Help with boolean simplification

Discussion in 'Homework Help' started by exothesis, Aug 31, 2010.

  1. exothesis

    Thread Starter New Member

    Aug 31, 2010
    3
    0
    Simplify XY'Z + X'Y'Z + XYZ

    So far I have got:
    Y'Z(X + X') + XYZ
    --> Y'Z + XYZ

    What do I do next? Any help would be appreciated.
     
  2. exothesis

    Thread Starter New Member

    Aug 31, 2010
    3
    0
    Never mind, I managed to solve the problem

    --> XY'Z + X'Y'Z + XYZ
    --> Y'Z(X + X') + XYZ
    --> Y'Z(1) + XYZ
    --> Y'Z + XYZ
    --> Z(Y' + XY)
    --> Z((Y' + X)(Y' + Y))
    --> Z((Y' + X)(1))
    --> Z(Y' + X)
    --> Y'Z + XZ
     
  3. Georacer

    Moderator

    Nov 25, 2009
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    1,266
    I' not quite sure this is correct. In step 6:
    (Y'+X)(Y'+Y)=Y'Y'+Y'Y+XY'+XY=Y'+XY'+XY
    wich is different from
    Y'+XY
     
  4. exothesis

    Thread Starter New Member

    Aug 31, 2010
    3
    0
    Complement law - (Y + Y') = 1
    Therefore it follows from Identites that
    (Y' + X)(Y' + Y)
    = (Y' + X)(1)
    = Y' + X
     
  5. Georacer

    Moderator

    Nov 25, 2009
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    1,266
    I don't argue about how (Y' + X)(Y' + Y) becomes (Y' + X).
    What troubles me is how (Y' + XY) becomes (Y' + X)(Y' + Y)
     
  6. AMIT_GOHEL

    Member

    Jul 13, 2010
    67
    7
    Hello Georacer,

    You may have confusion between math's we everyday use nd the boolean maths..

    What you have asked is Distributive low of Boolean algibra.

    Regards...
     
  7. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    You are correct indeed! It's been a while since I read my Boolean Algebra manual (Morris Mano). It seems in Boolean logic the * operation (and) is distributive for the + (or) operation
    i.e.  x \cdot (y + z)= x \cdot y + x \cdot z

    but also the + operation is distributive for the * operation
    i.e.  x+(y \cdot z)=(x+y)\cdot (x+z)

    There's todays minimum positive knowledge for me!
     
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