Lecture 3: Boolean Algebra The big picture Boolean algebra Part of the combinational logic topics (memoryless) Axioms Different from sequential logic (can store Useful laws and theorems information) Examples Axioms and theorems allow you to… … design logic functions … know how to combine different logic gates … simplify or optimize complex operations 1 2 Boolean algebra Boolean algebra axioms 1. Closure: 4. Identity: A Boolean algebra consists of… a+b is in B a+0 = a a set of elements B a•b is in B a•1 = a binary operators (+ , •) _ 2. Commutative: 5. Distributive: unary operator (' or ) a+b = b+a a+( b•c) = (a+b)•(a+c) a•b = b•a a•(b+c) = (a•b)+( a•c) 3. Associative: 6. Complementarity: a+( b+c) = ( a+b)+c a+a' = 1 a•(b•c) = ( a•b)•c a•a' = 0 3 4 Binary logic Logic gates and truth tables XYZ Axioms hold for binary logic where AND X•Y XY 0 0 0 0 1 0 B = {0, 1} X Y Z 1 0 0 1 1 1 • → AND + → OR OR X + Y XYZ 0 0 0 ' → NOT 0 1 1 X Z Y 1 0 1 1 1 1 _ A Boolean function maps some number of NOT X' X inputs over {0, 1} into an output set {0, 1} XY X Y 0 1 1 0 5 6 Boolean expressions Precedence Any logic function that is expressible as a 1. Parentheses truth table can be written in Boolean 2. NOT algebra. 3. AND XYZ Z=X•Y X Y X' Z Z=X'•Y 0 0 0 0 0 1 0 0 1 0 0 1 1 1 4. OR 1 0 0 1 0 0 0 1 1 1 1 1 0 0 ____ X Y X' Y' X •Y X' •Y' Z Z=(X•Y)+(X'•Y') Example: A+B •C = (A)+(B •(C)) 0 0 1 1 0 1 1 0 1 1 0 0 0 0 1 0 0 1 0 0 0 1 1 0 0 1 0 1 7 8 Duality Useful laws and theorems Duality (a meta-theorem—a theorem about Identity theorems ) X+0 = X Dual: X•1 = X All Boolean expressions have logical duals Null Any theorem that can be proved is also proved X+1 = 1 Dual: X•0 = 0 for its dual Idempotent Replace: • with +, + with •, 0 with 1, and 1 with 0 Leave the variables unchanged X+X = X Dual: X•X = X Involution (X')' = X Example: The dual of X+0= X is X•1= X Complementarity X + X' = 1 Dual: X•X' = 0 9 10 Useful laws and theorems Useful laws and theorems Commutative Absorption X+Y = Y+X Dual: X•Y = Y•X X+X•Y = X Dual: X•(X+Y) = X Associative (X+Y')•Y = X•Y Dual: (X•Y')+ Y=X+Y X+( Y+Z) = ( X+Y)+ Z Dual: X•(Y•Z) = ( X•Y)•Z Distributive Consensus X•(Y+Z) = ( X•Y)+( X•Z) Dual: X+( Y•Z) = ( X+Y )•(X+Z ) X•Y+Y•Z+X'•Z = X•Y+X'•Z Uniting Dual: ( X+Y)•(Y+Z)•(X'+ Z) = ( X+Y)•(X'+ Z) X•Y+X•Y' = X Dual: (X+Y)•(X+Y') = X Multiplying and factoring (X+Y)•(X'+ Z) = X•Z+X'•Y Dual: X•Y+X'•Z = ( X+Z)•(X'+ Y) 11 12 DeMorgan's law Proving theorems Procedure for complementing Boolean functions Example 1: Uniting theorem Replace: • with +, + with •, 0 with 1, and 1 with 0 X•Y+X•Y' = X•(Y+Y') Distributive Replace all variables with their complements = X•(1) Complementarity = X Identity Look familiar? Duality: "Leave the variables unchanged" Example 2: Absorption However, duality and DeMorgan's are NOT the same thing! X+X•Y = X•1+ X•Y Identity = X•(1+ Y) Distributive Example: ___ The complement of F = X•Y is F = X+Y = X •(1) Null = X Identity 13 14 Proving theorems Applying DeMorgan's Example 3: Consensus Find the complement of F=(A+B)•(A’+C). X•Y+Y•Z+X'•Z Answer: F’=(A’•B’)+(A•C’) = XY + (1) YZ + X'Z Identity A B C F A B C F’ = XY + ( X+X') YZ + X'Z Complementarity 0 0 0 0 0 0 0 1 = XY + XYZ + X'YZ + X'Z Distributive 0 0 1 0 0 0 1 1 = XY + X'YZ + X'Z Absorption 0 1 0 1 0 1 0 0 0 1 1 1 0 1 1 0 {AB + A = A} with A = XY and B = Z 1 0 0 0 1 0 0 1 = XY + X'Z Absorption 1 0 1 1 1 0 1 0 {AB + A = A} with A = X'Z and B = Y 1 1 0 0 1 1 0 1 1 1 1 1 1 1 1 0 15 16 Logic simplification Z = A'BC + AB'C' + AB'C + ABC' + ABC = A'BC + AB'(C' + C) + AB(C' + C) Distributive = A'BC + AB'(1) + AB(1) Complementarity = A'BC + AB' + AB Identity = A'BC + A(B' + B) Distributive = A'BC + A(1) Complementarity = A'BC + A Identity = BC + A Absorption {( X•Y')+ Y=X+Y} with X = BC and Y = A 17.