Dipper Semigroup is Commutative Semigroup

Theorem
The dipper semigroup is a commutative semigroup.

Proof
Recall the definition of the dipper semigroup:

Let $m, n \in \Z$ be integers such that $m \ge 0, n > 0$.

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 $+_{m, n}$ be 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:
 * $m + k n \le a + b$

Taking the semigroup axioms in turn:

Let $a, b \in \N_{< \paren {m \mathop + n} }$ be arbitrary.

From Dipper Operation is Closed on Initial Segment:
 * $a +_{m, n} b \in \N_{< \paren {m \mathop + n} }$

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

We have from Dipper Operation is Associative that $+_{m, n}$ is associative.

The semigroup axioms are thus seen to be fulfilled, and so $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ is a semigroup.

Then we note that Integer Addition is Commutative:
 * $a + b = b + a$

and so it follows directly that $a +_{m, n} b = b +_{m, n} a$.

Hence $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ is a commutative semigroup.