What is an axiomatization of the equality-free theory of antisymmetric relations?

Question

An antisymmetric relation is defined as a binary relation $R$ on a set $S$ such that $(xRy \land yRx) \rightarrow x=y$, for all $x,y$ in $S$. Certainly, they can't be defined in first-order logic without equality. However, what is an axiomatization of the equality-free theory of antisymmetric relations? I conjecture that all you need is the sentence $(xRy \land yRx) \rightarrow xRx$. Is the conjecture true? Also, bonus question, can someone give an example of a binary relation that satisfies the equality-free theory of antisymmetric relations, but is not itself an antisymmetric relation?

Answer 1

Your proposed sentence is not strong enough. Consider, for example, the "distance-$<17$" relation on $\mathbb{R}$ with the usual metric.

The issue is that we need "transitivity within reflexivity regions:" if $x=y$, then $x$ and $y$ must be related to the same objects. The following pair of sentences on their own do the job: $$zRxRyRx\rightarrow zRy, \quad xRyRxRz\rightarrow yRz.$$ (Here I'm using the standard abbreviation of e.g. "$\alpha R\beta R\gamma$" for "$\alpha R\beta\wedge\beta R\gamma$.") The argument that this gives the full $\mathsf{FOL_{w/o=}}$-theory of asymmetric binary relations uses the standard "quasi-isomorphism" trick (I don't think it actually has a name), see the second highlighted fact in this old answer of mine. The point is that if $R$ is a relation on a set $X$ which satisfies the sentences above, then the relation $$x\sim y\quad\iff\quad xRyRx$$ is a congruence on the structure $(X;R)$, and the resulting quotient is an asymmetric binary relation.

This also shows how to get an example for your bonus question: take the complete binary relation $R=X^2$ on any set $X$ with at least two elements.