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.