Idempotent Semigroup/Properties/1

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Property of Idempotent Semigroup

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


Let $x \circ y = y$ and $y \circ x = x$.

Then for all $z \in S$:

$z \circ x \circ z \circ y = z \circ y$

and:

$z \circ y \circ z \circ x = z \circ x$


Proof

From Semigroup Axiom $\text S 1$: Associativity we take it for granted that $\circ$ is associative.


\(\ds \forall z \in S: \, \) \(\ds z \circ x \circ z \circ y\) \(=\) \(\ds z \circ x \circ z \circ x \circ y\) as $x \circ y = y$ by hypothesis
\(\ds \) \(=\) \(\ds z \circ x \circ y\) Definition of Idempotent Operation: $\paren {z \circ x} \circ \paren {z \circ x} = z \circ x$
\(\ds \) \(=\) \(\ds z \circ y\) as $x \circ y = y$ by hypothesis


\(\ds \forall z \in S: \, \) \(\ds z \circ y \circ z \circ x\) \(=\) \(\ds z \circ y \circ z \circ y \circ x\) as $y \circ x = x$ by hypothesis
\(\ds \) \(=\) \(\ds z \circ y \circ x\) Definition of Idempotent Operation: $\paren {z \circ y} \circ \paren {z \circ y} = z \circ y$
\(\ds \) \(=\) \(\ds z \circ x\) as $y \circ x = x$ by hypothesis

$\blacksquare$


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