Strict Ordering Preserved under Product with Cancellable Element

Theorem
Let $\left({S, \circ, \preceq}\right)$ be an ordered semigroup.

If: then:
 * $z$ is cancellable for $\circ$
 * $x \prec y$
 * $x \circ z \prec y \circ z$
 * $z \circ x \prec z \circ y$

If: then $x \prec y$.
 * $z$ is invertible, or if $\preceq$ is a total ordering
 * Either $x \circ z \prec y \circ z$ or $z \circ x \prec z \circ y$

If all the elements of $\left({S, \circ, \preceq}\right)$ are cancellable for $\circ$, and $\preceq$ is a total ordering, then:


 * $\forall x, y, z \in S: x \circ z \preceq y \circ z \iff x \preceq y$

Proof

 * If $z$ is cancellable and $x \prec y$, then by the definition of ordered structure, $x \circ z \preceq y \circ z$.

From the fact that $z$ is cancellable, $x \circ z = y \circ z \iff x = y$.

Thus as $x \circ z \ne y \circ z$ it follows that $x \circ z \prec y \circ z$ from Strictly Precedes.

Similarly, $z \circ x \prec z \circ y$ follows from $z \circ x \preceq z \circ y$.


 * Suppose $x \circ z \prec y \circ z$.

If $z$ is invertible, then because $S$ is a semigroup and therefore $\circ$ is associative, $z$ is cancellable from Invertible also Cancellable.

Thus:


 * $x = \left({x \circ z}\right) \circ z^{-1} \prec \left({y \circ z}\right) \circ z^{-1} = y$

If $\preceq$ is a total ordering, then:


 * $x \succeq y \implies x \circ z \succeq y \circ z$

which contradicts $x \circ z \prec y \circ z$.

So $x \prec y$.

For the final part:
 * $x \circ z \prec y \circ z \implies x \prec y$ from the above;
 * $x \circ z = y \circ z \implies x = y$ from the definition of cancellable.