# Cancellable Element is Cancellable in Subset

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## Theorem

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

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

Let $x \in T$ be cancellable in $S$.

Then $x$ is also cancellable in $T$.

## Proof

Let $x \in T$ be cancellable in $S$.

Then by definition, $x \in T$ is left cancellable in $S$.

It follows from Left Cancellable Element is Left Cancellable in Subset that $x \in T$ is left cancellable in $T$.

Again by definition, $x \in T$ is right cancellable in $S$.

It follows from Right Cancellable Element is Right Cancellable in Subset that $x \in T$ is right cancellable in $T$.

Thus $x$ is also both left cancellable and right cancellable in $T$.

So by definition, $x$ is cancellable in $T$.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets