Logic: Lecture 3, 11 August 2015
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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)
   |- 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
   
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