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Boolean Equation Simplification

  1. Aug 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Q1: Show that
    a) ~(~(A+B)~C.A).~(~(B.C)+A.B) = ~(~(C.~A)+~B.C)~(A.~B.~(C+~B))
    a-2) *same as above* ~(~(A+B).C'.A).~(~(B.C)+A.B) = ~(~(C.A')+B'.C).~(A.B'.~(C+B'))
    b) ~(A XOR B) = (~A) XOR B
    c) A + A'B = A + B

    Q2: Represent the following using only NAND gates, and only NOR gates
    a) A.B + ~(A.C).~(B+C)
    b) (A XOR B) + ~(A XOR B).(B + C)
    c) (A+B).(~(A+C)+~(B.C))

    2. Relevant equations
    3. The attempt at a solution
    LHS = A + (~A)BC = A + BC
    RHS = ~A + BC
    *where did i do wrong?
    b) Tried Truth Table
    A B || ~(A XOR B)
    0 0 || 1
    0 1 || 0
    1 0 || 0
    1 1 || 1

    A B ~A|| (~A) XOR B
    0 0 1 || 0
    0 1 1 || 1
    1 0 0 || 1
    1 1 0 || 0

    Is this suppose to be correct? i find it confusing...to me
    (~A) XOR B = A XOR B...but ofcourse the question is asking to prove
    ~(A XOR B) = (~A) XOR B!

    c) i'm quite sure this was distributive law...but how do i prove this WITHOUT using Karnaugh map?


    I : Trial and error solved the question backward!!!
    II: I first simplied the question using theorems & K-map, then with the result figured out the NOR/NAND equation.
    a) i got 2 solutions on this one..(???!!!!)
    I : ~(~A+~B~(~A+~C)+~(~B.~C)) or in different form...~(A'+B'~(A'+C')+~(B'C'))
    II: ~(A~BC) or in different format...~(AB'C)

    Method used
    b) dunno where i have to begin this since i ain't that great with XOR complements...If someone could give me any tips i'd appreciate it!
    c) ~(~A+B(~B)+C) or in different form...~(A'+BB'+C)
  2. jcsd
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