Max and Min Operations are Distributive over Each Other

Theorem
The Max and Min operations are distributive over each other:


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


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


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


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

Proof
To simplify our notation, let $$\max \left({x, y}\right)$$ be (temporarily) denoted $$x \overline \wedge y$$, and let $$\min \left({x, y}\right)$$ be (temporarily) denoted $$x \underline \vee y$$.

Note that, once we have proved:


 * $$x \overline \wedge \left({y \underline \vee z}\right) = \left({x \overline \wedge y}\right) \underline \vee \left({x \overline \wedge z}\right)$$


 * $$x \underline \vee \left({y \overline \wedge z}\right) = \left({x \underline \vee y}\right) \overline \wedge \left({x \underline \vee z}\right)$$

then the other results follow immediately by the fact that Min and Max are commutative.

There are the following cases to consider:
 * 1) $$x \le y \le z$$
 * 2) $$x \le z \le y$$
 * 3) $$y \le x \le z$$
 * 4) $$y \le z \le x$$
 * 5) $$z \le x \le y$$
 * 6) $$z \le y \le x$$


 * Let $$x \le y \le z$$. Then:

$$ $$ $$ $$ $$


 * Let $$x \le z \le y$$. Then:

$$ $$ $$ $$ $$


 * Let $$y \le x \le z$$. Then:

$$ $$ $$ $$ $$


 * Let $$y \le z \le x$$. Then:

$$ $$ $$ $$ $$


 * Let $$z \le x \le y$$. Then:

$$ $$ $$ $$ $$


 * Let $$z \le y \le x$$. Then:

$$ $$ $$ $$ $$

Thus in all cases it can be seen that the result holds.