Equivalence Relation is Circular

Theorem
Let $\RR \subseteq S \times S$ be an equivalence relation.

Then $\RR$ is also a circular relation.

Proof
Let $x, y, z \in S$ be arbitrary such that:
 * $\tuple {x, y} \in \RR$ and $\tuple {y, z} \in \RR$

We have that $\RR$ is transitive.

Hence:
 * $\tuple {x, z} \in \RR$

We also have that $\RR$ is symmetric.

Hence:
 * $\tuple {z, x} \in \RR$

We have demonstrated that:
 * $\tuple {x, y} \in \RR$ and $\tuple {y, z} \in \RR \implies \tuple {z, x} \in \RR$

and $\RR$ is circular by definition.