Equivalence Induced by Epimorphism is Congruence Relation

Theorem
Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures.

Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an epimorphism.

Let $\mathcal R_\phi$ be the equivalence induced by $\phi$.

Then the induced equivalence $\mathcal R_\phi$ is a congruence relation for $\circ$.

Proof
Let $x, x', y, y' \in S$ such that:
 * $x \, \mathcal R_\phi \, x' \land y \, \mathcal R_\phi \, y'$

By definition of induced equivalence:

Then:

Thus $\left({x \circ y}\right) \, \mathcal R_\phi \, \left({x' \circ y'}\right)$ by definition of induced equivalence.

So $\mathcal R_\phi$ is a congruence relation for $\circ$.