Logic: Lecture 13, 18 September 2012
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Algebraic characterization of elementary equivalence

Recall

  Elementary equivalence

  M == M' wrt L if, for every sentence phi in L, M |= phi iff M' |= phi

How do we show that M and M' satisfy the same set of sentences?

Fraisse's theorem for graphs

Notation:  (M,\bar{a}) |= phi(\bar{x})

Definition: quantifier rank

Definition: m-equivalence

  G = (V,E), H = (W,F), \bar{a} a tuple over V, \bar{b} a tuple over W

  (G,\bar{a}) =m= (H,\bar{b}) iff for every phi(\bar{x}) of
  quantifier rank <= m, 
  (G,\bar{a}) |= \phi(\bar{x}) iff (H,\bar{b}) |= \phi(\bar{x})
  
Note: \bar{x} and \bar{a} must match in length etc

Lemma:
  Suppose for every c in V, there is d in W such that
  (G,\bar{a}c) ={m-1}= (H,\bar{b}d) and vice versa.  Then
  (G,\bar{a}) =m= (H,\bar{b})

Definition: m-isomorphism
  (G,\bar{a}) ~0~ (H,\bar{b}) if there is a partial isomorphism

  (G,\bar{a}) ~m~ (H,\bar{b}) if

  - For each c in V, exists d in W, (G,\bar{a}c) ~(m-1)~
    (H,\bar{b}d)
  - Vice versa


Proposition:
  (G,\bar{a}) =0= (H,\bar{b}) iff (G,\bar{a}) ~0~ (H,\bar{b})

Lemma:
  Finitely many inequivalent formulas of quantifier depth <= m
  with at most k free variables.

Theorem:
  If (G,\bar{a}) =m= (H,\bar{b}) then (G,\bar{a}) ~m~ (H,\bar{b})

Other direction is easy

Definition:
  Finitely isomorphic:  (G,\bar{a}) ~f~ (H,\bar{b}) iff
  (G,\bar{a}) ~m~ (H,\bar{b}) for all m >= 0

Theorem (Fraisse)
  G ~f~ H iff G == H
  
Lifting the proof to

1. Arbitrary relations

   Relatively easy.  Only base case of induction needs to be adjusted.

2. Functions and constants

   Represent functions and constants in terms of relations.  See
   pdf lecture notes for details.

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