Logic: Lecture 4, 13 August 2015
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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.
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