# Operation is Right Operation iff Anticommutative with Left Cancellable Element

## Theorem

Let $\struct {S, \circ}$ be a semigroup.

Then:

$\circ$ is the right operation
$\circ$ is anticommutative and has a left cancellable element.

## Proof

### Sufficient Condition

Let $\circ$ be the right operation.

Then from Right Operation is Anticommutative we have that $\circ$ is anticommutative.

Let $x \in S$ be arbitrary.

Let $y, z \in S$ such that:

$x \circ z = x \circ y$

Then:

 $\ds x \circ z$ $=$ $\ds z$ Definition of Right Operation $\ds x \circ y$ $=$ $\ds y$ Definition of Right Operation $\ds \leadsto \ \$ $\ds z$ $=$ $\ds y$

That is, $x$ is a left cancellable element for all $x \in S$.

Thus:

$\circ$ is anticommutative and has a left cancellable element.

$\Box$

### Necessary Condition

Let $\circ$ be anticommutative and have a left cancellable element $z$.

As $\struct {S, \circ}$ is a semigroup it follows from Semigroup Axiom $\text S 1$: Associativity that $\circ$ is associative.

Hence from Associative and Anticommutative:

$\forall x, y, z \in S: z \circ x \circ y = z \circ y$

As $z$ is left cancellable:

$\forall x, y \in S: x \circ y = y$

$\blacksquare$