Max and Min Distributive

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:

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

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

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

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

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

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

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

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

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

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

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

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

$\blacksquare$