# Ordering in terms of Addition

## Theorem

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

Then $\forall m, n \in S$:

$m \preceq n \iff \exists p \in S: m \circ p = n$

## Proof

### Necessary Condition

From Naturally Ordered Semigroup Axiom $\text {NO} 3$: Existence of Product, we have:

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

$\Box$

### Sufficient Condition

Suppose that $m \circ p = n$.

 $\ds 0$ $\preceq$ $\ds p$ Definition of Zero of Naturally Ordered Semigroup $\ds \leadsto \ \$ $\ds m \circ 0$ $\preceq$ $\ds m \circ p$ $\preceq$ is compatible with $\circ$ $\ds \leadsto \ \$ $\ds m$ $\preceq$ $\ds n$ Zero is Identity in Naturally Ordered Semigroup

$\Box$

So $\forall m, n \in S$:

$m \preceq n \iff \exists p \in S: m \circ p = n$

$\blacksquare$