Natural Numbers without 1 fulfil Naturally Ordered Semigroup Axioms 1, 2 and 4

Theorem
Let $S \subseteq \N$ be the subset of the natural numbers defined as:


 * $S = \N \setminus \set 1 = \set {0, 2, 3, 4, \ldots}$

Then the algebraic structure:
 * $\struct {S, +, \le}$

is an ordered semigroup which fulfils the axioms:



but:
 * does not fulfil
 * $\struct {S, +}$ is not isomorphic to $\struct {\N, +}$.

Proof
Recall the axioms:

By the Well-Ordering Principle, $\struct {\N, \le}$ is a well-ordered set.

We have that $S \subseteq \N$.

By definition of well-ordered set, $S$ itself is well-ordered.

Hence holds.

By the construction of the natural numbers, $\struct {\N, +, \le}$ is a naturally ordered semigroup.

Hence holds for $\struct {\N, +, \le}$.

By Cancellable Element is Cancellable in Subset:
 * holds also for $\struct {S, +, \le}$.

We have that:
 * $0 \in M$

and:
 * $2 \in M$

and trivially holds.

We have that:

Consider $2, 3 \in \N$.

We have that:
 * $2 \le 3$

and:
 * $3 = 2 + 1$

But $1 \ne S$.

Hence:
 * $\exists 2, 3 \in S: \nexists p \in S: 2 + p = 3$

That is, $\struct {M, +, \le}$ does not fulfil.

Lack of Isomorphism
It remains to demonstrate that $S$ and $\N$ are not isomorphic.

there exists a (semigroup) isomorphism $\phi$ from $\struct {\N, +}$ to $\struct {S, +}$.

By definition of isomorphism:
 * $\phi$ is a homomorphism
 * $\phi$ is a bijection.

We have that:

That is:
 * $\forall n \in \N: \map \phi n = m n$

But let $p \in S: p = n + 1$.

Then:
 * $\nexists r \in \N: \map \phi r = p$

and so $\phi$ is not a surjection.

Hence, by definition, $\phi$ is not a bijection.

This contradicts our assertion that $\phi$ is an isomorphism.

Hence there can be no such semigroup isomorphism between $\struct {S, +}$ and $\struct {\N, +}$.