Element Commutes with Square in Semigroup

Theorem

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

Let $x \in S$.

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

Proof

Semigroup Axiom $\text S 0$: Closure is taken for granted.

\(\ds \forall x \in S: \, \)

\(\ds x \circ \paren {x \circ x}\)

\(=\)

\(\ds \paren {x \circ x}\circ x\)

Semigroup Axiom $\text S 1$: Associativity

$\blacksquare$