Need help quick boolean equation

Discussion in 'Homework Help' started by romo218, Feb 14, 2007.

  1. romo218

    Thread Starter New Member

    Feb 14, 2007
    2
    0
    Hey, i have this equation and i need to simplify it:
    x'yz + xy'z + xyz'


    i know it = xy + xz + yz
    i just dont know how to get there

    Its for when exactly 2 inputs are high

    i have tried many ways and i cant find out the right way.

    i need the way of getting it using laws and theorems, not the other grid way (cant remember name)

    thanks
     
  2. romo218

    Thread Starter New Member

    Feb 14, 2007
    2
    0
    please, if you have any ideas please tell me

    i need it by 10:00pm tonight

    :(
     
  3. chameleon

    New Member

    Feb 16, 2007
    1
    0
    but they are not equal!

    try x =1 , y=1 , z=1.
     
  4. beautifulmind

    New Member

    Mar 5, 2007
    4
    0
    Double check on this because i have done this ages ago since graduation

    x’yz + xy’z +xyz’

    =x’yz +x(y’z+yz’) => use truth table to prove that a’b+ab’= a +b or u may use Boolean algebra
    =x’yz+x(y+z)
    =x’yz + xy +xz
    =y(x’z+x) +xz applying a’b +a = a+b
    =y(x+z) +xz
    =yx+yz+xz
    :) smile
     
  5. Papabravo

    Expert

    Feb 24, 2006
    9,906
    1,723
    a'b + ab' is not a + b unless the second plus is exclusive OR but
    a'b + ab' + ab = a + b that I would believe
     
  6. beautifulmind

    New Member

    Mar 5, 2007
    4
    0
    yes i chose the sign +' i forgot the symbol usually we draw a circle around the + but u dont have amathematical editor in this forum and they are available u may add it

    i will put it in red color as well, the previouse solution was short

    Knowing shanon x+' y = xy' + x'y

    x'yz+xy'z+xyz'
    x(y'z+yz'+x'yz
    z(y+' z)+x'yz i will use +' for exlusive or
    yx+' xz+x'yz
    x'yz+yx+' xz
    y(x'z+x) +' xz
    y(x+z)+'xz
    xy+xz +' xz
    xy + z(x'y +xy') by shanon
    xy +zxy'+zx'y by expantion
    x(y+zy') + zx'y
    x(y+z) +zx'y
    xy +xz +zx'y
    xy+z(x+y)
    xy+zx+zy
     
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