Element Commutes with Square in Semigroup

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Theorem

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

Let $x \in S$.


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


Proof

\(\displaystyle x \circ \left({x \circ x}\right)\) \(=\) \(\displaystyle \left({x \circ x}\right) \circ x\) by definition of semigroup: $\circ$ is associative

$\blacksquare$