Natural Number Addition is Cancellable for Ordering

Theorem
Let $\N$ be the natural numbers.

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

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

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

That is, $+$ is cancellable on $\N$ for $<$.