Natural Number Addition Commutativity with Successor/Proof 1

Theorem
Let $\N$ be the natural numbers.

Then:
 * $\forall m, n \in \N: m^+ + n = \left({m + n}\right)^+$

Proof
Proof by induction:

From the definition of addition, we have that:

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * $\forall m \in \N: m^+ + n = \left({m + n}\right)^+$

Basis for the Induction
By definition, we have:

Thus $P \left({0}\right)$ is seen to be true.

This is our basis for the induction.

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

So this is our induction hypothesis $P \left({k}\right)$:
 * $\forall m \in \N: m^+ + k = \left({m + k}\right)^+$

Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$:
 * $\forall m \in \N: m^+ + k^+ = \left({m + k^+}\right)^+$

Induction Step
This is our induction step:

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

Therefore:
 * $\forall m, n \in \N: m^+ + n = \left({m + n}\right)^+$