Right Distributive Law for Natural Numbers

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

Proof
Using the axiom schema:

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $\forall a, b \in \N_{> 0}: \left({a + b}\right) \times n = \left({a \times n}\right) + \left({b \times n}\right)$

Basis for the Induction
$P \left({1}\right)$ is the case:

and so $P \left({1}\right)$ holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is our induction hypothesis:
 * $\forall a, b \in \N_{> 0}: \left({a + b}\right) \times k = \left({a \times k}\right) + \left({b \times k}\right)$

Then we need to show:
 * $\forall a, b \in \N_{> 0}: \left({a + b}\right) \times \left({k + 1}\right) = \left({a \times \left({k + 1}\right)}\right) + \left({b \times \left({k + 1}\right)}\right)$

Induction Step
This is our induction step:

The result follows by the Principle of Mathematical Induction.

Also see

 * Left Distributive Law for Natural Numbers