Metatheory - University of Cambridge

2. Substitution 14
2.2 Interpolation
The next result uses the idea of substitution to prove something rather deep.
When working with an argument, one can get the idea that there is an ‘active
core’ to the argument, which should be distinguished from some of the irrelevant
surrounding matter. The next result gives some formal content to this claim
I shall start with a little result which might be put as telling us the following:
if I am to draw a conclusion from some claim, there had better be something
in the original claim that my conclusion builds upon. More precisely:
Lemma 2.3. Suppose A ⊨ B, that A is not a contradiction, and that B is
not a tautology. Then at least one atomic sentence occurs in both A and B.
Proof. Supposing A is not a contradiction, there is a valuation, v, that makes
A true. Supposing B is not a tautology, there is a valuation, w, that makes
B false. Supposing also A and B have no atomic sentences in common, then
there is a valuation, x, which assigns the same values as v to all atomic sentence
letters in A, and the same values as w to all atomic sentence letters in B. By
hypothesis, x makes A true and B false. So A ⊭ B.
■
Our main result is going to employ a (fairly lengthy) induction to extend
Lemma 2.3 rather radically. First, we need a definition. Suppose that A ⊨ B.
Then an interpolant linking A to B is any sentence I meeting all three of
these conditions:
(int1) A ⊨ I
(int2) I ⊨ B
(int2) every atomic sentence in I occurs both in A and in B
Our result guarantees the existence of interpolants.
Theorem 2.4. Interpolation Theorem. Suppose A ⊨ B, that A is not
a contradiction, and that B is not a tautology. Then there is an interpolant
linking A to B.
Proof. Our proof consists in an (ordinary) induction on the number of atomic
sentences that occur in A but not B.
In the base case, there are no atomic sentences in A that are not also in
B. Hence we can let A itself be our interpolant, since A ⊨ A vacuously, and
A ⊨ B by hypothesis.
Now fix n, and suppose (for induction) that we can find an interpolant
whenever there are n atomic sentences that occur in the ‘premise’ but not the
‘conclusion’ of the entailment. Additionally, suppose that A ⊨ B, that A is
not a contradiction, that B is not a tautology, and that there are n + 1 atomic
sentences that occur in A but not in B. Lemma 2.3 tells us that there is at
least one atomic sentence, C , common to both A and B. Now, where:
• D is any sentence that occurs in A but not in B
• ⊤ is a tautology which contains no atomic sentence apart from C , e.g.
(C → C )
• ⊥ is a contradiction which contains no atomic sentence apart from C ,
e.g. (C ∧ ¬C )