craig - Details of Hunters Informal Proof of Craigs...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Details of Hunters Informal Proof of Craigs Interpolation Theorem for P Branden Fitelson 02/06/05 Hunters proof of Craigs Interpolation theorem for P is a bit opaque. Heres a more detailed version of his proof, which I sketched in class on Friday. Since this is our first (non-trivial) metatheorem, its worth doing a handout that proves it in some detail. Well 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 p A q be the set of propositional symbols occurring in a formula A , then we can express this case as the case in which S p A q S p B q . 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 p A q S p B q . Hence, S p A q S p C q S p A qX S p B q , which completes Case 1 ....
View Full Document

This note was uploaded on 08/01/2008 for the course PHIL 140A taught by Professor Fitelson during the Spring '07 term at University of California, Berkeley.

Page1 / 2

craig - Details of Hunters Informal Proof of Craigs...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online