C# Cookbook

3.12. Minimizing (Reducing) Your Boolean Logic

Problem

Many times a Boolean equation quickly becomes large, complex, and even unmanageable. You need a way to manage this complexity while at the same time verifying that your logic works as designed.

Solution

To fix this situation, try applying the theorems shown in Table 3-1 to minimize these types of equations.

Table 3-1. Boolean theorems

Theorem ID	Theorem definition
T0	!(!x) == x
T1	x | x == x
T2	x | !x == true
T3 (DeMorgan’s Theorem)	!x | !y == !(x & y)
T4	x & x == x
T5	x & !x == false
T6 (DeMorgan’s Theorem)	!x & !y == !(x | y)
T7 (Commutative Law)	x | y == y | x
T8 (Associative Law)	(x | y) | z == x | (y | z)
T9 (Distributive Law)	x & y | x & z == x & (y | z)
T10	x | x & y = x
T11	x & y | x & !y = x
T12	(x & y) | (!x & z) | (y & z) == (x & y) | (!x & z)
T13 (Commutative Law)	x & y == y & x
T14 (Associative Law)	(x & y) & z == x & (y & z)
T15 (Distributive Law)	(x | y) & (x | z) == x | (y & z)
T16	x & (x | y) = x
T17	(x | y) & (x | !y) = x
T18	(x | y) & (!x | z) & (y | z) == (x | y) & (!x | z)
T19	x | x | x | ... | x == x
T20	!(x | x | x | ... | x) == !x & !x & !x & ... & !x
T21	x & x & x & ... & x == x
T22	!(x & x & x & ... & x) == !x | !x | !x | ... | !x
T23	(x | y) & (w | z) == (x & w) | (x * z) | (y & w) | (y * z)
T24	(x & y) | (w & z) == (x | w) & (x | z) & (y | w) & (y | z)

In Table 3-1, assume that w, x, y, and z are all variables of type bool. The Theorem ID column ...