Natural Number Addition is Commutative/Proof 2

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


 * $\forall m, n \in \N: m + n = n + m$

Proof
Proof by induction.

Consider the natural numbers $\N$ defined as the elements of the minimal infinite successor set $\omega$.

From the definition of addition in $\omega$‎, we have that:

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

Basis for the Induction
From Natural Number Addition Commutes with Zero, we have:
 * $\forall m \in \N: m + 0 = m = 0 + m$

Thus $P \left({0}\right)$ is seen to be true.

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 $P \left({k^+}\right)$ is true.

So this is our induction hypothesis $P \left({k}\right)$:
 * $\forall m \in \N: m + k = k + m$

Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$:
 * $\forall m \in \N: m + k^+ = k^+ + m$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the Principle of Finite Induction.

Therefore:
 * $\forall m, n \in \N: m + n = n + m$