Logic: Lecture 4, 13 August 2015 -------------------------------- Proof of MCS properties 1. a in X iff not a not in X {a,not a} is inconsistent |- (a -> a) -> not (not (a -> a)) so (a and not a) is inconsistent If neither a nor not a are in X, find finite sets B and C so that B and a is inconsistent and C and not a is inconsistent. Rewrite to get |- a -> not B and |- (not a) -> not C |- (a -> b) -> (c -> d) -> (a or c) -> (b or d) Use a = a, c = not a, b = not B, d = not C to derive |- (not B or not C) Strong completeness Logical consequence: X |= a For all v, v |= X implies v |= a Want to show: X |-a iff X |= a Strong Soundness If X |- a then X |= a follows by induction, as usual. Strong completeness If X |= a then X |- a Indirect proof, via Compactness Compactness: X |= A iff there exists a finite subset Y of X such that Y |= A To prove Compactness: Finite Satisfiability: X is satisfiable iff every finite subset Y of X is satisfiable Assuming Finite Satisfiability, Proof of Compactness: (<==) is trivial (==>) For any Z,b, Z |= b iff Z + {~b} is unsatisfiable. Suppose X |= a. Then X + {~a} is unsatisfiable. Then there is a finite subset Y' of X + {~a} that is unsatisfiable. Set Y = Y' - {~a}. Clearly Y |= a, where Y is a finite subset of X. Assuming Compactness, Proof of Strong Completeness: If X |= a, by compactness there is a finite subset Y of X such that Y |= a. Let Y = {b1,b2,.., bm}. Verify that |= (b1 -> (b2 -> (... -> (bm -> a)))) By (normal) completeness, |- (b1 -> (b2 -> (... -> (bm -> a)))) Applying Deduction Theorem m times, b1,b2,...,bm |- A. Can also use strong completness to prove compactness: X |= a implies X |- a. Given a (finite) proof of A, collect formulas used in the proof as Y. Then Y |- A, whence Y |= a. ======================================================================