Logic: Lecture 3, 11 August 2015 -------------------------------- Derivability Assume X, prove a Formulas in X are concrete, not axiom schemes X |- a Redefine deriviations: A sequence where each formula is an instance of an axiom. obtained using MP or a member of X Deduction theorem: X,a |- b iff X |- a -> b Prove by inducion Examples 1. |- a -> (b -> c) -> (b -> (a -> c)) a -> (b -> c) |- a -> (b -> c) (Membership) a -> (b -> c),a |- b -> c (Deduction Theorem) a -> (b -> c),a,b |- c (Deduction Theorem) a -> (b -> c),b |- a -> c (Deduction Theorem) a -> (b -> c) |- b -> (a -> c) (Deduction Theorem) |- a -> (b -> c) -> (b -> (a -> c)) (Deduction Theorem) 2. |- ~~b -> b 1. |- (~b -> ~~b) -> ((~b -> ~b) -> b) (A3) 2. |- (~b -> ~~b) -> ((~b -> ~b) -> b) -> (~b -> ~b) -> ((~b -> ~~b) -> b) (Example 1 above) 3. |- (~b -> ~b) -> ((~b -> ~~b) -> b) (MP, 1,2) 4. |- (~b -> ~b) (Instance of a -> a) 5. |- (~b -> ~~b) -> b (MP, 3,4) 6. |- ~~b -> (~b -> ~~b) (A1) 7. ~~b |- ~b -> ~~b (Deduction Theorem) 8. ~~b |- (~b -> ~~b) -> b (From 5) 9. ~~b |- b (MP, 7,8) 10. |- ~~b -> b (Deduction Theorem) Completeness Prove contrapositive: not derivable implies not valid Consistency: not (|- ~a) Claim: a or b is consistent if either a is consistent or b is consistent a and b is consistent if both a and b are consistent (Converse fails!) Henkin-style proof of completeness: Show that every consistent formula is satisfiable Assuming this, completeness follows: Suppose b is not derivable. |- ~~b -> b, so ~~b is also not derivable (else, by MP, |- b) So ~b is consistent So ~b is satisfiable So b is not valid. Maximal consistent sets Finite set is consistent if conjunction is consistent Infinite set is consistent if all finite subsets are consistent MCS is a consistent set that cannot be extended Lindenbaum Lemma Every consistent set can be extended to an MCS Properties of MCS a in X iff ~a not in X a or b in X iff a in X or b in X MCS X defines a valuation v_X = all p in X Extends to all formulas: v_X |= A iff A in X Now, Henkin: a consistent => extend {a} to MCS X => v_X |= a => a is satisfiable ======================================================================