Definition:Cancellable Element

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

An element $$x \in \left ({S, \circ}\right)$$ is left cancellable iff:

$$\forall a, b \in S: x \circ a = x \circ b \Longrightarrow a = b$$

An element $$x \in \left ({S, \circ}\right)$$ is right cancellable iff:

$$\forall a, b \in S: a \circ x = b \circ x \Longrightarrow a = b$$

An element $$x \in \left ({S, \circ}\right)$$ is cancellable iff:


 * $$\forall a, b \in S: x \circ a = x \circ b \Longrightarrow a = b$$
 * $$\forall a, b \in S: a \circ x = b \circ x \Longrightarrow a = b$$

... that is, it is both left and right cancellable.

Some authors use "regular" to mean "cancellable", but this can be ambiguous so its use is not generally endorsed.

Obvious results
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 $$S$$.

$$\forall a, b \in S: x \circ a = x \circ b \Longrightarrow a = b$$.

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

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 $$T$$.