Element Commutes with Square in Semigroup
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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 |
$\blacksquare$