Cancellable Element is Cancellable in Subset

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

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

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 $T$.