Natural Number Addition is Commutative/Proof 2

Proof
Proof by induction.

Consider the natural numbers $\N$ defined as the elements of the minimally inductive st $\omega$.

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

For all $n \in \N$, let $\map P n$ 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 $\map P 0$ is seen to be true.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true.

So this is our induction hypothesis $\map P k$:
 * $\forall m \in \N: m + k = k + m$

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

Induction Step
This is our induction step:

So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction.

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