Associative Commutative Idempotent Operation is Self-Distributive

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Theorem

Let $\struct {S, \circ}$ be an algebraic structure, such that:

$(1): \quad \circ$ is associative
$(2): \quad \circ$ is commutative
$(3): \quad \circ$ is idempotent.

Then $\circ$ is self-distributive.


That is:

$\forall a, b, c \in S: \paren {a \circ b} \circ \paren {a \circ c} = a \circ b \circ c = \paren {a \circ c} \circ \paren {b \circ c}$


Proof

\(\ds \paren {a \circ b} \circ \paren {a \circ c}\) \(=\) \(\ds a \circ \paren {b \circ a} \circ c\) $\circ$ is associative
\(\ds \) \(=\) \(\ds a \circ \paren {a \circ b} \circ c\) $\circ$ is commutative
\(\ds \) \(=\) \(\ds \paren {a \circ a} \circ b \circ c\) $\circ$ is associative
\(\ds \) \(=\) \(\ds a \circ b \circ c\) $\circ$ is idempotent

$\Box$


\(\ds \paren {a \circ c} \circ \paren {b \circ c}\) \(=\) \(\ds a \circ \left({c \circ b}\right) \circ c\) $\circ$ is associative
\(\ds \) \(=\) \(\ds a \circ \paren {b \circ c} \circ c\) $\circ$ is commutative
\(\ds \) \(=\) \(\ds a \circ b \circ \paren {c \circ c}\) $\circ$ is associative
\(\ds \) \(=\) \(\ds a \circ b \circ c\) $\circ$ is idempotent

$\blacksquare$