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:
 * $x \mathop {\mathcal R_\phi} x' \land y \mathop {\mathcal R_\phi} y'$

By definition of induced equivalence:

Then:

Thus $\paren {x \circ y} \mathop {\mathcal R_\phi} \paren {x' \circ y'}$ by definition of induced equivalence.

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