Q:

How does $0$ being an identity element of naturally ordered semigroup follow from its definition?

In the context of naturally ordered semigroups, $0$ is defined as the semigroup's smallest element (in other words, ${\forall}x\,{\in}\,S:0\,{\preceq}\,x$ where $ S$ is our semigroup).
Natural numbers under addition are a naturally ordered semigroup. The source I'm using says that $0$ being the indentity element of $\mathbb{\ N}$ stems from the definition of $0$. How?

A:

Let us agree what a naturally ordered semigroup is first. A naturally ordered semigroup is a triple $(X,+,\leq)$  where $X$ is a set $+$ is a binary operation and $\leq$ is a relation such that:

$(X,+)$ is a commutative semigroup (i.e $+$ is commutative and associative);
$(X,\leq)$ is a preorder (i.e $\leq$ is reflexive, transitive and anti-symmetric);
For all $x,y,z \in X$, we have $x \leq y \Rightarrow x+z \leq y +z$;
For all $(X,\leq)$ is well ordered;
For all $x,y,z\in X$, we have $x+z=y+z \Rightarrow x=y$;
For all $x,y \in X$, we have $x \leq y \Rightarrow $ there exists $w$ such that w+x=y$;

Suppose $X$ is not empty.
Since $(X,\leq)$ is well ordered it has a least element $0$. We will prove that $0$ is the identity element of $(X,+)$. Since $0 \leq 0$ it follows by (6) that there exists $w$ such that $0+w=0$. However $0 \leq w$ and hence (by (3)) $0+0 \leq 0+w=0$. This means that $0+0=0$ (since $0\leq 0+0$ and $\leq$ is anti-symmetric). Next since $0 \leq x$ it follows by (6) that there exists $u$ such that $u+0=x$, and so we have $x +0=(u+0)+0=u+(0+0)=u+0=x$ as required.

