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 $T$ is a commutative monoid.

Proof
From Structure of Inverse Completion of Commutative Semigroup:
 * $T = S \circ' C^{-1}$

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

First it is demonstrated that $T = S \circ' C^{-1}$ is a semigroup.

Let $x, z \in S$.

Let $y, w \in C$.

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

So:

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 $S \circ' C^{-1}$ is a subsemigroup of $\left({T, \circ'}\right)$.

Proof of Commutative Subsemigroup
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 $\circ$.

Then:

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

It follows by definition that $S \circ' C^{-1}$ is a commutative subsemigroup of $\left({T, \circ'}\right)$.