Construction of Inverse Completion/Equivalence Relation/Members of Equivalence Classes

Theorem
Then: $\forall x, y \in S, a, b \in C:$


 * $(1): \quad \left({x \circ a, a}\right) \boxtimes \left({y \circ b, b}\right) \iff x = y$


 * $(2): \quad \left[\!\left[{\left({x \circ a, y \circ a}\right)}\right]\!\right]_\boxtimes = \left[\!\left[{\left({x, y}\right)}\right]\!\right]_\boxtimes$

where $\left[\!\left[{\left({x, y}\right)}\right]\!\right]_\boxtimes$ is the equivalence class of $\left({x, y}\right)$ under $\boxtimes$.

Proof
From Cross-Relation is Equivalence Relation we have that $\boxtimes$ is an equivalence relation.

Hence the equivalence class of $\left({x, y}\right)$ under $\boxtimes$ is defined for all $\left({x, y}\right) \in S \times C$.

From Semigroup is Subsemigroup of Itself, $\left({S, \circ}\right)$ is a subsemigroup of $\left({S, \circ}\right)$.

Also from Semigroup is Subsemigroup of Itself, $\left({C, \circ {\restriction_C}}\right)$ is a subsemigroup of $\left({C, \circ {\restriction_C}}\right)$.

The result follows from Elements of Cross-Relation Equivalence Class.