Naturally Ordered Semigroup forms Peano Structure

Theorem
Let $\left({S, \circ, \preceq}\right)$ be a naturally ordered semigroup.

Let $0 \in S$ be the zero of $S$.

Let $1 \in S$ be the one of $S$.

Let $s: S \to S$ be the mapping defined as:


 * $s \left({n}\right) := n \circ 1$

Then $\left({S, 0, s}\right)$ is a Peano structure.

Proof
We verify Peano's axioms in turn.

First, suppose that $s \left({m}\right) = s \left({n}\right)$ for some $m, n \in S$.

That is:


 * $m \circ 1 = n \circ 1$

By Axiom $(NO2)$, it follows that $m = n$.

Hence Axiom $(P3)$ holds.

Next, suppose that $s \left({n}\right) = 0$ for some $n \in S$.

That is:


 * $n \circ 1 = 0$

By Sum with One is Immediate Successor in Naturally Ordered Semigroup, this would imply:


 * $n \prec 0$

However, this contradicts the definition of $0$.

Consequently, Axiom $(P4)$ is satisfied.

Finally, let $A \subseteq S$ be such that $0 \in A$ and, for all $n \in S$:


 * $n \in A \implies s \left({n}\right) \in A$

Define $T := S \setminus A$.

Suppose that $T \ne \varnothing$.

Then by Axiom $(NO3)$, $T$ has a minimal element $n$.

By assumption, $0 \in A$, so that $n \ne 0$.

Therefore, by definition of $1$:


 * $1 \preceq n$

By Ordering in terms of Addition, there exists a $p \in S$ such that:


 * $p \circ 1 = n$

since $\circ$ is commutative.

By Sum with One is Immediate Successor in Naturally Ordered Semigroup:


 * $p \prec n$

Since $n$ is the minimal element of $T$, it follows that $p \notin T$.

Therefore, by definition, $p \in A$.

But by assumption on $A$, this means that $n = p \circ 1 \in A$.

This is a contradiction, hence $T = \varnothing$.

That is to say, $A = S$, and Axiom $(P5)$ holds as well.

Having verified all the axioms, we conclude that $\left({S, 0, s}\right)$ is a Peano structure.