Theorem: Ring of Integers has no Zero Divisors

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

Proof:
Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxminus$ 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 $\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$.
In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested.
From the method of construction, $\eqclass {c, c} {}$, where $c$ is any element of the natural numbers $\N$, is the identity of $\struct {\Z, +}$.
To ease the algebra, we will take $\eqclass {0, 0} {}$ as a canonical instance of this equivalence class.
We need to show that:
:$\forall a, b, c, d \in \N: \eqclass {a, b} {} \times \eqclass {c, d} {} = \eqclass {0, 0} {} \implies \eqclass {a, b} {} = \eqclass {0, 0} {} \lor \eqclass {c, d} {} = \eqclass {0, 0} {}$
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:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {} \times \eqclass {c, d} {}
      | r = \eqclass {0, 0} {}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \eqclass {a c + b d, a d + b c} {}
      | r = \eqclass {0, 0} {}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a c + b d + 0
      | r = a d + b c + 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a c + b d
      | r = a d + b c
      | c = 
}}
{{end-eqn}}
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.
{{WLOG}}, suppose that $\eqclass {c, d} {} \ne \eqclass {0, 0} {}$.
Then $c \ne d$.
{{WLOG}}, suppose also that $c > d$.
{{wtd|Ordering in terms of Addition should have a version for $\N$}}
From Ordering in terms of Addition, $\exists p \in \N: d + p = c$ where $p > 0$.
Then:
{{begin-eqn}}
{{eqn | l = a c + b d
      | r = a d + b c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a \paren {d + p} + b d
      | r = a d + b \paren {d + p}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a d + a p + b d
      | r = a d + b d + b p
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a p
      | r = b p
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a
      | r = b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \eqclass {a, b} {}
      | r = \eqclass {0, 0} {}
      | c = Construction of Inverse Completion: Equivalence Class of Equal Elements
}}
{{end-eqn}}
Similarly for when $c < d$.
Thus:
:$\eqclass {c, d} {} \ne \eqclass {0, 0} {} \implies \eqclass {a, b} {} = \eqclass {0, 0} {}$
A similar argument shows that:
:$\eqclass {a, b} {} \ne \eqclass {0, 0} {} \implies \eqclass {c, d} {} = \eqclass {0, 0} {}$
The equivalence between the two forms of the statement of this theorem follows from De Morgan's Laws: Conjunction of Negations and the Rule of Transposition.
{{qed}}