Natural Numbers with Extension fulfil Naturally Ordered Semigroup Axioms 1, 3 and 4/Lemma 2

Construction
There exists a unique total ordering $\preccurlyeq$ on $M$ such that:
 * the restriction of $\preccurlyeq$ to $\N$ is the given total ordering $\le$ on $\N$
 * $0 \prec \beta \prec 1$

This total ordering we will rename $\le$ to overload the notation for $\N$.

Proof
We define $\preccurlyeq$ on $M$ as follows:


 * $a \preccurlyeq b \iff \begin {cases} a, b \in \N: a \le b \\ a = 0, b = \beta \\ a = \beta, b \in \N_{>0} \\ a = \beta, b = \beta \end {cases}$

Existence
The restriction of $\preccurlyeq$ to $\N$ is seen to be the total ordering $\le$ on $\N$.

We have that:
 * $0 \preccurlyeq \beta$ and $0 \ne \beta$

from which:
 * $0 \prec \beta$

and:
 * $\beta \preccurlyeq 1$ and $\beta \ne 1$

from which:
 * $\beta \prec 1$

and:
 * $\forall a \in \N_{>0}: \beta \preccurlyeq a$

Finally:
 * $\beta = \beta$

and so:
 * $\beta \preccurlyeq \beta$

This demonstrates that $\preccurlyeq$ is total.

Hence existence of the total ordering in question has been proven.

Uniqueness
Let $\preccurlyeq'$ be another total ordering on $M$ such that:
 * the restriction of $\preccurlyeq'$ to $\N$ is the given total ordering $\le$ on $\N$
 * $0 \prec' \beta \prec' 1$

We have by definition of $\le$ on $\N$ that:
 * $\forall a, b \in \N: a \le b \implies a \preccurlyeq' b$

From $0 \prec' \beta \prec' 1$ we have:
 * $0 \preccurlyeq' \beta$
 * $\beta \preccurlyeq' 1$

Because $\beta \preccurlyeq' 1$ and an ordering is transitive:
 * $\forall a \in \N: 1 \le a \implies \beta \preccurlyeq' a$

Then as an ordering is reflexive:
 * $\beta \preccurlyeq' \beta$

Hence $\preccurlyeq'$ is defined as:


 * $a \preccurlyeq' b \iff \begin {cases} a, b \in \N: a \le b \\ a = 0, b = \beta \\ a = \beta, b \in \N_{>0} \\ a = \beta, b = \beta \end {cases}$

which is seen to be exactly the same as $\preccurlyeq$.

Hence the result.