boolean algebra problem

Discussion in 'Homework Help' started by randb, Dec 29, 2012.

  1. randb

    Thread Starter New Member

    Dec 29, 2012
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    Trying to simplify ab'c(bd+cde)+ac'. book answer indicates a(c'+b'de) I get a(c"+b'cde). How was the c eliminated? need help

    randb
     
  2. justtrying

    Active Member

    Mar 9, 2011
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    irrelevant, misread the question.
     
    Last edited: Dec 30, 2012
  3. WBahn

    Moderator

    Mar 31, 2012
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    But there's no point simplifying it further if it is wrong to begin with.

    So first determine if

    a(c'+b'de) =? a(c"+b'cde)

    (c'+b'de) =? (c"+b'cde)

    On the left hand side, if c is false, then the expression is true regardless of the other values. Is that also the case for the right hand side? If not, then you know that one of them is wrong. So ask the same sanity check question of the original expression.

    ab'c(bd+cde)+ac' = a(b'c(bd+cde)+c')

    If c is false, then the second factor on the right hand side is true regardless of the other values. So the book answer at least has a chance of being correct, but your's does not. So step back and redo your work until you pass all the sanity checks you can think of.
     
  4. randb

    Thread Starter New Member

    Dec 29, 2012
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    Sorry I just realized when checking replies I had made a typo on the answer I came up with in my post. Should be a(c'+b'cde) not c''.
     
  5. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    In boolean algebra, when you have (X'+XY), it can be simplified to X'+Y. Now, by applying this rule to the inside of the parentheses in your answer, you have c' on one side of the "OR" and c on the other. So because of that, you can eliminate the c on the right-hand side.
     
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  6. randb

    Thread Starter New Member

    Dec 29, 2012
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    DerStrom8

    Thank you
     
  7. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    Anytime randb. Glad to help!

    By the way, welcome to AAC!
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    But be sure you understand why you can make this simplification.

    In words, if X is False, then the expression is True because of the first term irrespective of the second. But if X is True, then the first term doesn't matter (we know it is False) and in the second term we know that X is True so it really doesn't matter there, either.

    Using Boolean Algebra:

    X'+ XY = X'(Y+Y')+ XY
    = X'Y + X'Y' + XY
    = (X'Y + X'Y) + X'Y' + XY
    = (X'Y + X'Y') + (X'Y + XY)
    = X'(Y+Y') + (X'+X)Y
    = X' + Y
     
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  9. randb

    Thread Starter New Member

    Dec 29, 2012
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    WBahn

    Thank you for the info.
     
  10. WBahn

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    Mar 31, 2012
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    You are more than welcome. Look forward to seeing you around AAC!
     
  11. randb

    Thread Starter New Member

    Dec 29, 2012
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    I also read that any OR can be distributed over an AND operation?

    a+(a'*b)= (a+a')(a+b)
    1*(a+b)
    a+b

    could this also be used?

    thank you
     
  12. WBahn

    Moderator

    Mar 31, 2012
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    Yes. Actually that's a much cleaner way to do it, but distributing the OR over AND is a property that few people are comfortable with because it is a property that does not hold for normal addition over multiplication and therefore violates the properties that we normally associated with those operator symbols. So, to use it, you may need to convince your audience that you really can do it (depending, of course, on your audience).

    It's actually a good little exercise to see if you can prove it via algebraic manipulation (as opposed to exhaustive brute force via a truth table).
     
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