Metatheory - University of Cambridge

2. Substitution 15
I make the following claim:
Claim 1: (A[ ⊤ /D] ∨ A[ ⊥ /D]) ⊨ B
To prove Claim 1, let v be any valuation that makes (A[ ⊤ /D] ∨ A[ ⊥ /D]) true.
If v makes A[ ⊤ /D] true, let w differ from v just by making D true; otherwise,
let w differ from v just by making D false. Either way, w makes A true, so
w makes B true too (since A ⊨ B). Finally, since D does not occur in B, v
makes B true too. This establishes the Claim.
Next, because there were n + 1 atomic sentences in A but not B, and D
was one of these atomic sentences, there must be exactly n atomic sentences in
A[ ⊤ /D] ∨ A[ ⊥ /D] but not B. Hence, by Claim 1 and the induction hypothesis,
we can find an interpolant, I , linking (A[ ⊤ /D] ∨ A[ ⊥ /D]) to B, i.e.:
1. (A[ ⊤ /D] ∨ A[ ⊥ /D]) ⊨ I
2. I ⊨ B
3. every atomic sentence in I occurs in both (A[ ⊤ /D] ∨ A[ ⊥ /D]) and in B
I shall now show that I is an interpolant linking A itself to B. Point (2), above,
tells us that I ⊨ B. Moreover, given point (3), it is obvious that every atomic
sentence in I occurs in both A and in B. So it remains to show that A ⊨ I .
Given that (A[ ⊤ /D] ∨ A[ ⊥ /D]) ⊨ I , it suffices, by the transitivity of entailment,
to prove the following claim:
Claim 2. A ⊨ (A[ ⊤ /D] ∨ A[ ⊥ /D])
To prove Claim 2, let v be any valuation that makes A true. If v makes D
true, then since ⊤ is a tautology, v makes A[ ⊤ /D] true. If v makes D false,
then v makes A[ ⊥ /D] true. Either way, any valuation that makes A true makes
(A[ ⊤ /D] ∨ A[ ⊥ /D]) true. This establishes Claim 2.
Hence I is an interpolant linking A to B. This completes the proof of the
induction case. The entire Theorem now follows by induction.
■
This is easily the longest proof that we have seen so far. So it is worth pausing,
to make sure both that you understand every step in the proof, and also how
the proof is put together.
In fact, this proof merits particular scrutiny since, implicitly, the proof
does more than just prove the Theorem. The Interpolation Theorem states
that, when A ⊨ B, an interpolant exists. If this were all we knew, then we
might have no idea about how to find an interpolant. But, viewed correctly,
the proof yields an algorithm for finding an interpolant. An algorithm is a
step-by-step process, followed ‘thoughtlessly’ (i.e. without any need for insight,
just by implementing very simple little rules), that is guaranteed to terminate
in finitely many steps. I shall describe the algorithm for finding an interpolant;
and having done so, it should be clear how it relates to the proof.
Given that A ⊨ B, linking A to B as follows:
Step 1: Designate A as the input sentence
Step 2: Let D be the first atomic sentence (alphabetically) that appears in
the input sentence, but that does not appear in B. Rewrite the input
sentence twice, once uniformly substituting a tautology in place of D
and once uniformly substituting a contradiction in place of D. Let the
disjunction of these two rewritten sentences be your new input sentence.