Equivalence iff Diagonal and Inverse Composite

Theorem
Let $$\mathcal R$$ be a relation on $$S$$.

Then $$\mathcal R$$ is an equivalence relation on $$S$$ iff $$\Delta_S \subseteq \mathcal R$$ and $$\mathcal R = \mathcal R \circ \mathcal R^{-1}$$.

Necessary Condition
Let $$\mathcal R$$ be an equivalence relation.

Then by definition, it is reflexive, symmetric and transitive.

As $$\mathcal R$$ is reflexive, we have $$\Delta_S \subseteq \mathcal R$$ from Reflexive contains Diagonal Relation.

As $$\mathcal R$$ is transitive, we have $$\mathcal R \circ \mathcal R \subseteq \mathcal R$$ from Transitive Relation contains Composite with Self.

But as $$\mathcal R$$ is symmetric, we also have $$\mathcal R = \mathcal R^{-1}$$ from Relation equals Inverse iff Symmetric.

Thus $$\mathcal R \circ \mathcal R^{-1} \subseteq \mathcal R$$.

Now we need to show that $$\mathcal R \subseteq \mathcal R \circ \mathcal R^{-1}$$:

Let $$\left({x, y}\right) \in \mathcal R$$.

Then as $$\mathcal R$$ is reflexive, $$\left({x, y}\right) \in \mathcal R \and \left({y, y}\right) \in \mathcal R$$ and so $$\left({x, y}\right) \in \mathcal R \circ \mathcal R$$.

As $$\mathcal R = \mathcal R^{-1}$$, it follows that $$\mathcal R \subseteq \mathcal R \circ \mathcal R^{-1}$$.

So the first part has been shown: $$\Delta_S \subseteq \mathcal R$$ and $$\mathcal R = \mathcal R \circ \mathcal R^{-1}$$.

Sufficient Condition

 * Now, let $$\Delta_S \subseteq \mathcal R$$ and $$\mathcal R = \mathcal R \circ \mathcal R^{-1}$$.

From Reflexive contains Diagonal Relation, $$\mathcal R$$ is reflexive.

Suppose $$\left({x, y}\right) \in \mathcal R$$.

Then $$\left({x, y}\right) \in \mathcal R \circ \mathcal R^{-1}$$.

So $$\exists z \in S: \left({x, z}\right) \in \mathcal R, \left({z, y}\right) \in \mathcal R^{-1}$$, and so $$\left({y, z}\right) \in \mathcal R$$.

Thus it follows that $$\mathcal R$$ is transitive.

And as $$\left({x, z}\right) \in \mathcal R$$, it follows that $$\left({z, x}\right) \in \mathcal R^{-1}$$ and so it follows that $$\left({y, x}\right) \in \mathcal R$$.

So it follows that $$\mathcal R$$ is symmetric.

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