Simplifying Boolean Expressions

Boolean Operators (T/F) xyx AND y FFF FTF TFF TTT xyx OR y FFF FTT TFT TTT xyx XOR y FFF FTT TFT TTF xNOT x FT TF

Boolean Operators (1/0) xyx AND y xyx OR y xyx XOR y xNOT x 01 10

Boolean Operators Symbols OperatorSymbol NOTā (overbar), a’, ~a AND· (mult. dot) OR+ XOR  (plus sign with circle around it)

Simplifying Boolean Expressions Commutative laws A + B = B + A A · B = B · A Associative laws A + (B + C) = (A + B) + C A · (B · C) = (A · B) · C Distributive laws A · (B + C) = A · B + A · C A + (B · C) = (A + B) · (A + C)

Simplifying Boolean Expressions Tautology laws A · A = A A + A = A A + ~A = 1 A · ~A = 0 Absorption Law A + (A · B) = A A · (A + B) = A

Simplifying Boolean Expressions Identities 0 · A = A = A A + 1 = 1 1 · A = A A = A Complement A + ~A · B = A + B

Examples A + A + A + A = A Using the Tautology law

A Bigger Example Simplify ~A · B + A · ~B + ~A · ~B ~A · B + A · ~B + ~A · ~B ~A · B + (A · ~B + ~A · ~B)  Associative ~A · B + (~B · (A + ~A))  Distributive ~A · B + ~B & Tautology ~A + ~B  Complement Verify with a truth table!

Practice Show that A + B · C = (A + B) · (A + C) is true using a truth table.

Practice Show that A + ~A · B = A + B

Practice Simplification Simplify A + AB + ~B and verify with a truth table

De Morgan’s Laws ~(A · B) = ~A + ~B ~A · ~B = ~(A+B) 1. Take a term ~A · ~B 2. NOT the individual members of the term A · B 3. Change the operator i.e. · to +, or + to · A + B 4. NOT the entire term ~(A+B)

De Morgan’s Law Example f = ~A · ~B + (~A + ~B) = ~~( ~A · ~B + (~A + ~B) )  NOT NOT = ~( (A + B) · ~(~A + ~B) )  De Morgan’s = ~( (A + B) · (A·B) )  De Morgan’s = ~( A·(A·B) + B·(A·B) )  Distributive = ~( A·B + A·B )  Tautology = ~(A·B)  Tautology