simplifying boolean expressions

Discussion in 'Homework Help' started by mikee, Aug 5, 2010.

  1. mikee

    Thread Starter New Member

    Aug 4, 2010
    4
    0
    hi
    can someone help me in simplifying this expression:

    (v'+u+w)(wx+y+uz')+(wx+uz'+y)

    and another one:

    (w'+x)yz'+(w''+x')yz'

    ...coz i'm not sure with my answers..
    i badly needed it now... tnx :)
     
  2. mps

    New Member

    Feb 4, 2010
    8
    0
    the second 1 is damn simple
    yz'(w+w'+x+x')
    2yz'

    for the 1st one open the bracket and apply properties of boolean
     
  3. vvkannan

    Active Member

    Aug 9, 2008
    138
    11
    Hi,
    1+1 = 1 and hence it must be yz'.
    And for the first one expand the terms and try simplifying applying the rule
    A+A' =1
     
  4. somenos

    New Member

    Jan 21, 2010
    2
    0
    wxv' + yv' + uz'c' + wxu + uy + uz' + wx + wy + wuz' + wx + uz' + y

    y(1+v'+u+w) + wx(1+v'+u) + z'u(1+v'+w)

    y + wx + z'u
     
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