Associative Idempotent Anticommutative

Theorem
Let $\circ$ be a binary operation on a set $S$.

Let $\circ$ be associative.

Then $\circ$ is anticommutative :
 * $(1): \quad \circ$ is idempotent

and:
 * $(2): \quad \forall a, b \in S: a \circ b \circ a = a$

Proof
Let $\circ$ be an associative operation on $S$.

Necessary Condition
Suppose $\circ$ is anticommutative.

Let $a \circ a = x$ for some $a, x \in S$. Then:

So $\circ$ being associative and anticommutative implies that $\circ$ is idempotent.

Now, let $a \circ b \circ a = x$ for some $a, b, x \in S$.

So $\circ$ being associative and anticommutative implies, via the fact (also proved) that $\circ$ is idempotent, that $a \circ b \circ a = a$.

Sufficient Condition
Now, suppose $\circ$ (which we take to be associative) is idempotent and:
 * $\forall a, b \in S: a \circ b \circ a = a$

It remains to be shown that $\circ$ is anticommutative.

Suppose $a \circ b = b \circ a$.

Then:

Similarly:

Also:

Then:

Hence:

So:
 * $\circ$ being idempotent
 * $a \circ b \circ a = a$

together imply that $\circ$ is anticommutative.