Ordering on Natural Numbers is Compatible with Addition

Theorem
Let $m, n, k \in \N$ where $\N$ is the set of natural numbers.

Then:
 * $m < n \iff m + k < n + k$

Proof
Proof by induction:

For all $k \in \N$, let $P \left({k}\right)$ be the proposition:
 * $m < n \iff m + k < n + k$

$P \left({0}\right)$ is true, as this just says $m + 0 = m < n = n + 0$ from Definition by Induction of Natural Number Addition‎.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({j}\right)$ is true, where $j \ge 0$, then it logically follows that $P \left({j^+}\right)$ is true.

So this is our induction hypothesis:
 * $m < n \iff m + j < n + j$

Then we need to show:
 * $m < n \iff m + j^+ < n + j^+$

Induction Step
This is our induction step:

Let $m < n$.

Then:

This gives:

So $P \left({j}\right) \implies P \left({j^+}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall m, n, k \in \N: m < n \implies m + k < n + k$