Natural Number Addition is Associative

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


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

Proof 1
Follows directly from Natural Numbers under Addition is Commutative Monoid.

A monoid by definition is a semigroup.

Again by definition, the operation in a semigroup is associative.

Proof 2
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, 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:
 * $\left({x + y}\right) + k^+ = x + \left({y + k^+}\right)$

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 Finite Induction.