Product is Right Identity Therefore Right Cancellable

Theorem

Let $\struct {S, \circ}$ 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 $\struct {S, \circ}$.

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

Then:

$\ds x \circ a$

$=$

$\ds y \circ a$

$\ds \leadsto \ \ $

$\ds \paren {x \circ a} \circ b$

$=$

$\ds \paren {y \circ a} \circ b$

$\ds \leadsto \ \ $

$\ds x \circ \paren {a \circ b}$

$=$

$\ds y \circ \paren {a \circ b}$

$\ds \leadsto \ \ $

$\ds x \circ e_R$

$=$

$\ds y \circ e_R$

$\ds \leadsto \ \ $

$\ds x$

$=$

$\ds y$

Definition of Right Identity

The result follows by definition of right cancellable.

$\blacksquare$