Natural Number Multiplication is Commutative/Proof 1

Proof
Natural number multiplication is recursively defined as:


 * $\forall m, n \in \N: \begin{cases}

m \times 0 & = 0 \\ m \times \left({n + 1}\right) & = m \times n + m \end{cases}$

From the Principle of Recursive Definition, there is only one mapping $f$ satisfying this definition; that is, such that:


 * $\forall n \in \N: \begin{cases}

f \left({0}\right) = 0 \\ f \left({n + 1}\right) = f \left({n}\right) + m \end{cases}$

Consider now $f'$ defined as $f' \left({n}\right) = n \times m$.

Then by Zero is Zero Element for Natural Number Multiplication:


 * $f' \left({0}\right) = 0 \times m = 0$

Furthermore:

showing that $f'$ also satisfies the definition of $m \times n$.

By the Principle of Recursive Definition it follows that:


 * $m \times n = n \times m$