# Associative Commutative Idempotent Operation is Distributive over Itself

## 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

$\blacksquare$