Natural Number Multiplication Distributes over Addition/Proof 2

Proof
We are to show that:
 * $\forall x, y, z \in \N: \paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$

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

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

For all $z \in \N$, let $\map P z$ be the proposition:
 * $\forall x, y \in \N: \paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$

Basis for the Induction
$\map P 0$ is the case:

and so $\map P 0$ holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true.

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

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

Induction Step
This is our induction step:

So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction:
 * $\forall x, y, z \in \N: \paren {x + y} \times n = \paren {x \times z} + \paren {y \times z}$

Next we need to show that:
 * $\forall x, y, z \in \N: z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$

So:

Thus we have proved:


 * $\forall x, y, z \in \N: z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$