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 $$\circ$$.

$$ $$ $$ $$ $$

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
It is also easily shown that $$x \circ' y^{-1}$$ commutes with $$z \circ' w^{-1}$$, so $$S \circ' C^{-1}$$ is shown to be a commutative subsemigroup of $$\left({T, \circ'}\right)$$:

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 $$\circ$$.

$$ $$ $$ $$ $$ $$ $$

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 $$T = S \circ' C^{-1}$$.