I am new to boolean algebra so having these two question to simplify hope you guys will help me, Q- Using Demorgan's theorem, show that a. (A + B)' (A' + B')' = 0 b. A + A'B + A'B' = 1
hi, as far as part b is concerned here is the proof: A+ A'B +A'B' = A + A' ( B + B' ) = A + A' = 1 because by demorgan laws, A + A' =1 part a is not clear what is that emoticon doing in between? B)
DeMorgans theorem says: (x AND y)' = x' OR y' (x OR y)' = x' AND y' For your example a: a. (A + B )' (A' + B')' = 0 This is composed of two expressions ANDed together Exp1 is (A + B )' Exp2 is (A' + B')' Convert each expression using DeMorgans Theorem Exp1 becomes A'B' Exp2 becomes AB So, does ( A'B' )( AB ) = 0 ? ( A'B' )( AB ) = A'B'AB -> associative property = A'AB'B -> associative = (A'A)(B'B') -> associative = (0)(0) -> complements identity = 0 -> Done So, (A + B )' (A' + B')' = 0 Perion
well sikander , one of the best ways to solve boolean algebra probs is to use Venn diagrams , u can really solve problems very quickly . |--------|--|------| |--------|--|------| |----1-- |3|---2--| |--------|- |-------| 1 is A\B 2 is B\A 3 is A intersection B a+a'b=the total area which is a+b now a' wud include all the space including B b' all space include all space including A their intersection wud include all the space except a and B so a+b + all space not including a+b=all space = 1 well ,the figure looks complicated but this is really easy .