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

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

Proof
Note that by definition of inverse completion, $\left({T, \circ'}\right)$ is a semigroup.

Thus $\circ'$ is associative.

First it is demonstrated that $S \circ' C^{-1}$ is a semigroup.

Let $x, z \in S$.

Let $y, w \in C$.

Then:

Thus:
 * $\left({x \circ z}\right)\circ' \left({w \circ y}\right)^{-1} \in S \circ' C^{-1}$

proving that $S \circ' C^{-1}$ is closed.

Therefore by Subsemigroup Closure Test:
 * $S \circ' C^{-1}$ is a subsemigroup of $\left({T, \circ'}\right)$

and thus a semigroup.

It remains to be shown that $\circ'$ is a commutative operation.

Let $\left({x \circ' y^{-1} }\right)$ and $\left({z \circ' w^{-1} }\right)$ be two arbitrary elements of $S \circ' C^{-1}$.

By Associativity and Commutativity Properties, $x, y, z, w$ all commute with each other under $\circ$.

As $\circ'$ is an extension of $\circ$, it follows that $x, y, z, w$ also all commute with each other under $\circ'$.

Then:

So $x \circ' y^{-1}$ commutes with $z \circ' w^{-1}$.

It follows by definition that $S \circ' C^{-1}$ is a commutative semigroup.