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Distributivity Theorem in Boolean Algebra

  1. Sep 1, 2011 #1
    Hello. I am wondering if the distributivity theorem works for factoring out more than a single term.

    Distributivity Theorem:
    [tex] BC + BD = B(C + D) [/tex]

    But can I do this:
    [ tex ] ABC + ABD = AB(C + D) [ /tex ]

    Thanks.
     
    Last edited: Sep 1, 2011
  2. jcsd
  3. Sep 1, 2011 #2
    Yes, essentially your just using X(Y+Z)=XY+XZ with X=AB, Y=C and Z=D...
     
  4. Sep 1, 2011 #3
    Awesome. Thanks.
     
  5. Sep 1, 2011 #4
    Okay, so I am using this concept in the following problem but am getting stuck.

    I am trying to simplify the following expression (A' is the complement of A) :

    Y= A'B'C' + A'B'C + A'BC + AB'C + ABC

    Y= A'B'(C' + C) + AB'C + A'BC + ABC

    Y= A'B' + A'BC + AB'C + ABC

    Y = A'B' + AB'C + BC(A + A')

    Y = A'B' + AB'C + BC

    Y = A'B' + C(AB' + B)

    Y = A'B' + C(A + B)

    Y = A'B' + CA + CB


    But when I work out a Karnaugh Map I get Y = A'B' + C

    How does CA + CB reduce to C?
     
  6. Sep 1, 2011 #5
    Stop here. Now continue as

    Y=A'B'+A'B'C+AB'C+BC
     
  7. Sep 1, 2011 #6
    Okay, I'll try that.

    But is there a way to simplify:
    A'B' + CA + CB

    to A'B' + C?

    Or can you just get stuck when simplifying Boolean expressions?

    Because when I work out the K-map on:
    A'B' + CA + CB

    I end up with:
    A'B' + C.
     
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