# Associative Idempotent Anticommutative

## Theorem

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

Let $\circ$ be associative.

Then $\circ$ is anticommutative if and only if:

$(1): \quad \circ$ 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:

 $\ds \paren {a \circ a} \circ b \circ a$ $=$ $\ds a \circ b \circ a$ $\circ$ is idempotent, proved above $\ds$ $=$ $\ds a \circ b \circ \paren {a \circ a}$ $\circ$ is idempotent $\ds \leadsto \ \$ $\ds a \circ \paren {a \circ b \circ a}$ $=$ $\ds \paren {a \circ b \circ a} \circ a$ $\circ$ is associative $\ds \leadsto \ \$ $\ds a \circ b \circ a$ $=$ $\ds a$ $\circ$ is anticommutative $\ds \leadsto \ \$ $\ds a \circ b \circ a$ $=$ $\ds a$ by hypothesis

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

$\Box$

### 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:

 $\ds \leadsto \ \$ $\ds a \circ \paren {b \circ a}$ $=$ $\ds a \circ \paren {a \circ b}$ by hypothesis: $a \circ b \circ a = a\in S$ $\ds a \circ b \circ a$ $=$ $\ds \paren {a \circ a} \circ b$ $\circ$ is associative $\ds$ $=$ $\ds a \circ b$ $\circ$ is idempotent $\ds$ $=$ $\ds a$ by hypothesis

Similarly:

 $\ds a \circ b$ $=$ $\ds b \circ a$ $\ds \leadsto \ \$ $\ds \paren {a \circ b} \circ a$ $=$ $\ds \paren {b \circ a} \circ a$ by hypothesis: $a \circ b \circ a \in S$ $\ds a \circ b \circ a$ $=$ $\ds b \circ \paren {a \circ a}$ $\circ$ is associative $\ds$ $=$ $\ds b \circ a$ $\circ$ is idempotent $\ds$ $=$ $\ds a$ by hypothesis

Also:

 $\ds a \circ b$ $=$ $\ds b \circ a$ $\ds \leadsto \ \$ $\ds b \circ \paren {a \circ b}$ $=$ $\ds b \circ \paren {b \circ a}$ by hypothesis: $b \circ a \circ b = b \in S$ $\ds b \circ a \circ b$ $=$ $\ds \paren {b \circ b} \circ a$ $\circ$ is associative $\ds$ $=$ $\ds b \circ a$ $\circ$ is idempotent $\ds$ $=$ $\ds b$ by hypothesis

Then:

 $\ds b \circ a$ $=$ $\ds a \circ b$ $\ds \leadsto \ \$ $\ds \paren {b \circ a} \circ b$ $=$ $\ds \paren {a \circ b} \circ b$ by hypothesis: $b \circ a \circ b = b \in S$ $\ds \leadsto \ \$ $\ds b \circ a \circ b$ $=$ $\ds a \circ \paren {b \circ b}$ $\circ$ is associative $\ds$ $=$ $\ds a \circ b$ $\circ$ is idempotent $\ds$ $=$ $\ds b$ by hypothesis

Hence:

 $\ds a$ $=$ $\ds a \circ b = b$ $\ds a$ $=$ $\ds b \circ a = b$ $\ds \leadsto \ \$ $\ds a$ $=$ $\ds b$

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.

$\blacksquare$