Ordering on Natural Numbers Compatible with Addition

Theorem
Let $\N_{> 0}$ be the 1-based natural numbers.

Let $+$ denote addition on $\N_{>0}$.

Let $<$ be the strict ordering on $\N_{>0}$.

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

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