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

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


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


 * $(2): \quad \eqclass {\tuple {x \circ a, y \circ a} } \boxtimes = \eqclass {\tuple {x, y} } \boxtimes$

where $\eqclass {\tuple {x, y} } \boxtimes$ is the equivalence class of $\tuple {x, y}$ under $\boxtimes$.

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

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

From Semigroup is Subsemigroup of Itself, $\struct {S, \circ}$ is a subsemigroup of $\struct {S, \circ}$.

Also from Semigroup is Subsemigroup of Itself, $\struct {C, \circ {\restriction_C} }$ is a subsemigroup of $\struct {C, \circ {\restriction_C} }$.

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