Ring of Integers has no Zero Divisors

Theorem
The integers have no zero divisors:
 * $$\forall x, y, \in \Z: x \times y = 0 \implies x = 0 \or y = 0$$

This can alternatively be expressed:
 * $$\forall x, y, \in \Z: x \ne 0 \and y \ne 0 \implies x \times y \ne 0$$

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.

From the method of construction, $$\left[\!\left[{c, c}\right]\!\right]$$, where $$c$$ is any element of the natural numbers $$\N$$, is the identity of $$\left({\Z, +}\right)$$.

To ease the algebra, we will take $$\left[\!\left[{0, 0}\right]\!\right]$$ as a canonical instance of this equivalence class.

We need to show that:


 * $$\forall a, b, c, d \in \N: \left[\!\left[{a, b}\right]\!\right] \times \left[\!\left[{c, d}\right]\!\right] = \left[\!\left[{0, 0}\right]\!\right] \implies \left[\!\left[{a, b}\right]\!\right] = \left[\!\left[{0, 0}\right]\!\right] \or \left[\!\left[{c, d}\right]\!\right] = \left[\!\left[{0, 0}\right]\!\right]$$

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

So:

$$ $$ $$ $$

We have to be careful here, and bear in mind that $$a, b, c, d$$ are natural numbers, and we have not defined (and, at this stage, will not define) subtraction on such entities.

Suppose WLOG that $$\left[\!\left[{c, d}\right]\!\right] \ne \left[\!\left[{0, 0}\right]\!\right]$$.

Then $$c \ne d$$.

Suppose also WLOG that $$c > d$$.

The Natural Numbers are a Naturally Ordered Semigroup.

So from the definition of Unique Minus in a naturally ordered semigroup, $$\exists p \in \N: d + p = c$$ where $$p > 0$$.

Then:

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

Similarly for when $$c < d$$.

Thus $$\left[\!\left[{c, d}\right]\!\right] \ne \left[\!\left[{0, 0}\right]\!\right] \implies \left[\!\left[{a, b}\right]\!\right] = \left[\!\left[{0, 0}\right]\!\right]$$.

A similar argument shows that $$\left[\!\left[{a, b}\right]\!\right] \ne \left[\!\left[{0, 0}\right]\!\right] \implies \left[\!\left[{c, d}\right]\!\right] = \left[\!\left[{0, 0}\right]\!\right]$$.

The alternative statement of this theorem follows from De Morgan and the Rule of Transposition.