Logic: Lecture 3, 14 August 2012
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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)

2. |- ~~B -> B

   1. |- (~B -> ~B) -> ((~B -> ~~B) -> B)   (A3)
   2. |- (~B -> ~B)                         ("Lemma", already proved)
   3. |- (~B -> ~~B) -> B                   (MP, 1,2)
   4. |- ~~B -> (~B -> ~~B)                 (A1)
   5. ~~B |- ~B -> ~~B                      (Deduction Theorem)
   6. ~~B |- (~B -> ~~B) -> B               (From 3)
   7. ~~B |- B                              (MP, 5,6)
   8. |- ~~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 A 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
   
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