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

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

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


 * $\paren {m + 1} + 1 = \paren {m + 1} + 1$

which holds trivially.

Thus $\map P 1$ is seen to be true.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis $\map P k$:
 * $\forall m \in \N: \paren {m + 1} + k = \paren {m + k} + 1$

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

Induction Step
This is our induction step:

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

Therefore:
 * $\forall m, n \in \N_{> 0}: \paren {m + 1} + n = \paren {m + n} + 1$