C# 3.0 Cookbook, 3rd Edition

3.7. 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
TI	x | x == x
T2	x | !x == true
T3	!x | !y == !(x & y)
T4	x & x == x
T5	x & !x == false
T6	!x & !y == !(x | y)
T7	x | y == y | x
T8	(x | y) | z == x | (y | z)
T9	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	x & y == y & x
T14	(x & y) & z == x & (y & z)
T15	(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 IDs allow easy identification of which theorems are being used in the Boolean equations that are being minimized in the Discussion section.

Discussion

Simplifying your Boolean logic will benefit your code by making it less cluttered and making its logic clearer and more readily understood. This simplification ...