Cancellability by Cayley Table

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

Let $\mathcal T$ be the Cayley table for $\left({S, \circ}\right)$.

Let $a \in S$ be an element of $S$.

Then $a$ is cancellable for $\circ$ :
 * $(1): \quad$ no element of $S$ is repeated in $\mathcal T$ in the row headed by $a$

and:
 * $(2): \quad$ no element of $S$ is repeated in $\mathcal T$ in the column headed by $a$.

Necessary Condition
Let $a \in S$ be cancellable for $\circ$.

Suppose there exists $x \in S$ which appears twice in a row in $\mathcal T$ headed by $a$.

Thus by definition of the structure of a Cayley table:
 * $\exists y_1, y_2 \in S: a \circ y_1 = x = a \circ y_2$

such that $y_1 \ne y_2$.

That contradicts the stipulation that $a$ is cancellable for $\circ$.

So no element of $S$ is repeated in $\mathcal T$ in the row headed by $a$.

Similarly, suppose there exists $x \in S$ which appears twice in a column in $\mathcal T$ headed by $a$.

Thus by definition of the structure of a Cayley table:
 * $\exists y_1, y_2 \in S: y_1 \circ a = x = y_2 \circ a$

such that $y_1 \ne y_2$.

That contradicts the stipulation that $a$ is cancellable for $\circ$.

So no element of $S$ is repeated in $\mathcal T$ in the column headed by $a$.

Sufficient Condition
Let $a$ be such that:
 * $(1): \quad$ no element of $S$ is repeated in $\mathcal T$ in the row headed by $a$

and:
 * $(2): \quad$ no element of $S$ is repeated in $\mathcal T$ in the column headed by $a$.

Thus from $(1)$:
 * $\forall y_1, y_2 \in S: y_1 \ne y_2 \implies a \circ y_1 \ne a \circ y_2$

So by the Rule of Transposition:
 * $\forall y_1, y_2 \in S: a \circ y_1 = a \circ y_2 \implies y_1 = y_2$

Also from $(2)$:
 * $\forall y_1, y_2 \in S: y_1 \ne y_2 \implies y_1 \circ a \ne y_2 \circ a$

So by the Rule of Transposition:
 * $\forall y_1, y_2 \in S: y_1 \circ a = y_2 \circ a \implies y_1 = y_2$

Thus it follows that $a$ is cancellable for $\circ$.