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 minimal element of $$S'$$;
 * $$1'$$ be the minimal element of $$S' - \left\{{0'}\right\} = S'^*$$.

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
Let $$T' = \mathrm{Rng} \left({g}\right)$$.

By the definition of Zero, $$0'$$ is the identity for $$\circ'$$.

Thus $$g \left({0}\right) = \circ'^0 1' = 0' \Longrightarrow 0' \in T'$$.

Now $$x' \in T' \Longrightarrow g \left({n}\right) = x'$$. Then:

So $$x' \in T' \Longrightarrow x' \circ' 1' \in T'$$.

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

So, $$\forall a' \in S': \exists a \in S: g \left({a}\right) = a'$$, thus $$g$$ is surjective.

We also have from Recursive Mapping to Semigroup that $$g \left({a \circ b}\right) = \left({a}\right) \circ' \left({b}\right)$$ and therefore $$g$$ is a homomorphism from $$\left({S, \circ}\right)$$ to $$\left({S', \circ'}\right)$$.

Now:

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

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

Suppose $$p \in T$$.

Then $$a \prec p \circ 1 \Longrightarrow a \preceq p$$.

Either of these is the case:


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

In either case, we have $$p \in T \Longrightarrow p \circ 1 \in T$$, and by induction $$T = S$$.

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

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


 * 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 $$1' \notin \mathrm{Rng} \left({f}\right)$$.

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

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

that $$f = g$$.

Thus the isomorphism $$g$$ is unique.