Ordering on Natural Numbers Compatible with Addition

Theorem
Let $\N$ be the natural numbers.

Let $+$ denote addition on $\N$.

Let $<$ be the strict ordering on $\N$.

Then:
 * $\forall a, b, n \in \N: a < b \implies a + n < b + n$

That is, $<$ is compatible with $+$ on $\N$.