Natural Number Addition is Associative/Proof 2

Theorem
The operation of addition on the set of natural numbers $\N$ is associative:


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

Proof
We are to show that:
 * $\left({x + y}\right) + n = x + \left({y + n}\right)$

for all $x, y, n \in \N$.

From Definition by Induction of Natural Number Addition/Corollary‎, we have by definition that:

Let $x, y \in \N$ be arbitrary.

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

Basis for the Induction
$P \left({0}\right)$ is the case:

and so $P \left({0}\right)$ holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, 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.