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