Dipper Operation is Closed on Initial Segment

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

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

Let $\N_{< \paren {m \mathop + n} }$ denote the initial segment of the natural numbers:
 * $\N_{< \paren {m \mathop + n} } := \set {0, 1, \ldots, m + n - 1}$

Let $+_{m, n}$ denote the dipper operation on $\N_{< \paren {m \mathop + n} }$:
 * $\forall a, b \in \N_{< \paren {m \mathop + n} }: a +_{m, n} b = \begin{cases}

a + b & : a + b < m \\ a + b - k n & : a + b \ge m \end{cases}$ where $k$ is the largest integer satisfying:
 * $m + k n \le a + b$

Then $+_{m, n}$ is closed on $\N_{< \paren {m \mathop + n} }$.

Proof
Let $a + b < m$.

Then:
 * $0 \le a +_{m, n} b < m$

and so:
 * $a +_{m, n} b \in \N_{< \paren {m \mathop + n} }$

Let $a + b > m$.

Then:
 * $a +_{m, n} b = a + b - k n$

where $k$ is the largest such that $m + k n < a + b$

Hence:

That is:
 * $a +_{m, n} b \in \N_{< \paren {m \mathop + n} }$

That is, $+_{m, n}$ is closed on $\N_{< \paren {m \mathop + n} }$.