Natural Number Addition Commutativity with Successor/Proof 2

Theorem
Let $\N$ be the natural numbers.

Then:
 * $\forall m, n \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$

Proof
Using the following axioms:

Proof by induction:

From Axiom:Axiomatization of 1-Based Natural Numbers, we have by definition that:

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $\forall m \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$

Basis for the Induction
When $n = 1$, we have:


 * $\left({m + 1}\right) + 1 = \left({m + 1}\right) + 1$

which holds trivially.

Thus $P \left({1}\right)$ 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 $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is our induction hypothesis $P \left({k}\right)$:
 * $\forall m \in \N: \left({m + 1}\right) + k = \left({m + k}\right) + 1$

Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$:
 * $\forall m \in \N: \left({m + 1}\right) + \left({k + 1}\right) = \left({m + \left({k + 1}\right)}\right) + 1$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall m, n \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$