Product is Right Identity Therefore Right Cancellable

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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.


\(\ds x \circ a\) \(=\) \(\ds y \circ a\) Semigroup Axiom $\text S 0$: Closure
\(\ds \leadsto \ \ \) \(\ds \paren {x \circ a} \circ b\) \(=\) \(\ds \paren {y \circ a} \circ b\) Semigroup Axiom $\text S 0$: Closure
\(\ds \leadsto \ \ \) \(\ds x \circ \paren {a \circ b}\) \(=\) \(\ds y \circ \paren {a \circ b}\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \leadsto \ \ \) \(\ds x \circ e_R\) \(=\) \(\ds y \circ e_R\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds y\) Definition of Right Identity

The result follows by definition of right cancellable.


Also see