Boolean simplification(1.4.3 c) Flashcards
Double negation
¬(¬A) = A
Commutation
the order of variables for an operator doesn’t matter
A∧B = B∧A
A∨B = B∨A
Association
allows us to fiddle with/remove brackets
OR
A∨(B∨C) = (A∨B)∨C = A∨B∨C
AND
A∧(B∧C) = (A∧B)∧C = A∧B∧C
Distribution
OR
A∧(B∨C) = (A∧B)∨(A∧C)
AND
A∨(B∧C) = (A∨B)∧(A∨C)
Absorption
applies when a variable is referenced inside and outside a bracket
OR
A∨(A∧B) = A
A∧(A∨B) = A
De Morgans first law
¬(A∨B) = ¬A∧¬B
the operator changes
De Morgans second law
¬(A∧B) = ¬A∨¬B
the operator changes
AND gate rules
if an input is zero the input will always output zero(X∧0 = 0)
if an input is always one the output will will always be the other input(X∧1 = X)
if x is zero than both inputs are zero if x is one both inputs are one (X∧X = X)
if an input is one then another must be zero and vice versa(X∧¬X = 0)
OR gate rules
if an input is zero the output will be what the other input is(X∨0 = X)
if an input is always one the output will always be one(X∨1 = 1)
if x is zero then both inputs are zero if x is one then both inputs are one(X∨X = X)
if an input is one then the other leg must be zero and vice versa(X∨¬X = 1)
X∧0 = 0 AND gate
if an input is zero the input will always output zero
X∧X = X AND gate
if x is zero than both inputs are zero if x is one both inputs are one
X∧1 = X AND gate
if an input is always one the output will will always be the other input
X∧¬X = 0 AND gate
if an input is one then another must be zero and vice versa
X∨0 = X OR gate
if an input is zero the output will be what the other input is
X∨1 = 1 OR gate
if an input is always one the output will always be one