Theorem (Fagin 1976, Glebskii et al. 1969): First-order logic admits the 0-1 law
Most finite model theorists equate the admission of 0-1 law with the inability to do "counting". Consider the following properties EVEN = { G : G has even number of vertices }. We see that μn(EVEN) = 1 iff n is even, and hence μ(EVEN) does not exist. I am yet to see a property whose asymptotic probability μ converges to something other than 0 or 1, or diverges, but has nothing to do with counting. Consider the following properties:
- μ(BIPARTITE) = 0
- μ(HAMILTONIAN) = 1
- μ(CONNECTED) = 1
- μ(TREE) = 0
On the other hand, if you admit constant or function symbols, it is possible to have first-order properties whose asymptotic probability converges to something other than 0 or 1. So, it is crucial to assume that our vocabulary σ be purely relational.
What is the use of 0-1 law in finite model theory apart from the amusement of some theoreticians? [I will have to confess many proofs about the 0-1 law, including Fagin's proof of the above theorem, are usually of great beauty.] Well, a primary concern of finite model theory is to determine how expressive a logic is over a given class of finite structures. By proving that a logic admits the 0-1 law, we show that any property whose asymptotic probability does not converge to either 0 or 1 is not expressible in the logic. For example, an immediate corollary of the above theorem is that first-order logic cannot express EVEN.
Let me briefly outline the proof ideas of the above theorem which recurs in 0-1 law proofs. We only need to show that, for each natural number k, there exists a property Pk of graphs such that
- μ(Pk) = 1, and
- Any two models of Pk cannot be distinguished by a first-order sentence of quantifier rank k. [Recall that the quantifier rank of a first-order sentence is the maximum depth of any nesting of quantifiers of the sentence.]
Why is this sufficient? Let f be a first-order sentence of quantifier rank k. There are two cases. First, there exists a model of Pk that is also a model of f. In this case, it is not hard to show that any model of Pk is also a model of f, which implies that μ(f) ≥ μ(Pk) = 1. The second case is that all models of Pk are not models of f, i.e., they are models of "not f". Therefore, μ("not f") ≥ μ(Pk) = 1. That is, we have μ(f) = 0.
The property Pk is commonly referred to as an extension axiom. Note that Pk does not have to be L-definable (where L is the logic that is to admit 0-1 law).
Two excellent expositions of the basics of 0-1 laws include Leonid Libkin's Elements of Finite Model Theory chapter 12, and James Lynch DIMACS 1995-1996 Tutorial notes on "Random Finite Models".