Naturally Ordered Semigroup Axioms imply Commutativity

Theorem
Consider the naturally ordered semigroup axioms:

Axioms $\text {NO} 1$, $\text {NO} 2$ and $\text {NO} 3$ together imply the commutativity of the naturally ordered semigroup $\struct {S, \circ, \preceq}$.

Proof
From, $\struct {S, \circ, \preceq}$ has a smallest element.

This is identified as zero: $0$.

From Zero is Identity in Naturally Ordered Semigroup, $0$ is the identity element of $\struct {S, \circ, \preceq}$,

It may be the case that $S$ is a singleton such that $S = \set 0$.

Then $\struct {S, \circ}$ degenerates to the trivial group.

From Trivial Group is Abelian, it follows that $\circ$ is commutative.

Let $S^* := S \setminus \set 0$ denote the complement of $\set 0$ in $S$.

From, $S^*$ also has a smallest element.

This we will call $1$.

It will be shown that $1$ commutes with every element of $S$.

Let $T \subseteq S$ be the set of element of $S$ which commute with $1$.

We have that $0$ is the identity element of $\struct {S, \circ, \preceq}$.

Hence:
 * $1 \circ 0 = 1 = 0 \circ 1$

and it is seen that $0 \in T$.

Now suppose $n \in T$.

That is:
 * $1 \circ n = n \circ 1$

We have:

So
 * $n \in T \implies n \circ 1 \in T$.

It follows from Principle of Mathematical Induction for Naturally Ordered Semigroup that:
 * $T = S$

That is, all the element of $S$ commute with $1$.

Hence the result.