craig_2 - A Proper Inductive Proof of the Interpolation...

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A Proper Inductive Proof of the Interpolation Theorem for P Branden Fitelson 02/14/07 Theorem . Let A and B be formulas of P , such that (1) they share at least one propositional symbol in com- mon, and (2) ± P A B . For any two such formulas of P , there exists a formula C (called the P -interpolant of the formulas A and B ) such that (3) ± P A C , (4) ± P C B , and (5) C contains only propositional symbols that occur in both A and B ( i.e. , only propositional symbols shared by A and B ). Setup for an inductive proof . Let S(A) be the set of propositional symbols occurring in A , S(B) be the set of propositional symbols occurring in B , and q be some propositional symbol that is shared by A and B ( i.e. , q S(A) S(B) ). We will focus on the set X = S(A) - S(B) of propositional symbols that occur in A but not in B . We will prove the interpolation theorem by strong mathematical induction on the cardinality of X . That is, we will prove that the following property of natural numbers holds for all n 0: S (n) : The interpolation theorem (above) holds when X = S(A) - S(B) = n . Proof. As always, a proper inductive proof involves a Basis Step and an Inductive step. Basis Step . Prove S ( 0 ) . That is, we must prove the interpolation theorem for the case in which there are zero propositional symbols occurring in A that do not also occur in B ( X = 0). In this case, the set of propositional symbols in A is a subset of those in B [ S(A) S(B) ]. Let C = A . Then, obviously, (3) ± P A C , since ± P A A . And, since the assumption of the theorem is that ± P A B , we also know that ± P C B . All we need to show is that (5) C contains only propositional symbols that occur in both A and B . But, this follows from the fact that
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craig_2 - A Proper Inductive Proof of the Interpolation...

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