TITLE: How can I formally prove that undirected graphs cannot be axiomatised in First Order Logic with only unary predicates?
QUESTION [5 upvotes]: All the models of the following FOL formula are simple undirected graphs and all the simple undirected graphs are a model of this formula [source]:
$$\forall x. \neg R(x,x)$$
$$\forall x \forall y. R(x,y) \rightarrow R(y,x)$$
Now, can I express this formula somehow with unary predicates alone in a function free FOL language ?  If not then how can I prove that it's not possible ?

REPLY [3 votes]: So far, we have an answer that generalizes to a large class of logics, and a more specific answer that treats first-order logic. Let me give an even more specific answer: one that focuses on the finite case.
Assume that there was a predicate formula $\varphi$ (given in terms of unary predicates and perhaps equality) whose models are simple undirected graphs, and that all the simple undirected graphs are a model of this formula $\varphi$. So the $n$-element models of $\varphi$  should be precisely the simple undirected graphs with $n$ vertices.
Since the formula $\varphi$ contains only finitely many (say $k$) unary predicates, and each unary predicate determines a single subset of these $n$ vertices, we would get that there are at most $2^{kn}$ different graph structures on $n$ vertices. However, it can be shown (see Flajolet and Sedgewick: Analytic Combinatorics, p. 105, available on the author's website) that as $n$ goes to infinity, the number of simple graphs (up to isomorphism) grows faster than $2^{n\left(\frac{n-1}{2} - \log n\right)}$. Needless to say, this rapidly outgrows any function of the form $2^{kn}$ for fixed $k$.