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How to simplify the expression -- Boolean algebra

  1. Apr 13, 2015 #1
    1. prove that:
    X'Y'Z + X'YZ' + XY'Z' + XYZ = (X⊕Y)⊕Z


    2. Relevant equations
    Use postulates and theorems.

    3. The attempt at a solution
    X'Y'Z + X'YZ' + XY'Z' + XYZ (original expression)
    X'Y'Z + X'YZ' + X(Y'Z' + YZ) (distributive)
    X'Y'Z + X'YZ' + X.1 (complement)
    X'Y'Z + X'YZ' + X (identity)


    need help.
    also may you expand the expression
    : (X⊕Y)⊕Z

    I think it is like this: (XY' + X'Y)(ZXY' +Z'X'Y)
    but i do not know if it's right
     
  2. jcsd
  3. Apr 13, 2015 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi,

    Not familiar with the ' notation, but I suppose X' means ##\neg X##

    When you expand ##(X\oplus Y) \oplus Z## you just write it out using ##(A\oplus B) = A'B + AB' ## twice.

    For me the easiest part is the second term ## (X'Y + XY') Z' = X'Y Z'+ XY' Z'## which are the 2nd and 3rd in the left hand side of your original staement.

    Leaves you to prove ## (X'Y + XY')' Z = X'Y' Z + XY Z##; not that hard, I hope ?
     
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