Product is Left Identity Therefore Left Cancellable

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


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


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

The result follows by definition of left cancellable.


Also see