Antiassociative Operation is not Commutative

Theorem
Let $\left({S, \circ}\right)$ be an algebraic structure and $ \circ$ be antiassociative on $S$.

Then $ \circ$ is not commutative on $S$.

Proof
We will show there are two elements in $S$ that do not commute.

Let $a \in S$

From Antiassociative implies No Idempotent Elements, $a \circ a \ne a$

So for some $b \in S$:


 * $a \circ a = b$

Then:


 * $\left({a \circ a}\right) \circ a = b \circ a$

and


 * $a \circ \left({a \circ a}\right) = a \circ b$

From our assumption, $\circ$ is anti-associative so:


 * $\left({a \circ a}\right) \circ a \ne a \circ \left({a \circ a}\right)$

Hence for some $a, b \in S$:


 * $a \circ b \ne b \circ a$