Natural Number Multiplication Distributes over Addition/Proof 2

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

Proof
We are to show that:
 * $\left({x + y}\right) \times n = \left({x \times n}\right) + \left({y \times n}\right)$

for all $x, y, n \in \N$.

From the definition of natural number multiplication, we have by definition that:

Let $x, y \in \N$ be arbitrary.

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $\forall x, y \in \N: \left({x + y}\right) \times n = \left({x \times n}\right) + \left({y \times n}\right)$

Basis for the Induction
$P \left({0}\right)$ is the case:

and so $P \left({0}\right)$ 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 $P \left({k^+}\right)$ is true.

So this is our induction hypothesis:
 * $\forall x, y \in \N: \left({x + y}\right) \times k = \left({x \times k}\right) + \left({y \times k}\right)$

Then we need to show:
 * $\forall x, y \in \N: \left({x + y}\right) \times k^+ = \left({x \times k^+}\right) + \left({y \times k^+}\right)$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the Principle of Finite Induction:


 * $\forall x, y, n \in \N: \left({x + y}\right) \times n = \left({x \times n}\right) + \left({y \times n}\right)$

Next we need to show that:
 * $n \times \left({x + y}\right) = \left({n \times x}\right) + \left({n \times y}\right)$

for all $x, y, n \in \N$.

So:

Thus we have proved:


 * $\forall x, y, n \in \N: n \times \left({x + y}\right) = \left({n \times x}\right) + \left({n \times y}\right)$