Boolean Algebra algebraic system now called Boolean Algebra . In - PDF document

Chapter 4 History � In 1854 George Boole introduced systematic treatment of logic and developed for this purpose an Boolean Algebra algebraic system now called Boolean Algebra . � In 1938 C. E. Shannon introduced a two-valued Boolean Algebra called Switching Algebra , in Computer Application which he demonstrated that this algebra can represented by electrical switching. Flaxer Eli - ComputerAppl Ch4 - 1 Flaxer Eli - ComputerAppl Ch4 - 2 Chapter Outline Boolean Algebra Definitions � Boolean (Switching) Algebra structure consist: � Boolean Algebra ( Switching Algebra ) – Set of elements (constant) B = {0, 1}. - Definitions – Binary operation O = {+, .}. - Basic Axioms – Unary operation {‘}. - Basic Theorems – The following axiom: - Representation of Boolean Functions Closure: a + b in B a . b in B Commutative: a + b = b + a a . b = b . a Associative: a + ( b + c) = (a +b) + c a . ( b . c) = (a .b) . c Identity: a + 0 = a a . 1 = a a + 1 = 1 a . 0 = 0 Distributive: a + (b . c) = (a + b) . (a + c) a . (b + c) = (a . b) + (a . c) Complement: a + a’ = 1 a . a’ = 0 Flaxer Eli - ComputerAppl Ch4 - 3 Flaxer Eli - ComputerAppl Ch4 - 4 Definitions Basic Axioms � Boolean Algebra : An algebraic structure defined with � A variable can take only one of two values{0,1} a set of elements B={ 0,1 }, a set of binary operators ( A 1 ) X= 0 if X ≠ 1 ( A 1 ’ ) X=1 if X ≠ 0 ( + , . , ‘ ), and a number of unproved axioms. � Symbolic Variables such as X, Y , Z represent the � NOT operation ( The complement Operation ) : elements. A variable can take the value “0” or “1” ( A 2 ) If X=0 then X’=1 ( A 2 ’ ) If X =1 then X’=0 which corresponds to the condition of a logic signal. � AND and OR operations ( Multiplication and Addition) : � Algebraic Operators : ( A 3 ) 0 . 0 = 0 ( A 3 ’) 0 + 0 = 0 - Addition operator ( + ) ( A 4 ) 1 . 1 = 1 ( A 4 ’) 1 + 1 = 1 - Multiplication operator ( . ) ( A 5 ) 0 . 1 = 1 . 0 = 0 ( A 5 ’) 1 + 0 = 0 + 1 = 1 - Complement operator ( ‘ ) Flaxer Eli - ComputerAppl Ch4 - 5 Flaxer Eli - ComputerAppl Ch4 - 6 1

Induction Proof Theorems - Single Variable � Null elements : ( T 1 ) X + 0 = X ( T 1 ’ ) X . 1 = X � T 3 : X+X=X � Identity elements : X X X+X 0 0 0 ( T 2 ) X + 1 = 1 ( T 2 ’ ) X . 0 = 0 1 1 1 � Idempotency : ( T 3 ) X + X = X ( T 3 ’ ) X . X = X � T 5 : X+X’=1 � Involution : X X’ X+X’ ( T 4 ) ( X’ )’ = X 0 1 1 � Complements : 1 0 1 ( T 5 ) X + X’ = 1 ( T 5 ’ ) X . X’ = 0 � Induction Proof : Show that the theorems are true for both X=0 and X=1 Flaxer Eli - ComputerAppl Ch4 - 7 Flaxer Eli - ComputerAppl Ch4 - 8 Theorems - Multiple Variables Theorems - Multiple Variables � Commutativity : � Covering : (T 6 ) X + Y = Y + X (T 6 ’) X . Y= Y . X (T 9 ) X+X.Y = X (T 9 ’) X.(X+Y)= X - The inputs of AND and OR gates can be interchanged. � Proof : - T 9 : X+X.Y=X.1+X.Y ( theorem T 1 ’ ) � Associativiy : =X.(1+Y) ( theorem T 8 - Distributivity ) (T 7 ) (X+Y)+Z = X+(Y+Z) (T 7 ’) (X.Y).Z= X.(Y.Z) =X.1 (theorem T 2 ) - The order of the input variables could be rearranged. =X ( theorem T 1 ’ ) � Distributivity : - T 9 ’: X.(X+Y) = (X+0).(X+Y) ( theorem T 1 ) (T 8 ) X.(Y+Z)= X.Y+X.Z (T 8 ’) X+Y.Z=(X+Y).(X+Z) = X+(0.Y) ( theorem T 8 ’- Distributivity ) - Multiplication distributes over addition = X+0 ( theorem T 2 ’ ) - Addition distributes over multiplication !!! = X ( theorem T 1 ) Flaxer Eli - ComputerAppl Ch4 - 9 Flaxer Eli - ComputerAppl Ch4 - 10 Duality DeMorgan � Duality : Every Boolean expression remains valid if the � DeMorgans Theorems (AND, OR) operators and {0,1} elements are interchanged. (T 13 ) (X 1 .X 2 . ... . X n )’ = X 1 ’+X 2 ’+...+X n ’ � Mathematical definition : if F is a Boolean Function then FD, (T 13 ’) (X 1 +X 2 + ... +X n )’= X 1 ’.X 2 ’. ... . X n ’ the dual, function is : � FD(X 1 ,X 2 , ... , X n , + , . ,’ ) def as F( X 1 ,X 2 ,....,X n , . ,+ ,’) Example : two-variable case � FD ≠ F ≡ X1 X1 ( X1.X2 )’ (X1’+X2’) � Example : X2 X2 - F(X 1 ,X 2 ,X 3 ) = X 1 +X 2 .X 3 FD(X 1 ,X 2 ,X 3 ) = X 1 .(X 2 +X 3 ) ≡ X1 X1 ( X1+X2 )’ (X1’.X2’) X2 X2 Flaxer Eli - ComputerAppl Ch4 - 11 Flaxer Eli - ComputerAppl Ch4 - 12 2

Exercise : Demorgan/Duality � F(A,B,C) = AB+AB’C+BC’ Find : F’ , FD � F’ = [AB+AB’C+BC’]’= (A’+B’)(A’+B+C’)(B’+C) FD = (A+B)(A+B’+C)(B+C’) FD(A’,B’,C’) = (A’+B)(A’+B+C’)(B’+C) FD’(A’,B’,C’ ) = [(A’+B’)(A’+B+C’)(B’+C)]’ = AB+AB’C+BC’ Flaxer Eli - ComputerAppl Ch4 - 13 3