Need Confirmation on one of the Boolean Laws

Discussion in 'Homework Help' started by KinkyBlok, Oct 29, 2010.

  1. KinkyBlok

    Thread Starter New Member

    Oct 29, 2010
    1
    0
    I just need to know if the boolean law
    A+BA' = A+B

    can be reversed to

    A' + B'A = A'+B'

    and if you want to help extra. i derived this equation from a truth table and was wondering if you could tell me if i have simplified it wrong in anyway:

    L1=A'B'C+A'BC'+A'BC+AB'C'+AB'C+ABC'

    L1= A'(B'C+BC'+BC) + A(B'C'+B'C+BC')

    L1=A'(B'C+B(C'+C)) + A(B'(C'+C)+BC')

    L1=A'(B'C+B(1)) + A(B'(1)+BC')

    L1=A'(B+C) + A(B'+BC')

    Any help would definitely be appreciated. Thank You
     
  2. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Of course you can use the formula in the form you described.

    You did a fine job in reforming the equation in the second part, but you messed up in the last step.

    L1=A'(B'C+B)+A(B'+BC')
    L1=A'(B+C)+A(B'+C')

    Which, incidentally isn't the same as, A XOR (B + C)
     
  3. zgozvrm

    Member

    Oct 24, 2009
    115
    2
    To answer your 1st question: "Yes"

    In Boolean algebra, addition is distributive over multiplication and multiplication is distributive over addition.

    That is, A(B + C) = AB + AC and A + BC = (A + B)(A + C)

    So, A' + B'A = (A' + B')(A' + A) = (A' + B')(1) = A' + B'
     
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