Help in Boolean algebra simplification:(W/ solution)

Discussion in 'Homework Help' started by FullMetalEngineer, Jan 15, 2009.

  1. FullMetalEngineer

    Thread Starter New Member

    Jul 22, 2008
    4
    0
    Simplify the ff. Boolean function using De morgan's identity and or boolean identity and or special logic functions:
    note ' is the complemented value

    a.)F(w,x,y,z) = [(y+x)(w+z)']' ==> Ans: (w'x' + w + z)
    = (y+x)' + (w+z) de morgan
    = y'x' + w + z ????

    b.)F(X,Y,Z) = [(X + Y' + Z')(X + Z')]' ==>Ans: X'YZ
    = (X + Y' + Z')' + (X + Z')' de morgan
    = X'(Y'+Z')' + (X'Z) de morgan
    = X'YZ + X'Z
    = X'Z(Y+1)
    = X'Z ????

    c.) F(x,y,z) = x'yz' + x'yz + (xyz' + xyz)' ==> Ans: y
    = x'y(z' + z) + (xy(z' + z))'
    = x'y + (xy)'
    = x'y + x' + y'
    = x' + y' ?????

    are the answers obtainable through simplifying or is it just a typo?
     
  2. vvkannan

    Active Member

    Aug 9, 2008
    138
    11
    Hello FullMetalEngineer,
    I too get the same answers as you have got.Might be an error
     
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