Natural Number Commutes with 1 under Addition

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


 * $\forall n \in \N_{> 0}: n + 1 = 1 + n$

Proof
Using the axiom schema:

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $n + 1 = 1 + n$

Basis for the Induction
Setting $n = 1$ we have that:
 * $1 + 1 = 1 + 1$

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

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:
 * $k + 1 = 1 + k$

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

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.