# Equivalence Class of Equal Elements of Cross-Relation

## Theorem

Let $\left({S, \circ}\right)$ be a commutative semigroup with cancellable elements.

Let $\left({C, \circ {\restriction_C}}\right) \subseteq \left({S, \circ}\right)$ be the subsemigroup of cancellable elements of $\left({S, \circ}\right)$, where $\circ {\restriction_C}$ denotes the restriction of $\circ$ to $C$.

Let $\left({S_1, \circ {\restriction_1}}\right) \subseteq \left({S, \circ}\right)$ be a subsemigroup of $S$.

Let $\left({S_2, \circ {\restriction_2}}\right) \subseteq \left({C, \circ {\restriction_C}}\right)$ be a subsemigroup of $C$.

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 $\boxtimes$ be the cross-relation on $S_1 \times S_2$, defined as:

$\left({x_1, y_1}\right) \boxtimes \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

Then:

$\forall c, d \in S_1 \cap S_2: \tuple {c, c} \boxtimes \tuple {d, d}$

## Proof

Note that in order for $\tuple {c, c}$ and $\tuple {d, d}$ to be defined, $c$ and $d$ must be in both $S_1$ and $S_2$.

Hence the restriction given:

$\forall c, d \in S_1 \cap S_2$

Then:

 $\displaystyle \forall c, d \in S_1 \cap S_2: c \circ d$ $=$ $\displaystyle d \circ c$ Commutativity of $\circ$ $\displaystyle \leadsto \ \$ $\displaystyle \tuple {c, c}$ $\boxtimes$ $\displaystyle \tuple {d, d}$ Definition of $\boxtimes$

Hence the result.

$\blacksquare$