Natural Number Multiplication Distributes over Addition/Proof 3

Theorem
The operation of multiplication is distributive over addition on the set of natural numbers $\N_{> 0}$:
 * $\forall x, y, n \in \N_{> 0}:$
 * $\left({x + y}\right) \times n = \left({x \times n}\right) + \left({y \times n}\right)$
 * $n \times \left({x + y}\right) = \left({n \times x}\right) + \left({n \times y}\right)$

Proof
Using the following axioms:

Left Distributive Law for Natural Numbers
First we show that:
 * $n \times \left({x + y}\right) = \left({n \times x}\right) + \left({n \times y}\right)$

Right Distributive Law for Natural Numbers
Then we show that:
 * $\left({x + y}\right) \times n = \left({x \times n}\right) + \left({y \times n}\right)$

The result follows.