Metatheory - University of Cambridge

2. Substitution 16
Step 3: Repeat step 2, until there are no atomic sentences that appear in the
input sentence but not in B. At that point, stop. The last sentence you
wrote down is an interpolant linking A to B.
It is worth seeing this in action. Suppose we start with the entailment:
((A ∧ (¬B ∨ C)) ∧ (¬C ∧ D)) ⊨ ((B ∨ D) ∧ (¬B ∨ E))
Running the algorithm, our initial input sentence is:
((A ∧ (¬B ∨ C)) ∧ (¬C ∧ D))
We would first eliminate A, obtaining (using ‘⊤’ to abbreviate some tautology
containing ‘B’, and ‘⊥’ to abbreviate some contradiction containing ‘B’):
(((⊤ ∧ (¬B ∨ C)) ∧ (¬C ∧ D)) ∨ ((⊥ ∧ (¬B ∨ C)) ∧ (¬C ∧ D)))
This would become our new input sentence; from which we would eliminate C:
((((⊤ ∧ (¬B ∨ ⊤)) ∧ (¬⊤ ∧ D)) ∨ ((⊥ ∧ (¬B ∨ ⊤)) ∧ (¬⊤ ∧ D))) ∨
(((⊤ ∧ (¬B ∨ ⊥)) ∧ (¬⊥ ∧ D)) ∨ ((⊥ ∧ (¬B ∨ ⊥)) ∧ (¬⊥ ∧ D))))
And this is our interpolant! But, as the example illustrates, the output of
this algorithm can be quite a mouthful. To simplify it, we can apply some
‘clean-up’ rules. Where ‘⊤’ abbreviates any tautology, and ‘⊥’ abbreviates any
contradiction, note, for example, that:
⊤ ∨ A
⊥ ∧ A
¬⊤
¬⊥
Using these equivalences and Lemma 2.2, we can simplify the subsentences of
our interpolant. 1 First, applying the rules for negation:
⊨
⊨
⊨
⊨
⊨ ⊤
⊨ ⊥
⊨ ⊥
⊨ ⊤
((((⊤ ∧ (¬B ∨ ⊤)) ∧ (⊥ ∧ D)) ∨ ((⊥ ∧ (¬B ∨ ⊤)) ∧ (⊥ ∧ D))) ∨
(((⊤ ∧ (¬B ∨ ⊥)) ∧ (⊤ ∧ D)) ∨ ((⊥ ∧ (¬B ∨ ⊥)) ∧ (⊤ ∧ D))))
Next, applying the rules for conjunction and disjunction:
((((⊤ ∧ ⊤) ∧ ⊥) ∨ (⊥ ∧ ⊥)) ∨ (((⊤ ∧ ¬B) ∧ D) ∨ (⊥ ∧ D)))
And, then, applying them again and again, we get, successively:
Which is much simpler!
((⊥ ∨ ⊥) ∨ ((¬B ∧ D) ∨ ⊥))
(⊥ ∨ (¬B ∧ D))
(¬B ∧ D)
1 Recall that a sentence A is a subsentence of B iff A is a part of B. We here read ‘part’
in a generous sense, so that every sentence is a part of itself.