Ordering on Natural Numbers is Compatible with Addition

Theorem

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

Then:

$m < n \iff m + k < n + k$

Corollary

Let $a, b, c, d \in \N$ where $\N$ is the set of natural numbers.

Then:

$a > b, c > d \implies a + c > b + d$

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:

 $\displaystyle m$ $<$ $\displaystyle n$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle m$ $\subsetneq$ $\displaystyle n$ Element of Finite Ordinal iff Subset‎ $\displaystyle \leadstoandfrom \ \$ $\displaystyle m^+$ $\subset$ $\displaystyle n$ Definition of Successor Set $\displaystyle \leadstoandfrom \ \$ $\displaystyle m^+$ $\subsetneq$ $\displaystyle n^+$ Definition of Successor Set $(1):\quad$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle m^+$ $<$ $\displaystyle n^+$ Element of Finite Ordinal iff Subset‎

This gives:

 $\displaystyle m + j$ $<$ $\displaystyle n + j$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle \paren {m + j}^+$ $<$ $\displaystyle \paren {n + j}^+$ from $(1)$ above $\displaystyle \leadstoandfrom \ \$ $\displaystyle m + j^+$ $<$ $\displaystyle n + j^+$ Definition of Addition in Minimal Infinite Successor Set

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

Therefore:

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

$\blacksquare$