Cancellable Element is Cancellable in Subset

Theorem
Let $\left ({S, \circ}\right)$ be an algebraic structure.

Let $\left ({T, \circ}\right) \subseteq \left ({S, \circ}\right)$.

If $x \in T$ such that $x$ is left or right cancellable in $S$, then it is also left or right cancellable in $T$.

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

That is, $\forall a, b \in S: x \circ a = x \circ b \implies a = b$.

Therefore, $\forall c, d \in T: x \circ c = x \circ d \implies c = d$.

Thus $x$ is left cancellable in $T$.

The same argument applies to $x \in T$ being right cancellable.

It follows that if $x$ is both right and left cancellable in $S$, it is also both right and left cancellable, i.e. cancellable, in $T$.