Structure of Inverse Completion of Commutative Semigroup

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

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

Let $\left({T, \circ'}\right)$ be an inverse completion of $\left({S, \circ}\right)$.

Then:
 * $T = S \circ' C^{-1}$

where:
 * $C^{-1}$ is the inverse of $C$ in $T$
 * $S \circ' C^{-1}$ is the subset product of $S$ with $C^{-1}$.

Proof
Let $a \in C$.

Then:

Thus, as:
 * $C^{-1} \subseteq S \circ' C^{-1}$

and:
 * $S \subseteq S \circ' C^{-1}$

by Union is Smallest Superset it follows that:
 * $S \cup C^{-1} \subseteq S \circ' C^{-1}$

From Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup:
 * $S \circ' C^{-1}$ is a commutative semigroup.

So $S \circ' C^{-1}$ is a semigroup which contains $S \cup C^{-1}$.

By definition of generator of semigroup, it follows that:
 * $\left\langle{S \cup C^{-1} }\right\rangle \subseteq S \circ' C^{-1}$

Let $z = x \circ' y^{-1} \in S \circ' C^{-1}$.

Then by definition:
 * $x \in S$

and:
 * $y^{-1} \in C^{-1}$

and so by definition of generator of semigroup:
 * $x \circ' y^{-1} \in \left\langle{S \cup C^{-1} }\right\rangle$

Thus: $S \circ' C^{-1} \subseteq \left\langle{S \cup C^{-1} }\right\rangle$

By definition of set equality:
 * $S \circ' C^{-1} = \left\langle{S \cup C^{-1} }\right\rangle$

and so by definition of the inverse completion:
 * $T = S \circ' C^{-1}$