Finite Model Theory/FO EFM

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The method for employing Ehrenfeucht-Fraisse-Games for (in-)expressibility-proofs is given by the following:


Let P be a property of finite σ-structures. Then the following are equivalent

  • P is not expressible in FO
  • for every k \in \mathbb{N} there exist two finite σ-structures \mathfrak{A}_k and \mathfrak{B}_k, such that the following are both satisfied
    • \mathfrak{A}_k\equiv_k \mathfrak{B}_k
    • \mathfrak{A}_k has P and \mathfrak{B}_k does not have P


  • Thus using the EFM works roughly as follows:
    • choose a k
    • construct two structures - one with the property, one without - that are big enough s.t. the duplicator wins the k-ary EFG
    • show that this can be expanded with k
  • So, a non-expressible property (i.e. the effort to check it) must be somehow 'expandable' with k


  • To begin pick two linear orders say A ={1, 2, 3, 4} and B ={1, 2, 3, 4, 5}. For a two-move Ehrenfeucht game D is to win, obviously. This gives us two structures that satisfy the above conditions for k = 2 and the Property having even cardinality (that A has and B doesn't). Now we have to expand this over all k \in \mathbb{N}. From the above example we adopt that in a linear order of cardinality 2^k or higher D has a winning strategy. Thus we choose the cardinalities depending on k as |A| = 2^k and |B| = 2^k+1. So we have found an even A and an odd B for every k, where D has a winning strategy. Thus (by the corollary) having even/odd cardinality is a property that can not be expressed in FO for finite σ-structures of linear order.