Logic: Lecture 5, 18 August 2015
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Strong completeness

  If X |= A then X |- A

Compactness:

   X |= A iff there exists a finite subset Y of X such that Y |= A

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.

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.

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.

Exercise:

Consider an infinite graph G=(V,E) and a finite set of colours
C.  A colouring of G is a function c: V -> C such that for any
edge (x,y) in E, c(x) != c(y).  Show that G has a valid colouring
iff every finite subgraph of G has a valid colouring.

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First order logic

Reasoning about mathematical structures: 
  A set S with relations, functions defined on S
    Rings, groups, graphs, linear orders, ...

Example: 
  A group is a set G with 
     a group operation + (binary function) and
     a distinguished identity element 0 (constant) 

  We can capture properties of groups using abstract names op for
  + and e for 0

  op is associative   : (G1) Ax Ay Az op(op(x,y),z)) = op(x,op(y,z))
  e is right identity : (G2) Ax op(x,e) = x
  right inverse exists: (G3) Ax Ey op(x,y) = e

  Note that the meaning of = is fixed as equality in the
  underlying set.  
 
  Any set S with a binary function f and an element s that
  satisfies the axioms for op and e is a group.  Any group (G,+,0)
  satisfies the axioms by mapping op to + and e to 0

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