# Product is Left Identity Therefore Left Cancellable

## Theorem

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

Let $e_L \in S$ be a left identity of $S$.

Let $a \in S$ such that:

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

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

## Proof

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

Then:

 $\displaystyle a \circ x$ $=$ $\displaystyle a \circ y$ $\displaystyle \implies \ \$ $\displaystyle b \circ \left({a \circ x}\right)$ $=$ $\displaystyle b \circ \left({a \circ y}\right)$ $\displaystyle \implies \ \$ $\displaystyle \left({b \circ a}\right) \circ x$ $=$ $\displaystyle \left({b \circ a}\right) \circ y$ as $\left({S, \circ}\right)$ is a semigroup, $\circ$ is associative $\displaystyle \implies \ \$ $\displaystyle e_L \circ x$ $=$ $\displaystyle e_L \circ y$ by hypothesis $\displaystyle \implies \ \$ $\displaystyle x$ $=$ $\displaystyle y$ Definition of Left Identity

The result follows by definition of left cancellable.

$\blacksquare$