Can you treat incongruences as you would congruent relations? i.e. do the following theorems still apply?
As an example, let's say you wanted to prove Euclid's Lemma which states:
If a prime $p$ divides the product $ab$ of two integers $a$ and $b$, then $p$ must divide at least one of those integers $a$ or $b$.
If we can show that $p\not\mid a,b$ implies that $p\not\mid ab$ then we have proven the contrapositive and thus our original statement. So, assume that $p\not\mid a,b$. Then by the definition of congruences,
$$a\not\equiv 0\ (mod\ p)$$ $$b\not\equiv 0\ (mod\ p)$$
If (iii) from Theorem 4.6 still holds then it follows that $$ab\not\equiv 0\ (mod\ p)$$
hence $p\not\mid ab$ as desired.
This is weak. (iii) is a case of selection. in any base ( even greater than 4) $$2\cdot 1 \equiv (-1)\cdot (-2)$$ Despite, no parts being congruent. The theorem above is simply the inverse of the 0 product rule. In the absence of zero divisors ( divisor of 0 in a ring etc forming a product of 0 with another non zero element) it is true, in the presence of 0 divisors it is not.