Antiassociative Operation is not Commutative

Theorem

Let $\struct {S, \circ}$ be an algebraic structure.

Let $\circ$ 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 Operation has no Idempotent Elements:

$a \circ a \ne a$

So for some $b \in S$:

$a \circ a = b$

Then:

$\paren {a \circ a} \circ a = b \circ a$

and:

$a \circ \paren {a \circ a} = a \circ b$

From our assumption, $\circ$ is antiassociative.

So:

$\paren {a \circ a} \circ a \ne a \circ \paren {a \circ a}$

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

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

$\blacksquare$