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The theory of equivalence relations can be axiomatized by 3 equality-free universal sentences, namely:

1.$xRx$

2.$xRy \rightarrow yRx$

3.$(xRy \land yRz) \rightarrow xRz$.

Now, certainly, we can add axioms to the theory of equivalence relations, so that, to give just one arbitrary example, the cardinality of the underlying universe is constrained to be between 5 and 9 elements. But I am interested in adding only equality-free universal sentences to the theory. I conjecture that only 3 possibilities can happen if we add an equality-free universal sentence $S$ to the theory of equivalence relations. Either $S$ is already part of the deductive closure of the theory of equivalence relations, in which case $S$ is redundant and there is no change, or $S$ narrows the class of models to just total relations, a total relation being a relation that holds between every pair of elements, and the third possibility is that $S$ narrows the class of models all the way down to just the empty set with the empty relation. (If you disallow empty models, then just consider $S$ to make the theory inconsistent in the third case). Let me give examples of the latter two cases: $xRy$ and $\neg xRy$, respectively. Anyway, is my conjecture true, that there are only these 3 possibilities? Or is there some weird intermediate case?

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  • $\begingroup$ "$R$ has at least two equivalence classes" can be phrased this way, can't it? "there exist $x,y$ with $\neg xRy$" is "$\neg (\forall x)(\forall y)(x Ry)$". (Or does the presence of the leading $\neg$ forbid that?) $\endgroup$ Commented Jul 7 at 14:08
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    $\begingroup$ @PatrickStevens Indeed, the presence of the leading $\lnot$ makes your sentence existential, not universal. Its negation "there is at most one equivalence class" is universal, but this is the OP's example $\forall x\forall y (xRy)$, which gives the class of total relations. However, you can generalize your example to get nontrivial counterexamples to the OP's conjecture. $\endgroup$ Commented Jul 7 at 15:02
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    $\begingroup$ "There exist at least three equivalence classes" is $\exists x\exists y\exists z(\lnot xRy\land \lnot xRz\land \lnot yRz)$. Its negation is $\forall x\forall y\forall z(xRy \lor xRz\lor yRz)$, which expresses that there are at most two equivalence classes. From here it's easy to see how to replace two by arbitrary finite $n$, obtaining infinitely many universal equality-free sentences which are pairwise non-equivalent relative to the theory of equivalence relations. $\endgroup$ Commented Jul 7 at 15:04

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As pointed out by Alex, you can write for each $n$ a universal sentence that says that the number of classes is at most $n$. In fact, I believe those are all the possibilities.

I think the easiest way to resolve this classification problem is to simply classify equivalence relations up to their (equality-free) theory.

Note that if $(X,R)$ is a set with an equivalence relation, then the canonical map $X\to X/R$ is elementary (in the language without equality). So it suffices to classify the equivalence relations with singleton classes.

Alex's sentences show that two such relations with different finite number of classes are not equivalent. On the other hand, once we fix the number of classes, even full first order theory is fixed (and indeed, in the finite case, even the isomorphism type). Thus, the possible universal theories are the theories of $(n,=)$ for $n\in \omega$ and $(\omega,=)$.

Now, given any universal sentence $\varphi$, the set of $n$ such that $\varphi$ is true in $(n,=)$ is $\omega$ (in which case, $(\omega,=)$ also satisfies it, because direct limits preserve universal sentences), or an initial segment of $\omega$ (because $(n,=)$ embeds into $(n+1,=)$ and $(\omega,=)$). Thus, $\varphi$ is equivalent to one of Alex's sentences (because it has the same models).

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