On $\mathbb R$, we have the usual $\min$ and $\max$ operators. They are distributive over each other. In other words,
$$
\displaylines{
\max(x,\min(y,z))=\min(\max(x,y),\max(x,z)), \\
\min(x,\max(y,z))=\max(\min(x,y),\min(x,z)).
}
$$
This result is proven here. However, I don’t like the idea of having to list all the cases. Here is an alternative way to prove it.
We first give two lemmas.
Lemma 1. For any $a, b \in \mathbb R$, we have $\min (a, b) \le a$.
This is obvious.
Lemma 2. For a fixed $x \in \mathbb R$, the functions $f(t) = \min(x, t)$ and $g(t) = \max(x, t)$ on $\mathbb R$ are non-decreasing. In other words, $\min(x, -)$ and $\max(x, -)$ preserves the $\le$ relation.
This is obtained by breaking the cases of $\min$ and $\max$ and drawing the function graphs.
Now we can prove the main proposition: $\max(x,\min(y,z))=\min(\max(x,y),\max(x,z))$.
First notice that $y$ and $z$ are symmetric, so without loss of generality, we may assume $y \le z$. Then $\min(y, z) = y$ and $\text{LHS} = \max(x, y)$.
By Lemma 1, $\text{RHS} \le \max(x, y) = \text{LHS}$.
By Lemma 2, $\max(x, z) \ge \max(x, y)$. Then $\text{RHS} \ge \min( \max(x, y), \max(x, y)) = \max(x, y) = \text{LHS}$ by using Lemma 2 again.
Thus $\text{LHS} = \text{RHS}$. QED.
The other proposition $\min(x,\max(y,z))=\max(\min(x,y),\min(x,z))$ can be proved in a similar fashion.
In fact, this is a special case of the following general statement:
Every totally ordered set is a distributive lattice with max as join and min as meet.
The above proof can be easily adapted for this general statement.