Min Operation on Toset forms Semigroup

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

Let $\min \left({x, y}\right)$ denote the min operation on $x, y \in S$.

Then $\left({S, \min}\right)$ is a semigroup.

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

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

So $\min$ is closed on $S$.

Then we have that the min operation is associative:
 * $\forall x, y, z \in S: \min \left({x, \min \left({y, z}\right)}\right) = \min \left({\min \left({x, y}\right), z}\right)$

Hence the result, by definition of semigroup.

Also see

 * Max Operation on Toset is Semigroup