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:
 * $(1): \quad x \le y \le z$
 * $(2): \quad x \le z \le y$
 * $(3): \quad y \le x \le z$
 * $(4): \quad y \le z \le x$
 * $(5): \quad z \le x \le y$
 * $(6): \quad z \le y \le x$

$(1): \quad $ Let $x \le y \le z$. Then:

$(2): \quad $ Let $x \le z \le y$. Then:

$(3): \quad $ Let $y \le x \le z$. Then:

$(4): \quad $ Let $y \le z \le x$. Then:

$(5): \quad $ Let $z \le x \le y$. Then:

$(6): \quad $ Let $z \le y \le x$. Then:

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