Inverse Completion of Commutative Semigroup is Inverse Completion of Itself

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 $\struct {T, \circ'}$ is its own inverse completion.

Proof
Let $x \circ' y^{-1}$ be cancellable for $\circ'$, where $x \in S$ and $y \in C$.

We have that $y$ is invertible for $\circ'$.

So by Invertible Element of Associative Structure is Cancellable:
 * $y$ is cancellable for $\circ'$.

Now by definition of inverse element:
 * $x = \paren {x \circ' y^{-1} } \circ' y$

Thus $x$ is also cancellable for $\circ'$.

By Cancellable Elements of Semigroup form Subsemigroup, $x$ is cancellable for $\circ$.

So:
 * $x \in S \implies x \in C$

Thus $x$ is invertible for $\circ'$.

Hence by Inverse of Product in Associative Structure:
 * $x \circ' y$ is invertible for $\circ'$.

So every cancellable element of $\struct {T, \circ'}$ is invertible.

So, by definition, $\struct {T, \circ'}$ is its own inverse completion.