Operations with Identities which Distribute over each other are Idempotent

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Theorem

Let $S$ be a set.

Let $\odot$ and $\otimes$ be operations on $S$ which have identity elements $e$ and $u$ respectively.

Let $\odot$ and $\otimes$ be distributive over each other.


Then $\odot$ and $\otimes$ are both idempotent.


Proof

\(\ds e \otimes \paren {u \odot e}\) \(=\) \(\ds \paren {e \otimes u} \odot \paren {e \otimes e}\) as $\otimes$ is distributive over $\odot$
\(\ds \leadsto \ \ \) \(\ds e \otimes u\) \(=\) \(\ds \paren {e \otimes u} \odot \paren {e \otimes e}\) as $e$ is the identity of $\odot$
\(\ds \leadsto \ \ \) \(\ds e\) \(=\) \(\ds e \odot \paren {e \otimes e}\) as $u$ is the identity of $\otimes$
\(\ds \leadsto \ \ \) \(\ds e\) \(=\) \(\ds e \otimes e\) as $e$ is the identity of $\odot$


In the same way:

\(\ds u \odot \paren {e \otimes u}\) \(=\) \(\ds \paren {u \odot e} \otimes \paren {u \odot u}\) as $\odot$ is distributive over $\otimes$
\(\ds \leadsto \ \ \) \(\ds u \odot e\) \(=\) \(\ds \paren {u \odot e} \otimes \paren {u \odot u}\) as $u$ is the identity of $\otimes$
\(\ds \leadsto \ \ \) \(\ds u\) \(=\) \(\ds u \otimes \paren {u \odot u}\) as $e$ is the identity of $\odot$
\(\ds \leadsto \ \ \) \(\ds u\) \(=\) \(\ds u \odot u\) as $u$ is the identity of $\otimes$


Then we have:

\(\ds \forall a \in S: \, \) \(\ds a \otimes a\) \(=\) \(\ds \paren {a \odot e} \otimes \paren {a \odot e}\) as $e$ is the identity of $\odot$
\(\ds \) \(=\) \(\ds a \odot \paren {e \otimes e}\) as $\odot$ is distributive over $\otimes$
\(\ds \) \(=\) \(\ds a \odot e\) a priori
\(\ds \) \(=\) \(\ds a\) as $e$ is the identity of $\odot$


and:

\(\ds \forall a \in S: \, \) \(\ds a \odot a\) \(=\) \(\ds \paren {a \otimes u} \odot \paren {a \otimes u}\) as $u$ is the identity of $\otimes$
\(\ds \) \(=\) \(\ds a \otimes \paren {u \odot u}\) as $\otimes$ is distributive over $\odot$
\(\ds \) \(=\) \(\ds a \otimes u\) a priori
\(\ds \) \(=\) \(\ds a\) as $u$ is the identity of $\otimes$

Hence the result by definition of idempotent operation.

$\blacksquare$


Sources

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