Self-Distributive Quasigroup with at least Two Elements has no Identity

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

Let $S$ have at least $2$ elements.

Then $\struct {S, \odot}$ has no identity element.

Proof
$S$ has an identity element $e$ and another element $a$ such that $a \ne e$.

Recall the definition of quasigroup:


 * $\forall a, b \in S: \exists ! x \in S: x \circ a = b$

That is:

We have:

That is, there are two $x \in S$ such that $x \circ a = a$.

This contradicts our assertion that $\struct {S, \odot}$ is a quasigroup.

Hence there can be no such element.