Cross-Relation is Equivalence Relation

Theorem
Let $\left({S, \circ}\right)$ be a commutative semigroups.

Let $\left({S_1, \circ_{\restriction_1}}\right), \left({S_2, \circ_{\restriction_2}}\right)$ be subsemigroups of $S$, where $\circ_{\restriction_1}$ and $\circ_{\restriction_2}$ are the restrictions of $\circ$ to $S_1$ and $S_2$ respectively.

Let $\left({S_1 \times S_2, \oplus}\right)$ be the (external) direct product of $\left({S_1, \circ_{\restriction_1}}\right)$ and $\left({S_2, \circ_{\restriction_2}}\right)$, where $\oplus$ is the operation on $S_1 \times S_2$ induced by $\circ_{\restriction_1}$ on $S_1$ and $\circ_{\restriction_2}$ on $S_2$.

Let $\mathcal R$ be the cross-relation on $S_1 \times S_2$, defined as:
 * $\left({x_1, y_1}\right) \mathop {\mathcal R} \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

Then $\mathcal R$ is an equivalence relation on $\left({S_1 \times S_2, \oplus}\right)$.

Reflexivity

 * $x_1 \circ y_1 = x_1 \circ y_1 \implies \left({x_1, y_1}\right) \mathop {\mathcal R} \left({x_1, y_1}\right)$

So $\mathcal R$ is a reflexive relation.

Symmetry
So $\mathcal R$ is a symmetric relation.

Transitivity
So $\mathcal R$ is a transitive relation.

All the criteria are therefore seen to hold for $\mathcal R$ to be an equivalence relation.