Element Commutes with Square in Semigroup

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

Let $x \in S$.

Then $x$ commutes with $x \circ x$.

Proof
Since $S$ is non-empty, let $x \in S$. Then by, $x \circ x \in S$.

By, $\forall x, x \circ x \in S \implies x \circ \left({x \circ x}\right) \in S$

By, $\forall x \circ x, x \in S \implies \left({x \circ x}\right) \circ x \in S$

Hence,