Ordering on Natural Numbers Compatible with Addition

Theorem
Let $\N_{> 0}$ be the 1-based natural numbers:
 * $\N_{> 0} = \left\{{1, 2, 3, \ldots}\right\}$

Let $<$ be the (strict) ordering on $\N_{> 0}$ defined as Ordering on Natural Numbers:


 * $\forall a, b \in \N_{>0}: a < b \iff \exists c \in \N_{>0}: a + c = b$

Then:
 * $\forall a, b, n \in \N_{>0}: a < b \implies a + n < b + n$

That is, $>$ is compatible with $+$ on $\N_{>0}$.