Integers Modulo m under Max Operation form Ordered Semigroup

Theorem
Let $\Z_m$ denote the set of integers modulo $m$:
 * $\Z_m = \set {0, 1, \ldots, m - 1}$

Let $\vee_m$ be the operation on $\Z_m$ defined as:
 * $\forall a, b \in \Z_m: a \vee_m b = \max \set {a, b}$

Then the ordered algebraic structure $\struct {\Z_m, \vee_m, \le}$ is an ordered semigroup.

Proof
Taking the semigroup axioms in turn:

We have that:
 * $\forall a, b \in \Z_m: \max \set {a, b} \in \set {a, b}$

and so:
 * $\max \set {a, b} \in \Z_m$

Thus $\struct {\Z_m, \vee_m}$ is closed.

We have that Max Operation is Associative.

Thus $\vee_m$ is associative.

The semigroup axioms are thus seen to be fulfilled, and so $\struct {\Z_m, \vee_m}$ is a semigroup.

Ordered Semigroup
We now need to show that:
 * $(1): \quad x \le y \implies x \vee_m z \le y \vee_m z$

and:
 * $(2): \quad x \le y \implies z \vee_m x \le z \vee_m y$

Because Max Operation is Commutative, $(2)$ follows directly from $(1)$.

Hence we need demonstrate $(1)$ only.

Let $x, y, z \in S$.

Let $x \le y$.


 * Case $1$. $z \le x$


 * Case $2$. $x \le z \le y$


 * Case $3$. $y \le z$

Thus $\le$ is compatible with $\vee_m$.

The result follows.