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 $\map P k$ be the proposition:
 * $m < n \iff m + k < n + k$

$\map P 0$ is true, as this just says $m + 0 = m < n = n + 0$.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P j$ is true, where $j \ge 0$, then it logically follows that $\map P {j^+}$ 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 $\map P j \implies \map P {j^+}$ and the result follows by the Principle of Mathematical Induction.

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