Cancellable Elements of Semigroup form Subsemigroup

Theorem
Let $\left ({S, \circ}\right)$ be a semigroup.

Let $C$ be the set of cancellable elements of $\left ({S, \circ}\right)$.

Then $\left({C, \circ}\right)$ is a subsemigroup of $\left ({S, \circ}\right)$.

Proof
Now let $C$ be the set of cancellable elements of $\left ({S, \circ}\right)$.

Let $x, y \in C$.

Then $x$ and $y$ are both left cancellable and right cancellable.

Thus by Left Cancellable Elements of Semigroup form Subsemigroup:


 * $x \circ y$ is left cancellable

and by Right Cancellable Elements of Semigroup form Subsemigroup:


 * $x \circ y$ is right cancellable.

Thus $x \circ y$ is both left cancellable and right cancellable, and therefore cancellable.

Thus $x \circ y \in C$.

Thus $\left ({C, \circ}\right)$ is closed and is therefore by the Subsemigroup Closure Test a semigroup of $\left ({S, \circ}\right)$.