boolean algebra

Discussion in 'Homework Help' started by ann, Feb 22, 2008.

  1. ann

    Thread Starter Member

    Feb 15, 2008
    14
    0
    Can this boolean algebra be reduced further? f=/cd+/ab+/a/c+a/bcd
     
  2. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    144
    Yes.

    Double NOT the function (this doesn't change the function), so: f=//(/cd+/ab+/a/c+a/bcd).

    Now think about DeMorgan's theorems.

    Have a go and post up your efforts, I'll give you some feedback.

    Dave
     
  3. ann

    Thread Starter Member

    Feb 15, 2008
    14
    0
    I did the boolean algebra again this time using k maps and realized I had made a mistake in reducing my equation. The reduced one now is /CD+/AB+/A/C+A/BD and I don't think this can be reduced further. Can it?
     
  4. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    144
    Sorry I missed this one. Feel free to PM to remind me in future.

    Without your working out I cannot give you any feedback on your answer, but I simplified it to:

    f = (ABCD)'

    (Note I use ' to denote NOT/INVERTED)

    I used the two techniques I told you about above. How?

    f = (CD)' + (AB)' + (A'B') + (AB'CD)

    Double NOT (this doesn't change the function):

    f = ((CD)' + (AB)' + A'B' + AB'CD)''

    Take out a common factor in the last two terms:

    f = ((CD)' + (AB)' + B'( A' + ACD))''

    Apply DeMorgan's theorem:

    f = ((CD)''(AB)''(B'(A' + ACD))')'

    (CD)'' = CD; (AB)'' = AB; so:

    f = (CDAB(B'(A' + ACD))')'

    Apply DeMorgan's theorem on (B'(A' + ACD))':

    f = (CDAB(B'' + (A' + ACD)'))'

    B'' = B

    Apply DeMorgan's theorem on (A' + ACD)':

    f = (CDAB(B + (A''(ACD)')))'

    A'' = A

    Apply DeMorgan's theorem on (ACD)':

    f = (CDAB(B + (A(A' + C' + D'))))'

    Multiply out (A(A' + C' + D'):

    f = (CDAB(B + (AA' + AC' + AD')))'

    AA' = 0

    f = (CDAB(B + AC' + AD'))'

    Multiply everything out:

    f = (CDABB + CDABAC' + CDABAD')'

    BB = B; CC' = 0; DD' = 0; therefore

    f = (CDAB)' = (ABCD)'

    Not easy eh?!

    Dave
     
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