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# craig - Details of Hunter's"Informal Proof of Craig's...

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Details of Hunter’s “Informal” Proof of Craig’s Interpolation Theorem for P Branden Fitelson 02/06/05 Hunter’s proof of Craig’s Interpolation theorem for P is a bit opaque. Here’s a more detailed version of his proof, which I sketched in class on Friday. Since this is our first (non-trivial) metatheorem, it’s worth doing a handout that proves it in some detail. We’ll see similar kinds of proofs often in the course. Theorem . Let A and B be formulas of P , such that (1) they share at least one propositional symbol in common, 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 ). [Intuitively, if P A B (and A and B have some symbols in common!) it is always possible to reason from A to B via a formula C that has no propositional symbols not shared by A and B . This is sometimes called “linear reasoning” from A to B , since it takes no “detours” through “irrelevant” or “tangent” unshared propositional symbols.] Proof. Case 1 : There are zero propositional symbols occuring in A that do not also occur in B . That is, the set of propositional symbols in A is a subset of those in B . If we let S A be the set of propositional symbols occurring in a formula A , then we can express this case as the case in which S A S B . In this case, just 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 proposi- tional symbols that occur in both A and B . But, this follows from the fact that C A , and the assumption of this Case, which is that S A S B . Hence, S A S C S A S B , which completes Case 1 .

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craig - Details of Hunter's"Informal Proof of Craig's...

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