Naturally Ordered Semigroup is Unique

Theorem
Let $\left({S, \circ, \preceq}\right)$ and $\left({S', \circ', \preceq'}\right)$ be naturally ordered semigroups.

Let:
 * $0'$ be the smallest element of $S'$
 * $1'$ be the smallest element of $S' \setminus \left\{{0'}\right\} = S'^*$.

Then the mapping $g: S \to S'$ defined as:
 * $\forall a \in S: g \left({a}\right) = \circ'^a 1'$

is an isomorphism from $\left({S, \circ, \preceq}\right)$ to $\left({S', \circ', \preceq'}\right)$.

This isomorphism is unique.

Thus, up to isomorphism, there is only one naturally ordered semigroup.

Proof that Mapping is Isomorphism
Let $T' = \operatorname{Cdm} \left({g}\right)$, that is, the codomain of $g$.

By Zero is Identity in Naturally Ordered Semigroup, $0'$ is the identity for $\circ'$.

Thus:
 * $g \left({0}\right) = \circ'^0 1' = 0'$

and so $0' \in T'$

Suppose $x' \in T'$.

Then:
 * $g \left({n}\right) = x'$

and so:

So:
 * $x' \in T' \implies x' \circ' 1' \in T'$

Thus, by the Principle of Finite Induction applied to $S'$:
 * $T' = S'$

So:
 * $\forall a' \in S': \exists a \in S: g \left({a}\right) = a'$

and so $g$ is surjective by definition.

From Index Laws for Semigroup: Sum of Indices:
 * $g \left({a \circ b}\right) = g \left({a}\right) \circ' g \left({b}\right)$

and therefore $g$ is a homomorphism from $\left({S, \circ}\right)$ to $\left({S', \circ'}\right)$.

Now:

For $p \in S$, let $S_p$ be the initial segment of $S$:


 * $S_p = \left\{{x \in S: x \prec p}\right\}$

Let:
 * $T = \left\{{p \in S: \forall a \in S_p: \circ'^a 1' \prec' \circ'^p 1'}\right\}$

Now $S_0 = \varnothing \implies 0 \in T$.

Suppose $p \in T$.

Then:
 * $a \prec p \circ 1 \implies a \preceq p$

By Initial Segment of Sum with One, either of these is the case:


 * $(1): \quad a \prec p: p \in T \implies \circ'^a 1' \prec' \circ'^p 1' \prec' \circ'^{\left({p \circ 1}\right)} 1'$


 * $(2): \quad a = p: \circ'^a 1' = \circ'^p 1' \prec' \circ'^{\left({p \circ 1}\right)} 1'$

In either case, we have:
 * $p \in T \implies p \circ 1 \in T$, and by thePrinciple of Finite Induction:
 * $T = S$

So $n \prec p \implies \circ'^n 1' \prec' \circ'^p 1'$.

Thus $g$ is a surjective monomorphism and therefore is an isomorphism from $\left({S, \circ, \preceq}\right)$ to $\left({S', \circ', \preceq'}\right)$.

Proof that Isomorphism is Unique
Now we need to show that the isomorphism $g$ is unique.

Let $f: S \to S'$ be another isomorphism different from $g$.

Suppose $f \left({1}\right) \ne 1'$.

We show by induction that $1' \notin \operatorname{Cdm} \left({f}\right)$.

... Thus $1' \notin \operatorname{Cdm} \left({f}\right)$ which is a contradiction.

Thus $f \left({1}\right) = 1$ and it follows

that $f = g$.

Thus the isomorphism $g$ is unique.