Dipper Relation is Congruence for Addition

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 addition.

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.

So, let $x, y \in \N$ be such that:
 * $x \mathrel {\RR_{m, n} } y$

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.