Cancellable Elements of Semigroup form Subsemigroup

Theorem
The right cancellable elements of a semigroup $$\left ({S, \circ}\right)$$ form a subsemigroup of $$\left ({S, \circ}\right)$$.

Similarly, the left cancellable elements of a semigroup $$\left ({S, \circ}\right)$$ form a subsemigroup of $$\left ({S, \circ}\right)$$.

Consequently, the cancellable elements of a semigroup $$\left ({S, \circ}\right)$$ form a subsemigroup of $$\left ({S, \circ}\right)$$.

Proof
Let $$C_\rho$$ be the set of right cancellable elements of $$\left ({S, \circ}\right)$$, that is:

$$C_\rho = \left\{{x \in S: \forall a, b \in S: a \circ x = b \circ x \Longrightarrow a = b}\right\}$$

Let $$x, y \in C_\rho$$. Then:

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

A similar argument holds for the left cancellable elements.

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 and right cancellable.

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

Thus $$x \circ y$$ is both left 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 $$\left ({S, \circ}\right)$$.