Strict Ordering Preserved under Cancellability in Totally Ordered Semigroup

Theorem

Let $\struct {S, \circ, \preceq}$ be a totally ordered semigroup.

If either:

$x \circ z \prec y \circ z$

or

$z \circ x \prec z \circ y$

then $x \prec y$.

Proof

Let $x \circ z \prec y \circ z$.

Aiming for a contradiction, suppose $x \succeq y$.

As $\struct {S, \circ, \preceq}$ is an ordered semigroup, $\preceq$ is compatible with $\circ$.

Hence we have:

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

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

We have that $\preceq$ is a total ordering, and that it is not the case that $x \succeq y$.

Hence by the Trichotomy Law:

$x \prec y$

Similarly for $z \circ x \prec z \circ y$.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Theorem $15.1: \ 2^\circ$