Left Cancellable Commutative Operation is Right Cancellable

Theorem

Let $\struct {S, \circ}$ be an algebraic structure.

Let $\circ$ be left cancellable and also commutative.

Then $\circ$ is also right cancellable.

Proof

Let $\circ$ be both left cancellable and commutative on a set $S$.

Then:

\(\ds a \circ c\)

\(=\)

\(\ds b \circ c\)

\(\ds \leadsto \ \ \)

\(\ds c \circ a\)

\(=\)

\(\ds c \circ b\)

$\circ$ is Commutative

\(\ds \leadsto \ \ \)

\(\ds a\)

\(=\)

\(\ds b\)

$\circ$ is Left Cancellable

$\blacksquare$

Also see

• Right Cancellable Commutative Operation is Left Cancellable