Zero is Identity in Naturally Ordered Semigroup

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

Let $0$ be the zero of $\left({S, \circ, \preceq}\right)$.

Then $0$ is the identity for $\circ$.

That is:


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

Proof
By definition of an ordering:


 * $0 \preceq 0$

Thus from axiom $(NO 3)$:


 * $\exists p \in S: 0 \circ p = 0$

By the definition of zero:


 * $0 \preceq 0 \circ 0$ and $0 \preceq p$

Thus since $\preceq$ is compatible with $\circ$:


 * $0 \circ 0 \preceq 0 \circ p = 0$

Thus:


 * $0 \circ 0 \preceq 0$ and $0 \preceq 0 \circ 0$

Hence, as $\preceq$ is antisymmetric, 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 axiom $(NO 2)$:


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

Similarly:


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

meaning:


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

Thus:


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

and so $0$ is the identity for $\circ$.