Product is Right Identity Therefore Right Cancellable

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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:

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

The result follows by definition of right cancellable.

$\blacksquare$


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