# Product is Right Identity Therefore Right Cancellable

## Theorem

Let $\left({S, \circ}\right)$ be a semigroup.

Let $e_R \in S$ be a right identity of $S$.

Let $a \in S$ such that:

$\exists b \in S: a \circ b = e_R$

Then $a$ is right cancellable in $\left({S, \circ}\right)$.

## Proof

Let $x, y \in S$ be arbitrary.

Then:

 $\ds x \circ a$ $=$ $\ds y \circ a$ $\ds \implies \ \$ $\ds \left({x \circ a}\right) \circ b$ $=$ $\ds \left({y \circ a}\right) \circ b$ $\ds \implies \ \$ $\ds \left({x \circ a}\right) \circ b$ $=$ $\ds \left({y \circ a}\right) \circ b$ as $\left({S, \circ}\right)$ is a semigroup, $\circ$ is associative $\ds \implies \ \$ $\ds x \circ e_R$ $=$ $\ds y \circ e_R$ By hypothesis $\ds \implies \ \$ $\ds x$ $=$ $\ds y$ Definition of Right Identity

The result follows by definition of right cancellable.

$\blacksquare$