Logic: Lecture 16, 11 October 2012
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Quantifier elimination:

   T admits quantifier elimination if for every formula phi there
   is a quantifier free formula psi such that T |= phi <-> psi

   Aside: What does it mean to say T |= alpha for a formula alpha
          with free variables?  For every intpretation (M,s) such
          that (M,s) |= T, (M,s) |= alpha.  Of course, (M,s) |= T
          is independent of s, so effectively we have that for
          every M such that M |= T, for every s, (M,s) |= alpha.
          This is equivalent to implicit universal quantification
          of all free variables in alpha.


Suffices to consider

  Ex (a_1 and ... and a_n), each a_i atomic or negation of atomic

Proof:

  Convert Ex theta to Disjunctive Normal Form (DNF): 

  - Each clause is a conjunction of literals (atomic formula or
    negation), formula is disjunction of clauses

  - DNF can be constructed from truth table.  One clause per row
    that evaluate to tt

  Distribute Ex over or to get (Ex clause_1) or ... or (Ex clause_k)

  Eliminiate quantifiers "inside out" from innermost Ex, so each
    step works with formulas of the form above

Assume x appears in each a_i: otherwise Ex (a and b) = a and Ex b

Terms are of the form S^k(u) where u is either 0 or a variable.

Atomic formulas are of the form S^i(u) = S^j(v).

Thus, in 

  Ex (a_1 and ... and a_n), each a_k is S^i(x) = S^j(u) or its
  negation.

  Can assume that u is not x --- otherwise reduce S^i(x) = S^i(x)
  to 0=0 (i.e. tt) and S^i(x) = S^j(x), i =/= j to 0 =/= 0 (i.e. ff).

  Case 1:
    Each a_i is of the form ~(S^j(x) = u).    Can always find x
    that satisfies finitely many inequations, so reduce to 0=0.

  Case 2:
    At least one equality.  Wlog, assume a_1 is S^m(x) = u.

    - Add (u =/= 0 and u =/= S(0) and .. u =/= S^{m-1}(0)
      so that any assignment for variable in u yields a valid
      assignment for x.

    - Transform every other atomic formula of the form S^j(x) = v
      to S^{j+m}(x) = S^m(v) and substitute u to get S^j(u) = S^m(v).
  
    Now we have an equivalent formula in which x no longer occurs.

This yields an alternate proof that A_S is complete:

  Let phi be any sentence.  By quantifier elimination, there is a
  quantifier free sentence psi such that A_S |- phi <-> psi.
 
  Atomic formulas in psi are of the form S^k(0) = S^l(0).  Can
  effectively decide if psi is true or false, correponding to
  which phi is true or false, so A_S |= phi or A_S |= ~phi

Definability:

  Subsets of Nat:
    Only finite/co-finite subsets of Nat are definable

    - Verify this formally for youselves 

  Binary relations:

    Can define a set of pairs (x,y) only if it is covered by a
    finite set of lines (linear) or its complete is.  

  Corollary: < is not definable.

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