Dipper Relation is Equivalence Relation

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 an equivalence relation.

Proof
First let it be noted that $\RR_{m, n}$ can be written as:


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

where $\size {x - y}$ denotes the absolute value of $x - y$.

Checking in turn each of the criteria for equivalence:

Reflexivity
We have that:
 * $\forall x \in \N: x = x$

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

Thus $\RR_{m, n}$ is seen to be reflexive.

Symmetry
Let $x, y \in \N$ such that $x \mathrel {\RR_{m, n} } y$.

If $x = y$ then trivially $y = x$ and so:
 * $y \mathrel {\RR_{m, n} } x$

Otherwise note that:
 * $n \divides \size {x - y} \iff n \divides \size {y - x}$

and it follows directly that:
 * $y \mathrel {\RR_{m, n} } x$

Thus $\RR_{m, n}$ is seen to be symmetric.

Transitivity
Let $x, y, z \in \N$ such that $x \mathrel {\RR_{m, n} } y$ and $y \mathrel {\RR_{m, n} } z$.

First suppose $x = y$ or $y = z$ or $x = z$.

Then trivially $x \mathrel {\RR_{m, n} } z$.

Otherwise we have that:
 * $x \ne y \ne z \ne x$

and:
 * $x, y, z \ge m$

, let $x < y < z$.

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

Then:

Thus $\RR_{m, n}$ is seen to be transitive.

Thus $\RR_{m, n}$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.