Self-Distributive Quasigroup is Idempotent

Theorem
Let $\struct {S, \odot}$ be a self-distributive quasigroup.

Then $\odot$ is an idempotent operation.

Proof
Because $S$ is a quasigroup:
 * $\forall a, b \in S: \exists ! x \in S: x \odot a = b$
 * $\forall a, b \in S: \exists ! y \in S: a \odot y = b$

In particular:


 * $\exists ! x \in S: x \circ \paren {a \odot b} = y$

for arbitrary $a, b \in S$.

Let $y = a \odot \paren {a \odot b}$.

Because $\struct {S, \odot}$ is self-distributive:
 * $y = \paren {a \odot a} \odot \paren {a \odot b}$

Because $S$ is a quasigroup, we have that:
 * $a = a \odot a$

As $a$ is arbitrary, the result follows.