Definition:Zero (Number)

Theorem
Let $$\left({S, \circ; \preceq}\right)$$ be a naturally ordered semigroup.

Then $$\left({S, \circ; \preceq}\right)$$ has a minimal element.

This minimal element of $$\left({S, \circ; \preceq}\right)$$ is called zero and has the symbol $$0$$.

That is: $$\forall n \in S: 0 \preceq n$$.

This element $$0$$ is the identity for $$\circ$$. That is:

$$\forall n \in S: n \circ 0 = n = 0 \circ n$$

Proof
This follows immediately from naturally ordered semigroup: NO 1.

$$\left({S, \circ; \preceq}\right)$$ is well-ordered, so has a minimal element.

Now by the definition of the zero: $$0 \preceq 0$$ (as zero precedes everything).

Thus from Naturally Ordered Semigroup: NO 3, $$0 \preceq 0 \Longrightarrow \exists p \in S: 0 \circ p = 0$$

By the definition of the zero: $$0 \preceq 0 \circ 0$$ and $$0 \preceq p$$ (as zero precedes everything).

Thus from naturally ordered semigroup: NO 2: $$0 \circ 0 \preceq 0 \circ p = 0$$.

Thus $$0 \circ 0 \preceq 0 \land 0 \preceq 0 \circ 0$$, and by the antisymmetry of ordering, it follows that $$0 \circ 0 = 0$$.

Because $$\left({S, \circ; \preceq}\right)$$ is a semigroup, $$\circ$$ is associative.

So $$\forall n \in S: \left({n \circ 0}\right) \circ 0 = n \circ \left({0 \circ 0}\right) = n \circ 0$$.

Thus from naturally ordered semigroup: NO 2: $$\forall n \in S: n \circ 0 = n$$.

Finally, $$\forall n \in S: 0 \circ n = n$$ follows because a naturally ordered semigroup is commutative.

Thus $$\forall n \in S: n \circ 0 = n = 0 \circ n$$ and $$0$$ is the identity for $$\circ$$.