Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup

Theorem
Let $\left({S, \circ}\right)$ be a commutative monoid whose identity is $e$.

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 $T = S \circ' C^{-1}$, and is a commutative semigroup.

Proof of Subsemigroup
Let $x, z \in S, y, w \in C$.

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 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 $\left({T, \circ'}\right)$.

Proof of Equality
Let $a \in C$.

Then:

Thus, as $C^{-1} \subseteq S \circ' C^{-1}$ and $S \subseteq S \circ' C^{-1}$, we have $S \cup C^{-1} \subseteq S \circ' C^{-1}$ by Union Smallest.

By the definition of the inverse completion, we see that $T = S \circ' C^{-1}$.