Theorem: Max and Min Operations are Distributive over Each Other

The Max and Min operations are distributive over each other:
:$\map \max {x, \map \min {y, z} } = \map \min {\map \max {x, y}, \map \max {x, z} }$
:$\map \max {\map \min {x, y}, z} = \map \min {\map \max {x, z}, \map \max {y, z} }$
:$\map \min {x, \map \max {y, z} } = \map \max {\map \min {x, y}, \map \min {x, z} }$
:$\map \min {\map \max {x, y}, z} = \map \max {\map \min {x, z}, \map \min {y, z} }$

Proof:
To simplify our notation, let $\map \max {x, y}$ be (temporarily) denoted $x \overline \wedge y$, and let $\map \min {x, y}$ 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 from Max Operation is Commutative and Min Operation is 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:
{{begin-eqn}}
{{eqn | l = x \overline \wedge \paren {y \underline \vee z}
      | r = x \overline \wedge y = y
      | c = 
}}
{{eqn | l = \paren {x \overline \wedge y} \underline \vee \paren {x \overline \wedge z}
      | r = y \underline \vee z = y
      | c = 
}}
{{eqn | o = 
      | c = 
}}
{{eqn | l = x \underline \vee \paren {y \overline \wedge z}
      | r = x \underline \vee z = x
      | c = 
}}
{{eqn | l = \paren {x \underline \vee y} \overline \wedge \paren {x \underline \vee z}
      | r = x \overline \wedge x = x
      | c = 
}}
{{end-eqn}}
$(2): \quad $ Let $x \le z \le y$.
Then:
{{begin-eqn}}
{{eqn | l = x \overline \wedge \paren {y \underline \vee z}
      | r = x \overline \wedge z = z
      | c = 
}}
{{eqn | l = \paren {x \overline \wedge y} \underline \vee \paren {x \overline \wedge z}
      | r = y \underline \vee z = z
      | c = 
}}
{{eqn | o = 
      | c = 
}}
{{eqn | l = x \underline \vee \paren {y \overline \wedge z}
      | r = x \underline \vee y = x
      | c = 
}}
{{eqn | l = \paren {x \underline \vee y} \overline \wedge \paren {x \underline \vee z}
      | r = x \overline \wedge x = x
      | c = 
}}
{{end-eqn}}
$(3): \quad $ Let $y \le x \le z$.
Then:
{{begin-eqn}}
{{eqn | l = x \overline \wedge \paren {y \underline \vee z}
      | r = x \overline \wedge y = x
      | c = 
}}
{{eqn | l = \paren {x \overline \wedge y} \underline \vee \paren {x \overline \wedge z}
      | r = x \underline \vee z = x
      | c = 
}}
{{eqn | o = 
      | c = 
}}
{{eqn | l = x \underline \vee \paren {y \overline \wedge z}
      | r = x \underline \vee z = x
      | c = 
}}
{{eqn | l = \paren {x \underline \vee y} \overline \wedge \paren {x \underline \vee z}
      | r = y \overline \wedge x = x
      | c = 
}}
{{end-eqn}}
$(4): \quad $ Let $y \le z \le x$.
Then:
{{begin-eqn}}
{{eqn | l = x \overline \wedge \paren {y \underline \vee z}
      | r = x \overline \wedge y = x
      | c = 
}}
{{eqn | l = \paren {x \overline \wedge y} \underline \vee \paren {x \overline \wedge z}
      | r = x \underline \vee x = x
      | c = 
}}
{{eqn | o = 
      | c = 
}}
{{eqn | l = x \underline \vee \paren {y \overline \wedge z}
      | r = x \underline \vee z = z
      | c = 
}}
{{eqn | l = \paren {x \underline \vee y} \overline \wedge \paren {x \underline \vee z}
      | r = y \overline \wedge z = z
      | c = 
}}
{{end-eqn}}
$(5): \quad $ Let $z \le x \le y$.
Then:
{{begin-eqn}}
{{eqn | l = x \overline \wedge \paren {y \underline \vee z}
      | r = x \overline \wedge z = x
      | c = 
}}
{{eqn | l = \paren {x \overline \wedge y} \underline \vee \paren {x \overline \wedge z}
      | r = y \underline \vee x = x
      | c = 
}}
{{eqn | o = 
      | c = 
}}
{{eqn | l = x \underline \vee \paren {y \overline \wedge z}
      | r = x \underline \vee y = x
      | c = 
}}
{{eqn | l = \paren {x \underline \vee y} \overline \wedge \paren {x \underline \vee z}
      | r = x \overline \wedge z = x
      | c = 
}}
{{end-eqn}}
$(6): \quad $ Let $z \le y \le x$.
Then:
{{begin-eqn}}
{{eqn | l = x \overline \wedge \paren {y \underline \vee z}
      | r = x \overline \wedge z = x
      | c = 
}}
{{eqn | l = \paren {x \overline \wedge y} \underline \vee \paren {x \overline \wedge z}
      | r = x \underline \vee x = x
      | c = 
}}
{{eqn | o = 
      | c = 
}}
{{eqn | l = x \underline \vee \paren {y \overline \wedge z}
      | r = x \underline \vee y = y
      | c = 
}}
{{eqn | l = \paren {x \underline \vee y} \overline \wedge \paren {x \underline \vee z}
      | r = y \overline \wedge z = y
      | c = 
}}
{{end-eqn}}
Thus in all cases it can be seen that the result holds.
{{qed}}
Category:Max and Min Operations
Category:Examples of Distributive Operations
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