Ring of Integers has no Zero Divisors

Theorem
Integers have no zero divisors:

$$\forall x, y, \in \mathbb{Z}: x \times y = 0 \Longrightarrow z = 0 \lor y = 0$$

Proof
We need to show that:

$$\forall a, b, c, d \in \mathbb{N}: \left[\left[{\left({a, b}\right)}\right]\right]_\boxminus \times \left[\left[{\left({c, d}\right)}\right]\right]_\boxminus = \left[\left[{\left({0, 0}\right)}\right]\right]_\boxminus \Longrightarrow \left[\left[{\left({a, b}\right)}\right]\right]_\boxminus = \left[\left[{\left({0, 0}\right)}\right]\right]_\boxminus \lor \left[\left[{\left({c, d}\right)}\right]\right]_\boxminus = \left[\left[{\left({0, 0}\right)}\right]\right]_\boxminus$$

The element $$\left[\left[{\left({0, 0}\right)}\right]\right]_\boxminus$$ is a perfectly adequate member of the "zero" equivalence class, and makes the arithmetic easier.

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.

Thus:

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

Suppose, without loss of generality, that $$\left[\left[{\left({c, d}\right)}\right]\right]_\boxminus \ne \left[\left[{\left({0, 0}\right)}\right]\right]_\boxminus$$.

Then $$c \ne d$$. Suppose $$c > d$$, and let $$d + p = c, p > 0$$.

Then:

Similarly for when $$c < d$$.

Thus $$\left[\left[{\left({c, d}\right)}\right]\right]_\boxminus \ne \left[\left[{\left({0, 0}\right)}\right]\right]_\boxminus \Longrightarrow \left[\left[{\left({a, b}\right)}\right]\right]_\boxminus = \left[\left[{\left({0, 0}\right)}\right]\right]_\boxminus$$.

A similar argument shows that $$\left[\left[{\left({a, b}\right)}\right]\right]_\boxminus \ne \left[\left[{\left({0, 0}\right)}\right]\right]_\boxminus \Longrightarrow \left[\left[{\left({c, d}\right)}\right]\right]_\boxminus = \left[\left[{\left({0, 0}\right)}\right]\right]_\boxminus$$.