Natural Number Addition is Commutative/Proof 3

Theorem
The operation of addition on the set of natural numbers $\N_{> 0}$ is commutative:


 * $\forall x, y \in \N_{> 0}: x + y = y + x$

Proof
Using the following axioms:

Let $x \in \N_{> 0}$ be arbitrary.

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

Basis for the Induction
From Natural Number Commutes with 1 under Addition, we have that:
 * $\forall x \in \N_{> 0}: x + 1 = 1 + x$

and so $P \left({1}\right)$ holds.

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:
 * $x + k = k + x$

Then we need to show:
 * $x + \left({k + 1}\right) = \left({k + 1}\right) + x$

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.