Integer Multiplication Distributes over Addition

Theorem
Integer multiplication is distributive over addition:


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

Proof
Let us define $$\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus$$ as in the formal definition of integers.

That is, $$\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus$$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $$\boxminus$$.

$$\boxminus$$ is the congruence relation defined on $$\N \times \N$$ by $$\left({x_1, y_1}\right) \boxminus \left({x_2, y_2}\right) \iff x_1 + y_2 = x_2 + y_1$$.

In order to streamline the notation, we will use $$\left[\!\left[{a, b}\right]\!\right]$$ to mean $$\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\boxminus$$, as suggested.

We need to show that, $$\forall a, b, c, d, e, f \in \N$$:


 * $$\left[\!\left[{a, b}\right]\!\right] \times \left({\left[\!\left[{c, d}\right]\!\right] + \left[\!\left[{e, f}\right]\!\right]}\right) = \left({\left[\!\left[{a, b}\right]\!\right] \times \left[\!\left[{c, d}\right]\!\right]}\right) + \left({\left[\!\left[{a, b}\right]\!\right] \times \left[\!\left[{e, f}\right]\!\right]}\right)$$


 * $$\left({\left[\!\left[{c, d}\right]\!\right] + \left[\!\left[{e, f}\right]\!\right]}\right) \times \left[\!\left[{a, b}\right]\!\right] = \left({\left[\!\left[{c, d}\right]\!\right] \times \left[\!\left[{a, b}\right]\!\right]}\right) + \left({\left[\!\left[{e, f}\right]\!\right] \times \left[\!\left[{a, b}\right]\!\right]}\right)$$

From Natural Numbers form Semiring, we can take it for granted that:
 * addition and multiplication are commutative and associative on the natural numbers $$\mathbb{N}$$;
 * natural number multiplication is distributive over natural number addition.

So:

$$ $$ $$ $$ $$ $$

$$ $$ $$ $$ $$ $$