Equivalence Induced by Epimorphism is Congruence Relation

Theorem
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.

Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.

Let $\RR_\phi$ be the equivalence induced by $\phi$.

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

Proof
Let $x, x', y, y' \in S$ such that:
 * $x \mathrel {\RR_\phi} x' \land y \mathrel {\RR_\phi} y'$

By definition of induced equivalence:

Then:

Thus $\paren {x \circ y} \mathrel {\RR_\phi} \paren {x' \circ y'}$ by definition of induced equivalence.

So $\RR_\phi$ is a congruence relation for $\circ$.