Natural Number Addition is Associative/Proof 3

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:
 * $\paren {x + y} + n = x + \paren {y + n}$

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

and so $\map P 1$ holds.

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:
 * $\paren {x + y} + k = x + \paren {y + k}$

Then we need to show:
 * $\paren {x + y} + \paren {k + 1} = x + \paren {y + \paren {k + 1} }$

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.