Logic: Lecture 19, 30 October 2012
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Defn: A relation over N is recursive iff it is representable in
      some consistent finitely axiomatizable theory.

In A_E, we can represent all "recursive functions" (in the sense
of recursive function theory), so, in particular

      A relation is recursive iff it is representable in the
      theory Cn AE.

This is an alternative formulation of Church's Thesis.

Corollary: Any recursive relation is definable in Th(NT).

Given a set X, #X denotes {#alpha | alpha in X}.

A is an effective axiomatization if #A is recursive.  

Defn: An axiomatization A is decidable if #Con(A) is recursive.

#A recursive does not imply #Con(A) is recursive --- it may only
be recursively enumerable.  However, if A is a complete theory
--- for every sentence sigma, either sigma or ~sigma is in Con(A)
--- then #Con(A) is recursive.

Fixed Point Lemma:

  For any formula beta in which only v1 occurs free, we can
  construct a sentence sigma such that 

     A_E |- sigma <-> beta(#sigma)

Observe that

      NT |= sigma <-> beta(#sigma)

Theorem (Tarski Undefinability)
  The set #Th(NT) is not definable in NT.

Proof:
  
  Suppose beta defines #Th(NT).  Apply fixed point lemma to
  ~beta to get sigma such that:
     
      NT |= sigma <-> ~beta(#sigma)

  Case 1: sigma is true -> #sigma is not in #Th(NT)
  Case 2: sigma is false -> #sigma is in #Th(NT)

  Contradiction!

Corollary:
  #Th(NT) is not recursive.
Proof:
  Since it is not definable.
  
Theorem (Goedel Incompleteness)
  If A is a subset of Th(NT) and #A is recursive, then Con(A) is
  not complete.

Proof:
  If Con(A) is complete, Con(A) = Th(NT).  Also, #Con(A) would be
  recursive, but #Th(NT) is not.

A more informative proof:

  Suppose A subset of Th(NT) such that #Con(A) is recursive.
  Consider beta that represents #Con(A).  Apply fixed point
  theorem to ~beta as in proof of Tarski undecidability.

  The resulting sigma is true, but #sigma is not in #Con(A).

Lemma:
  If #Con(Sigma) is recursive, then for any  new axiom tau,
  #Con(Sigma;tau) is recursive.

Proof:
  alpha in Con(Sigma;tau) iff tau -> alpha in Con(Sigma).

Strong Undecidability:
  Let T be a theory such that T + A_E is consistent. Then #T is
  not recursive.

Proof:
  If #T is recursive, so is #(T + A_E) since A_E is finite.
  Again apply fixed point lemma.

Corollary:
  If #Sigma is recursive and Sigma+A_E is consistent, then
  Con(Sigma) is not complete.