Natural Numbers are Non-Negative Integers/Proof 1

Proof
Let $m \in \N$.

Then by definition of $0$:
 * $0 \le m$

Conversely, let $m \in \Z: 0 \le m$.

Then:
 * $\exists x, y \in \N: m = x - y$

Thus:
 * $y \le m + y = x$

By Naturally Ordered Semigroup: NO 3:
 * $\exists z \in \N: z + y = x = m + y$

From Naturally Ordered Semigroup: NO 2, $y$ is cancellable.

Hence: $m = z \in \N_{> 0}$

Thus $(1)$ holds.

$(2)$ follows from $(1)$.

We infer from $(1)$ that:
 * $m \notin \N \iff m < 0$

We infer from $(2)$ that:
 * $-m > 0 \iff -m \in \N_{> 0}$

But by Ordering of Inverses in Ordered Monoid:
 * $m < 0 \iff -m > 0$

Therefore $(3)$ also holds.