# Operation is Left Operation iff Anticommutative with Right Cancellable Element

## Theorem

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

Then:

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

## Proof

### Sufficient Condition

Let $\circ$ be the left operation.

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

Let $x \in S$ be arbitrary.

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

$z \circ x = y \circ x$

Then:

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

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

Thus:

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

$\Box$

### Necessary Condition

Let $\circ$ be anticommutative and have a right 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:

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

As $z$ is right cancellable:

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

$\blacksquare$