Logic: Lecture 4, 16 August 2012
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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.


Proof of finite satisfiability:

Koenig's lemma:

   A finitely branching tree is infinite iff it has an infinite
   path

Given this, to prove:

   X is satisfiable iff every finite subset Y of X is satisfiable

   (<==) is trivial

   (==>) 
   Contrapositive:  If X is not satisfiable, some finite subset Y
   is not satisfiable.

   Enumerate P = {p1, p2, ...}.  Let Pi = {p1,p2,...,pi}

   Construct a tree of valuations where nodes are level i are
   valuations over Pi and v{j+1} is a child of vj iff they agree
   on Pj.   Every infinite path in the tree corresponds to a
   valuation and vice versa.

   A valuation v is bad if v(B) = ff for some B in X.  Prune the
   tree of valuations by retaining only the first bad node along
   each path.

   Claim: This tree is finite

   If not, there is an infinite path pi with no bad nodes.  The
   corresponding valuation v must be such that v(B) = tt for
   every B in X.  This means X is satisfiable!

   Given that the pruned tree is finite, it has finitely many
   leaves.  Each leaf is "witnessed" by some unsatisfiable B in
   X.  Collect the finite set of witnesses to define Y.

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