Ordering on Positive Integers is Equivalent to Ordering on Natural Numbers

Theorem

Let $u, v \in \Z_{>0}$ be natural numbers.

Consider the mapping $\phi: \N_{>0} \to \Z_{>0}$ defined as:

$\forall u \in \N_{>0}: \map \phi u = u'$

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

Let $u', v' \in \Z_{>0}$ be strictly positive integers.

Then:

$u > v \iff u' > v'$

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

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

Then:

$\ds u'$

$>$

$\ds v'$

$\ds \leadstoandfrom \ \ $

$\ds \eqclass {b + u, b} {}$

$>$

$\ds \eqclass {c + v, c} {}$

$\ds \leadstoandfrom \ \ $

$\ds b + u + c$

$>$

$\ds c + v + b$

$\ds \leadstoandfrom \ \ $

$\ds u$

$>$

$\ds v$

$\blacksquare$