Robinson’s Joint Consistency Theorem

Let L be a first-order language with or without equality, and let L1 and L2 be two expansions of L such that L = L1 ∩ L2. Also, let S1 be a set of L1-sentences, S2 a set of L2-sentences, and let S = S1 ∩ S2. If S1 and S2 are both consistent, their union S1 S2 is not necessarily consistent. Indeed, if S is incomplete, that is, there is some L-sentence C such that neither S → C nor S → ¬C is provable, S1 could contain C and S2 could contain ¬C, and S1 S2 would be inconsistent. A concrete illustration of this phenomenon can be given using G¨odel’s incompleteness theorem for Peano’s arithmetic, which states that there is a sentence C of the language of arithmetic such that neither AP → C nor AP → ¬C is provable, where AP consists of the axioms of Peano’s arithmetic (see example 5.6.3). (For a treatment of G¨odel’s incompleteness theorems, see Enderton, 1972; Kleene, 1952; Shoenfield, 1967; or Monk, 1976.) Since Peano’s arithmetic is incomplete, then {AP , C} and {AP , ¬C} are both consistent, but their union is inconsistent.

A. Robinson’s theorem shows that inconsistency does not arise if S is

complete. Actually, one has to be a little careful about the presence in the

.

language of function symbols or of the equality predicate =.

Theorem 6.7.1 (Robinson’s joint consistency theorem) Let L be a firstorder language either without function symbols and without equality, or with equality (and possibly function symbols). Let L1 and L2 be two expansions of L such that L = L1 ∩ L2. Let S1 be a set of L1-sentences, S2 a set of L2- sentences, and let S = S1 ∩S2. If S1 and S2 are consistent and S is complete, that is, for every closed L-formula C, either S → C, or S → ¬C, then the union S1 S2 of S1 and S2 is consistent.

Proof : Assume that S1 S2 is inconsistent. Then |= S1 S2 →. By the completeness theorem (theorem 5.6.1), there is a finite subsequent of S1 S2 → that is provable. Let A1, ..., Am, B1, ..., Bn → be such a sequent, where A1, ..., Am S1, and B1, ..., Bn S2. It is immediate that

(A1 ... Am) ¬(B1 ... Bn).

Let L be a first-order language either without function symbols and without equality, or with equality (and possibly function symbols). Let L1 and L2 be two expansions of L such that L = L1 ∩ L2. Let S1 be a set of L1-sentences, and let S2 a set of L2-sentences. Assume that S1 and S2 are consistent. Then the union S1 S2 of S1 and S2 is consistent i for every closed L-formula C, either S1 → C is not provable, or S2 → ¬C is not provable.

6.7.2.Prove that the version of Robinson’s joint consistency theorem given in problem 6.7.1 implies Craig’s interpolation theorem.

Hint: Let A B be a provable formula. Let S1 = {C | A → C}, and S2 = {C | ¬B → C}. Then S1 S2 is inconsistent.

Notes and Suggestions for Further Reading

Gentzen’s cut elimination theorem is one of the jewels of proof theory. Originally, Gentzen’s motivation was to provide constructive consistency proofs, and the cut elimination theorem is one of the main tools.

Gentzen’s original proof can be found in Szabo, 1969, and other proofs are in Kleene, 1952, and Takeuti, 1975. A very elegant proof can also be found in Smullyan, 1968. The proof given in Section 6.4 is inspired by Schwichtenberg and Tait (Barwise, 1977, Tait, 1968).

Craig himself used Gentzen systems for proving his interpolation theorem. We have followed a method due to Maehara sketched in Takeuti, 1975, similar to the method used in Kleene, 1967. There are model-theoretic proofs of Craig’s theorem, Beth’s definability theorem, and Robinson’s joint consistency theorem. The reader is referred to Chang and Keisler, 1973, or Shoenfield, 1967.

The reader interested in proof theory should also read the article by Schwichtenberg in Barwise, 1977. For an interesting analysis of analytic versus nonanalytic proofs, the reader is referred to the article by Frank Pfenning, in Shostak, 1984a.

rem in the literature, thus causing a confusion. The Skolem-Herbrand-G¨odel theorem is similar in form to Herbrand’s theorem but deals with unsatisfiability (or validity), a semantic concept.

The reason for the confusion is probably that, by G¨odel’s completeness theorem, validity can be equated to provability. Hence, Herbrand’s original theorem can also be stated for unsatisfiability (or validity).

However, Herbrand’s original theorem is a deeper result, whose proof is significantly harder than the Skolem-Herbrand-G¨odel theorem, and Herbrand’s theorem also yields more information than the latter. More on this subject will be said at the end of Sections 7.5 and 7.6. For an illuminating discussion, the reader should consult Goldfarb’s introduction to Herbrand, 1971.

In this chapter, we shall give in Section 7.5 a version of Herbrand’s original theorem for prenex formulae, and in Section 7.6 a version of the Skolem-Herbrand-G¨odel theorem for formulae in NNF (actually, half of this theorem is more like Herbrand’s original theorem).

The Skolem-Herbrand-G¨odel theorem can be viewed as the theoretical basis of most theorem proving methods, in particular the resolution method, one of the best known techniques in automatic theorem proving. In fact, the completeness of the resolution method will be shown in Chapter 8 by combining the Skolem-Herbrand-G¨odel theorem and theorem 4.3.2.

A constructive version of Herbrand’s theorem can be obtained relatively easily from Gentzen’s sharpened Hauptsatz, which is obtained by analyzing carefully cut-free proofs of formulae of a special type.

Gentzen’s sharpened Hauptsatz shows that provided the formulae in the bottom sequent have a certain form, a proof can be reorganized to yield another proof in normal form such that all the quantifier inferences appear below all the propositional inferences. The main obstacle to the permutation of inferences is the possibility of having a quantifier rule applied to the side formula arising from a propositional rule as illustrated below:

P (a) xQ(f (x)) → Q(f (g(a)))

The obvious solution that consists in permuting the : lef t rule and the : lef t rule does not work since a quantifier rule is not allowed to apply to a subformula like xQ(f (x)) in P (a) xQ(f (x)).

There are at least two ways of resolving this di culty:

(1)Impose restrictions on formulae so that a quantifier inference cannot be applied to the side formula of a propositional inference.

(2)Allow more general quantifier rules applying to subformulae.

The standard approach has been to enforce (1) by requiring the formulae to be prenex formulae. A Prenex formula is a formula consisting of a (possibly empty) string of quantified variables followed by a quantifier-free formula, and it is easily seen that (1) holds.

The second approach is perhaps not as well known, but is possible. Smullyan has given quantifier rules (page 122 in Chapter 14 of Smullyan, 1968) that allow the permutation process to be performed for unrestricted formulae, and a general version of the extended Hauptsatz is also given (theorem 2, page 123). These rules are binary branching and do not seem very convenient in practice. We will not pursue this method here and refer the interested reader to Smullyan, 1968.

We have also discovered that Andrews’s version of the Skolem-Herbrand- G¨odel theorem (Andrews, 1981) suggests quantifier rules such that (2) holds for formulae in negation normal form. Such rules are quite simple, and since there is a refutation method based on Andrews’s version of Skolem-Herbrand- G¨odel’s theorem (the method of matings), we shall give a proof of the extended Hauptsatz for such a system. Since every formula is equivalent to a prenex formula and to a formula in negation normal form, there is no loss of generality in restricting our attention to such normal forms. In this Chapter, it is assumed that no variable occurs both free and bound in any sequent (or formula).

7.2 Prenex Normal Form

First, we define the concept of a prenex formula.

Definition 7.2.1 (Prenex form) A formula is a prenex formula (or in prenex form) i either it contains no quantifiers, or it is of the form Q1x1...QnxnB, where B is a formula with no quantifiers (quantifier-free), x1, ..., xn are (not necessarily distinct) variables, and Qi {, }, for i = 1, ..., n.

In order to show that every formula is equivalent to a prenex formula, the following lemma will be used.

Lemma 7.2.1 The following formulae are valid:

where x does not occur free in B.

x(A B) ≡ (A xB)

x(A B) ≡ (A xB)

where x does not occur free in A.

where x does not occur free in B.

where y is free for x in A, and y does not occur free in A unless y = x

(y / F V (A) − {x}).

Proof : Some of these equivalences have already been shown. We prove two new cases, leaving the others as an exercise. A convenient method is to construct a proof tree (in G).

We give proofs for ( xA B) ≡ x(A B) and ( xA B) ≡ x(A B), where x is not free in B. As usual, to prove X ≡ Y , we prove X Y and

YX.

(i)Proof of ( xA B) x(A B):

A[y/x], B → (A[y/x] B)

xA, B → (A B)[y/x]

for a new variable y

xA, B → x(A B)

( xA B) → x(A B)

→ ( xA B) x(A B)

We have used the fact that since x is not free in B, B[y/x] = B.

(ii) Proof of x(A B) ( xA B)

→ x(A B) ( xA B)

Again, y is a new variable and we used the fact that B[y/x] = B.

(iii) Proof of ( xA B) x(A B):

A[y/x] → A[y/x], B

( xA B) → A[y/x], B

( xA B) → A[y/x] B

( xA B) → x(A B)

→ ( xA B) x(A B)

As in (ii) y is a new variable and we used the fact that B[y/x] = B.

(iv) Proof of x(A B) ( xA B):

A[y/x] B → A[y/x], B

x(A B) → A[y/x], B

x(A B) → xA, B

x(A B) → ( xA B)

→ x(A B) ( xA B)

As in (iii) y is a new variable and we used the fact that B[y/x] = B.

Theorem 7.2.1 For every formula A, a prenex formula B can be constructed such that A ≡ B is valid (A and B are equivalent).

Proof : To simplify the proof, we will assume that Q1x1...QnxnB denotes the quantifier-free formula B when n = 0. The proof proceeds using the induction principle applied to A. The base case is trivial since an atomic formula is quantifier free. The induction step requires six cases. We cover three of these cases, leaving the others as exercises.