Associative Commutative Idempotent Operation is Distributive over Itself

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Theorem

Let $\left({S, \circ}\right)$ be an algebraic structure, such that:

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

Then $\circ$ is distributive over itself.


That is:

$\forall a, b, c \in S: \left({a \circ b}\right) \circ \left({a \circ c}\right) = a \circ b \circ c = \left({a \circ c}\right) \circ \left({b \circ c}\right)$


Proof

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

$\Box$

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

$\blacksquare$