Natural Number Multiplication Distributes over Addition

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

Proof 1
Follows directly from the fact that the Natural Numbers form Commutative Semiring.