Natural Numbers are Non-Negative Integers/Proof 2

Proof
Consider the formal definition of the integers: $x = \eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers.

Let $x \in \Z_{>0}$ be a (strictly) positive integer.

Thus by definition:
 * $x > 0$

This is equivalent to the condition that $a > b$.

Hence:
 * $x = \eqclass {b + u, b} {}$

for some $u \in \N_{>0}$.

It is immediate that:
 * $\forall c \in \N: \eqclass {b + u, b} {} = \eqclass {c + u, c} {}$

Consider the mapping $\phi: \N_{>0} \to \Z_{>0}$ defined as:
 * $\forall u \in \N_{>0}: \map \phi u = u'$

where $u' \in \Z$ be the (strictly) positive integer $\eqclass {b + u, b} {}$.

From the above we have that $\phi$ is well-defined.

Let $\eqclass {b + u, b} {} = \eqclass {c + v, c} {}$.

Then:
 * $b + u + c = b + c + v$

and so:
 * $u = v$

Hence $\phi$ is an injection.

Next note that all $u' \in \Z_{>0}$ can be expressed in the form:
 * $u' = \eqclass {b + u, b} {}$

for arbitrary $b$.

Hence $\phi$ is a surjection.

Hence $\phi$ is an isomorphism.

By defining $\map \phi 0 = \eqclass {b, b} {}$ we see that $\N$ and $\Z_{\ge 0}$ are isomorphic.