Non-Cancellable Elements of Semigroup form Subsemigroup

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

Let $T \subseteq S$ be the subset of $S$ containing the elements of $S$ which are specifically not cancellable in $\struct {S, \circ}$.

Then $\struct {T, \circ}$ forms a subsemigroup of $S$.

Proof
Recall the definition of cancellable element:

An element $x \in \struct {S, \circ}$ is cancellable :
 * $\forall a, b \in S: x \circ a = x \circ b \implies a = b$
 * $\forall a, b \in S: a \circ x = b \circ x \implies a = b$

From the Subsemigroup Closure Test it is sufficient to demonstrate that:
 * $\forall x, y \in T: x \circ y \in T$

Let $x, y \in T$.

Then:
 * $\exists a, b \in S: a \circ y = b \circ y \text { but } a \ne b$

or:
 * $\exists a, b \in S: y \circ a = y \circ b \text { but } a \ne b$

, suppose that:
 * $\exists a, b \in S: y \circ a = y \circ b \text { but } a \ne b$

Hence:

So we have:
 * $\paren {x \circ y} \circ a = \paren {x \circ y} \circ b$

but $a \ne b$.

Hence $x \circ y$ is not cancellable.

As $x$ and $y$ are arbitrary, the result follows.