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 |
$\blacksquare$