Reflexive Circular Relation is Equivalence

Theorem
Let $\mathcal R \subseteq S \times S$ be a reflexive and circular relation in $S$.

Then $\mathcal R$ is an equivalence relation.

Proof
To prove a relation is an equivalence, we need to prove it is reflexive, symmetric and transitive.

So, checking in turn each of the criteria for equivalence:

Reflexive
By hypothesis $\mathcal R$ is reflexive.

Symmetric
By reflexivity $\left({x, x}\right) \in \mathcal R$.

If $\left({x, y}\right) \in \mathcal R$ then by the definition of circular relation $\left({y, x}\right) \in \mathcal R$.

Hence $\mathcal R$ is symmetric.

Transitive
If $\left({x, y}\right), \left({y, z}\right) \in \mathcal R$ then $\left({z, x}\right) \in \mathcal R$.

By $\mathcal R$ being symmetric $\left({x, z}\right) \in \mathcal R$.

Hence $\mathcal R$ is transitive.

Thus is $\mathcal R$ is reflexive, symmetric and transitive, and therefore an equivalence.