Finite Model Theory/FO EFM

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

Theorem

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

P is not expressible in FO
for every k $\in \mathbb {N}$ $\in \mathbb {N}$ there exist two finite σ-structures ${\mathfrak {A}}_{k}$ ${\mathfrak {A}}_{k}$ and ${\mathfrak {B}}_{k}$ ${\mathfrak {B}}_{k}$ , 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

Remarks

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

Examples

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.