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

S0

Γ→ ∆, A, C

Γ→ ∆, A

replacing T1 by (T1, A) if necessary. Consider the first case, the other being

similar. By the induction hypothesis, we can find a proof T1 for Γ, Λ → ∆, Θ, B, such that depth(T1) < depth(T1) + depth(T2) and c(T1) ≤ |A|. By the inversion lemma, we can find a proof T2 of B, Λ → Θ such that depth(T2) ≤

depth(T2) and c(T2) ≤ |A|. The following proof T

is such that depth(T ) ≤ depth(T1) + depth(T2) and has cut-rank c(T ) ≤ |A|, since |B| < |A|.

Case 2.3: A is of the form (B C). This case is symmetric to case 2.2.

Case 2.4: A is of the form xB. As in case 2.2, we can assume that A belongs to the premise of the last inference in T1, so that T1 is of the form

S0

Γ→ ∆, A, B[t/x]

Γ→ ∆, A

By the induction hypothesis, we can find a proof tree T1 for Γ, Λ → ∆, Θ, B[t/x], such that depth(T1) < depth(T1) + depth(T2) and c(T1) ≤ |A|. By the inversion lemma, for any variable y not bound in T2 and distinct from

all eigenvariables in T2, there is a proof T2 for B[y/x], Λ → Θ, such that depth(T2) ≤ depth(T2) and c(T2) ≤ |A|. By the substitution lemma, we can

construct a proof T2 for B[t/x], Λ → Θ, also such that depth(T2 ) ≤ depth(T2) and c(T2 ) ≤ |A|. Since |B[t/x]| < |A|, the proof obtained from T1 and T2 by

applying a cut to B[t/x] has cut-rank ≤ |A|.

Case 2.5: A is of the form xB. This case is symmetric to case 2.4. This concludes all the cases.

Finally, we can prove the cut elimination theorem.

The function exp(m, n, p) is defined recursively as follows:

exp(m, 0, p) = p; exp(m, n + 1, p) = mexp(m,n,p).

This function grows extremely fast in the argument n. Indeed, exp(m, 1, p) = mp, exp(m, 2, p) = mmp , and in general, exp(m, n, p) is an iterated stack of exponentials of height n, topped with a p:

exp(m, n, p) = mmm···mp n

The Tait-Schwichtenberg’s version of the cut elimination theorem for G1nnf follows.

Theorem 6.4.1 (Cut elimination theorem for G1nnf ) Let T be a G1nnf - proof with cut-rank c(T ) of a sequent Γ → ∆. A cut-free proof T for Γ → ∆ such that depth(T ) ≤ exp(2, c(T ), depth(T )) can be constructed.

Proof : We prove the following claim by induction on the depth of proof

trees.

Claim: Let T be a G1nnf -proof with cut-rank c(T ) for a sequent Γ → ∆. If c(T ) > 0 then we can construct a proof T for Γ → ∆, such that

c(T ) < c(T ) and depth(T ) ≤ 2depth(T ).

Proof of Claim: If either the last inference of T is not a cut, or it is a cut and c(T ) > |A| + 1, where A is the cut formula of the last inference, we can apply the induction hypothesis to the immediate subtrees T1 or T2 (or T1) of T . We are left with the case in which the last inference is a cut and c(T ) = |A| + 1. The proof is of the form

By the induction hypothesis, we can construct a proof T1 for Γ → ∆, A and a proof T2 for A, Γ → ∆, such that c(Ti ) ≤ |A| and depth(Ti ) ≤ 2depth(Ti), for i = 1, 2. Applying the reduction lemma, we obtain a proof T such that, c(T ) ≤ |A| and depth(T ) ≤ depth(T1) + depth(T2). But

depth(T1) + depth(T2) ≤ 2depth(T1) + 2depth(T2) ≤

2max(depth(T1),depth(T2))+1 = 2depth(T ).

Hence, the claim holds for T .

The proof of the theorem follows by induction on c(T ), and by the definition of exp(2, m, n).

It is remarkable that theorem 6.4.1 provides an upper bound on the depth of cut-free proofs obtained by converting a proof with cuts. Note that the “blow up” in the size of the proof can be very large.

Cut-free proofs are “direct” (or analytic), in the sense that all inferences are purely mechanical, and thus require no ingenuity. Proofs with cuts are “indirect” (or nonanalytic), in the sense that the cut formula in a cut rule may not be a subformula of any of the formulae in the conclusion of the inference. Theorem 6.4.1 suggests that if some ingenuity is exercised in constructing proofs with cuts, the size of a proof can be reduced significantly. It gives a measure of the complexity of proofs. Without being very rigorous, we can say that theorem 6.4.1 suggests that there are theorems that have no easy proofs, in the sense that if the steps are straightforward, the proof is very long, or else if the proof is short, the cuts are very ingenious. Such an example is given in Statman, 1979.

We now add equality axioms to the system G1nnf .

6.4.5 The System G1nnf=

The axioms and inference rules of the system G1nnf= are defined as follows.

Definition 6.4.6 The system G1nnf= is obtained by adding to G1nnf the following sequents as axioms. All sequents of the form

.

(i) Γ → ∆, t = t;

Lemma 6.4.8 (Completeness of G1nnf= ) For the system G1nnf= , every valid sequent is provable.

Proof : The lemma can be proved using the completeness of LKe (theorem 6.3.1). First, we can show that in an LKe-proof, all weakenings can be moved above all other inferences. Then, we can show how such a proof can be simulated by a G1nnf= -proof. The details are rather straighforward and are left as an exercise.

282 6/Gentzen’s Cut Elimination Theorem And Applications

6.4.6 The Cut Elimination Theorem for G1nnf=

Gentzen’s cut elimination theorem also holds for G1nnf= if an inessential cut

the completeness of equational logic (see the problems). The proof requires a modification of lemma 6.4.6. The cut-rank of a proof is now defined by considering essential cuts only.

Lemma 6.4.9 (Reduction lemma for G1nnf= ) Let T1 be a G1nnf= -proof of Γ → ∆, A, and T2 a G1nnf= -proof of A, Λ → Θ, and assume that c(T1), c(T2) ≤

Proof : We proceed by induction on depth(T1) + depth(T2). The only new case is the case in which A is a literal of the form P t1...tn or ¬P t1...tn, and one of the two axioms Γ → ∆, A and A, Λ → Θ is an equality axiom. Assume that A = P t1...tn, the other case being similar. If Γ → ∆ or Λ → Θ is an axiom, then Γ, Λ → ∆, Θ is an axiom. Otherwise, we have three cases.

Case 1: Γ → ∆, A is an equality axiom, but A, Λ → Θ is not. Then, either ¬A is in Λ or A is in Θ, and Γ → ∆, A is obtained from some equality

Λ, the sequent Γ, Λ → ∆, Θ is an equality axiom also obtained from some sequent of the form

Case 2: Γ → ∆, A is not an equality axiom, but A, Λ → Θ is. This is similar to case 1.

Case 3: Both Γ → ∆, A and A, Λ → Θ are equality axioms. In this case, Γ → ∆, A is obtained from some equality axiom of the form

by application of ¬ : rules, and A, Λ → Θ is obtained from some equality axiom of the form

by applications of ¬ : rules.

Then, Γ, Λ → ∆, Θ is obtained by applying negation rules to the sequent

→ ∆ , Θ , P r1...rn.

A proof with only inessential cuts can be given using axioms derived by applying ¬ : rules to the provable (with no essential cuts) sequents

and n inessential cuts (the i-th cut with cut formula si = ri or ¬si = ri). The

depth of this proof is n, which is bounded by m. The rest of the proof is left as an exercise.

We obtain the following cut elimination theorem.

Theorem 6.4.2 (Cut elimination theorem for G1nnf= ) Let T be a G1nnf= - proof with cut-rank c(T ) for a sequent Γ → ∆, and let m be defined as in lemma 6.4.9. A G1nnf= -proof T for Γ → ∆ without essential cuts such that depth(T ) ≤ exp(m + 2, c(T ), depth(T )) can be constructed.

Proof : The following claim can be shown by induction on the depth of proof trees:

Claim: Let T be a G1nnf= -proof with cut-rank c(T ) for a sequent Γ → ∆. If c(T ) > 0 then we can construct a G1nnf= -proof T for Γ → ∆, such that

c(T ) < c(T ) and depth(T ) ≤ (m + 2)depth(T ).

The details are left as an exercise.

Note: The cut elimination theorem with inessential cuts (with atomic

.

cut formulae of the form s = t) also holds for LKe. The interested reader is refered to Takeuti, 1975.

As an application of theorem 6.3.1, we shall prove Craig’s interpolation theorem in the next section. The significance of a proof of Craig’s theorem using theorem 6.3.1 is that an interpolant can actually be constructed. In turn, Craig’s interpolation theorem implies two other important results: Beth’s definability theorem, and Robinson’s joint consistency theorem.

(a) Prove that for every deduction tree without essential cuts for a sequent Γ0 → ∆0, the formulae in the sets of formulae Γ {¬B | B

∆} for all sequents Γ → ∆ occurring in that tree, belong to S or are

.

equations of the form s = t, where S = Γ0 {¬B | B ∆0}.

(b) Deduce from (a) that not all G1nnf= -formulae are provable.

6.4.14. Let L be a first-order language with equality and with function symbols and constant symbols, but no predicate symbols. Such a language will be called equational .

Let e1, ..., em → e be a sequent where each ei is a closed formula of the

.

form x1... xn(s = t), called a universal equation, where {x1, ..., xn}

.

is the set of variables free in s = t, and e is an atomic formula of

.

the form s = t, called an equation. Using theorem 6.4.2, prove that if e1, ..., en → e is G1nnf= -provable, then it is provable using only axioms

(including the equality axioms), the cut rule applied to equations (of

.

the form s = t), weakenings, and the : lef t rule.

6.4.15. Let L be an equational language as defined in problem 6.4.14. A substitution function (for short, a substitution) is any function σ :

V → T ERML assigning terms to the variables in V. By theorem 2.4.1, there is a unique homomorphism σ : T ERML → T ERML extending σ and defined recursively as follows:

For every variable x V, σ(x) = s(x).

For every constant c, σ(c) = c.

For every term f t1...tn T ERML,

σ(f t1...tn) = f σ(t1)...σ(tn).

By abuse of language and notation, the function σ will also be called a substitution, and will often be denoted by σ.

The subset X of V consisting of the variables such that s(x) = x is called the support of the substitution. In what follows, we will be dealing with substitutions of finite support. If a substitution σ has finite support {y1, ..., yn} and σ(yi) = si, for i = 1, .., n, for any term t, the substitution instance σ(t) is also denoted as t[s1/y1, ..., sn/yn].

Let E =< e1, ..., en > be a sequence of equations, and E the sequence of their universal closures. (Recall that for a formula A with set of free variables F V (A) = {x1, ..., xn}, x1... xnA is the universal closure of A.)

We define the relation −→E on the set T ERML of terms as follows. For any two terms t1, t2,

t1 −→E t2