Natural Number Addition is Associative/Proof 3

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


 * $\forall x, y, n \in \N_{> 0}: x + \left({y + n}\right) = \left({x + y}\right) + n$

Proof
Using the following axioms:

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

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

Basis for the Induction
From Axiom $C$, we have by definition that:
 * $\forall x, y \in \N_{> 0}: \left({x + y}\right) + 1 = x + \left({y + 1}\right)$

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:
 * $\left({x + y}\right) + k = x + \left({y + k}\right)$

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

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.