Min Operation on Toset forms Semigroup

Theorem
Let $\left({S, \preceq}\right)$ be a totally ordered set.

Then $\left({S, \max}\right)$ and $\left({S, \min}\right)$ are semigroups.

Proof
By the definition of the max and min operations, either:
 * $\max \left({x, y}\right) = x$

or
 * $\max \left({x, y}\right) = y$

and similarly, either:
 * $\min \left({x, y}\right) = x$

or
 * $\min \left({x, y}\right) = y$

So both $\max$ and $\min$ are closed on $S$.

Then we have that Max and Min are Associative:
 * $\forall x, y, z \in S: \max \left({x, \max \left({y, z}\right)}\right) = \max \left({\max \left({x, y}\right), z}\right)$

and
 * $\forall x, y, z \in S: \min \left({x, \min \left({y, z}\right)}\right) = \min \left({\min \left({x, y}\right), z}\right)$

Hence the result, by definition of semigroup.