Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup

Theorem
Let $\struct {S, \circ}$ be a commutative semigroup.

Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.

Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.

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, $\struct {T, \circ'}$ 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:
 * $\paren {x \circ z} \circ' \paren {w \circ y}^{-1} \in S \circ' C^{-1}$

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

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

and thus a semigroup.

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

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

By Element Commutes with Product of Commuting Elements, $x, y, z, w$ 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.