Max and Min Operations are Distributive over Each Other

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


 * $\max \set {x, \min \set {y, z} } = \min \set {\max \set {x, y}, \max \set {x, z} }$


 * $\max \set {\min \set {x, y}, z} = \min \set {\max \set {x, z}, \max \set {y, z} }$


 * $\min \set {x, \max \set {y, z} } = \max \set {\min \set {x, y}, \min \set {x, z} }$


 * $\min \set {\max \set {x, y}, z} = \max \set {\min \set {x, z}, \min \set {y, z} }$

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

Note that, once we have proved:


 * $x \overline \wedge \paren {y \underline \vee z} = \paren {x \overline \wedge y} \underline \vee \paren {x \overline \wedge z}$


 * $x \underline \vee \paren {y \overline \wedge z} = \paren {x \underline \vee y} \overline \wedge \paren {x \underline \vee z}$

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.