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 $$T$$.