Dipper Relation is Congruence for Multiplication

Theorem
Let $m \in \N$ be a natural number.

Let $n \in \N_{>0}$ be a non-zero natural number.

Let $\RR_{m, n}$ be the dipper relation on $\N$:


 * $\forall x, y \in \N: x \mathrel {\RR_{m, n} } y \iff \begin {cases} x = y \\ m \le x < y \text { and } n \divides \paren {y - x} \\ m \le y < x \text { and } n \divides \paren {x - y} \end {cases}$

Then $\RR_{m, n}$ is a congruence relation for multiplication.

Proof
From Dipper Relation is Equivalence Relation we have that $\RR_{m, n}$ is an equivalence relation.

From Equivalence Relation is Congruence iff Compatible with Operation, it is sufficient to show that:


 * $\forall x, y, z \in \N: x \mathrel {\RR_{m, n} } y \implies \paren {x z} \mathrel {\RR_{m, n} } \paren {y z}$

and:


 * $\forall x, y, z \in \N: x \mathrel {\RR_{m, n} } y \implies \paren {z x} \mathrel {\RR_{m, n} } \paren {z y}$

Furthermore, because Natural Number Addition is Commutative, it is sufficient to demonstrate the first of these only.

First let $x = y$.

We have that:
 * $x = y \implies x z = y z$

and so:
 * $\paren {x z} \mathrel {\RR_{m, n} } \paren {y z}$

Otherwise, we have that:
 * $x \ne y$ and $x, y \ge m$

, let $x < y$.

If not, then as $\RR_{m, n}$ is symmetric we can rename $x$ and $y$ as necessary.

First note that if $z = 0$, then:
 * $\paren {x z} = \paren {y z}= 0$

and so:
 * $\paren {x z} \mathrel {\RR_{m, n} } \paren {y z}$

Otherwise we have that $z \ge 1$ and so both $x z \ge m$ and $y z \ge m$.

Then:

Hence the result.