Ordering of Naturally Ordered Semigroup is Strongly Compatible

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

Then $\preceq$ is strongly compatible with $\circ$:
 * $\forall m, n, p \in S: m \preceq n \iff m \circ p \preceq n \circ p$

Proof
The forward implication is immediate from $\preceq$ being compatible with $\circ$:


 * $\forall m, n, p \in S: m \preceq n \implies m \circ p \preceq n \circ p$

Conversely, suppose that $m \circ p \preceq n \circ p$.

Suppose that $n \prec m$.

Then as $\preceq$ is compatible with $\circ$:


 * $n \circ p \preceq m \circ p$

Since $\preceq$ is an ordering, this implies:


 * $n \circ p = m \circ p$

By axiom $(\text {NO} 2)$, it follows that:


 * $n = m$

contradicting our assumption that $n \prec m$.

Hence, since $\preceq$ is a total ordering:


 * $m \preceq n$

as desired.

Also see

 * Strict Ordering of Naturally Ordered Semigroup is Strongly Compatible