Associative Idempotent Anticommutative

Theorem
Let $\circ$ be a binary operation on a non-empty 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.

Since $S$ is non-empty, let $a\in S$ be arbitrary.

By definition, $\circ$ is anticommutative implies that:


 * $\forall a \in S: a \circ a = a \circ a \iff a = a$

By $\circ$ is a well-defined operation, $a \circ a$ is well-defined.

It remains to be shown that $a \circ a \in S$.

Let $a, w\in S$ be arbitrary. Since $\circ$ is anticommutative, then: $\paren {a \circ a} \circ w = w \circ \paren {a \circ a} \iff a \circ a = w$

Thus $a \circ a \in S$. That is:
 * $\forall \paren {a \circ a}, w \in S: \paren {a \circ a} \circ w = w \circ \paren {a \circ a} \iff a \circ a = w$

Let $w = a \in S$. Then:


 * $\forall \paren {a \circ a}, a \in S: \paren {a \circ a} \circ a = a \circ \paren {a \circ a} \iff a \circ a = a$

Hence:
 * $\forall a \in S: \paren {a \circ a} \circ a = a \circ \paren {a \circ a} \iff a \circ a = a$

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

Now, it remains to be shown that for any $a, b \in S$, if $\circ$ is associative and anticommutative, then $ a \circ b \circ a = a$.

In particular, it remains to be shown that $a \circ b \circ a \in S$.

By assumption:
 * $\forall a, b \in S: a \circ b = b \circ a \iff a = b$

By $\circ$ is a well-defined operation, $a \circ b$ and $b \circ a$ are well-defined.

Let $a, b, x \in S$, then:

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

This implies that:
 * $\forall \paren {a \circ b}, x \in S: \paren {a \circ b} \circ x = x \circ \paren {a \circ b} \iff a \circ b = x$

Hence there exists $\paren {a \circ b } \in S$ for some $a, b \in S$.

Similarly, let $b, a, y \in S$, then:

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

This implies that:
 * $\forall \paren {b \circ a}, y \in S: \paren {b \circ a} \circ y = y \circ \paren {b \circ a} \iff b \circ a = y$

Hence there exists $\paren {b \circ a} \in S$ for some $a, b \in S$.

Let $a, b, z \in S$, then:

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

Hence there exists $\paren {a \circ b \circ a} \in S$ for some $a, b \in S$.

Hence:

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

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

together imply that $\circ$ is anticommutative.