# Cross-Relation is Equivalence Relation

## Theorem

Let $\struct {S, \circ}$ be a commutative semigroup with cancellable elements.

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

Let $\struct {S_1, \circ {\restriction_1} } \subseteq \struct {S, \circ}$ be a subsemigroup of $S$.

Let $\struct {S_2, \circ {\restriction_2} } \subseteq \struct {C, \circ {\restriction_C} }$ be a subsemigroup of $C$.

Let $\left({S_1 \times S_2, \oplus}\right)$ be the (external) direct product of $\struct {S_1, \circ {\restriction_1} }$ and $\struct {S_2, \circ {\restriction_2} }$, 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:

$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$

Then $\boxtimes$ is an equivalence relation on $\struct {S_1 \times S_2, \oplus}$.

## Proof

### Reflexivity

$\forall \tuple {x_1, y_1} \in S_1 \times S_2: x_1 \circ y_1 = x_1 \circ y_1 \implies \tuple {x_1, y_1} \boxtimes \tuple {x_1, y_1}$

So $\boxtimes$ is a reflexive relation.

$\Box$

### Symmetry

 $\ds \tuple {x_1, y_1}$ $\boxtimes$ $\ds \tuple {x_2, y_2}$ $\ds \leadsto \ \$ $\ds x_1 \circ y_2$ $=$ $\ds x_2 \circ y_1$ $\ds \leadsto \ \$ $\ds x_2 \circ y_1$ $=$ $\ds x_1 \circ y_2$ $\circ$ is commutative $\ds \leadsto \ \$ $\ds \tuple {x_2, y_2}$ $\boxtimes$ $\ds \tuple {x_1, y_1}$

So $\boxtimes$ is a symmetric relation.

$\Box$

### Transitivity

 $\ds \tuple {x_1, y_1}$ $\boxtimes$ $\ds \tuple {x_2, y_2}$ $\, \ds \land \,$ $\ds \tuple {x_2, y_2}$ $\boxtimes$ $\ds \tuple {x_3, y_3}$ $\ds \leadsto \ \$ $\ds x_1 \circ y_2$ $=$ $\ds x_2 \circ y_1$ $\, \ds \land \,$ $\ds x_2 \circ y_3$ $=$ $\ds x_3 \circ y_2$ $\ds \leadsto \ \$ $\ds x_1 \circ y_3 \circ y_2$ $=$ $\ds x_1 \circ y_2 \circ y_3$ $\circ$ is commutative $\ds$ $=$ $\ds x_2 \circ y_1 \circ y_3$ substituting $x_2 \circ y_1$ for $x_1 \circ y_2$ $\ds$ $=$ $\ds x_2 \circ y_3 \circ y_1$ $\circ$ is commutative $\ds$ $=$ $\ds x_3 \circ y_2 \circ y_1$ substituting $x_3 \circ y_2$ for $x_2 \circ y_3$ $\ds$ $=$ $\ds x_3 \circ y_1 \circ y_2$ $\circ$ is commutative $\ds \leadsto \ \$ $\ds x_1 \circ y_3$ $=$ $\ds x_3 \circ y_1$ as $y_2 \in S_2$, therefore $y_2 \in C$, and so is cancellable for $\circ$ $\ds \leadsto \ \$ $\ds \tuple {x_1, y_1}$ $\boxtimes$ $\ds \tuple {x_3, y_3}$

So $\boxtimes$ is a transitive relation.

$\Box$

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

$\blacksquare$