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Boolean Expression Simplification

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Simplify the following expressions
    (x'y'+xy'+x'y)

    (p+q'p)(p+qr)


    2. Relevant equations
    Laws of boolean algebra

    3. The attempt at a solution
    For the first one I've found many ways to solve it... something isn't right here
    (x'y'+xy'+x'y)
    =x'(y'+y)+xy'
    =x'+xy'
    =x'+x+y'
    =y'

    OR

    =y'(x'+x)+x'y
    =y'+x'y
    =y'+x'+y
    =x'
    I'm definantly doing something wrong here... probably making up laws :biggrin:

    For the second I'm not sure if the last line is even possible
    (p+q'p)(p+qr)
    =pp+pqr+ppq'+q'qr
    =p+pqr+pq'+0
    =p+pqr
    =p? Can I do that with pqr having 3 variables?
     
  2. jcsd
  3. Oct 24, 2011 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member


    That is wrong. Apply:

    a+a'b=a+b

    For the second question, replace the fist factor by p+q'p=p. You made a mistake in the derivation, but the result is p.

    ehild
     
  4. Oct 24, 2011 #3
    Ahh so the first is
    =x'+y'

    and with the second if I change it to:
    p(p+qr)
    won't I end up with
    =p+pqr in the end anyway?

    Also how do you know what rule you're supose to use? Is there a specific order or is it just whatever you notice first?
     
  5. Oct 24, 2011 #4

    ehild

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    Homework Helper
    Gold Member

    I said the result was correct. You made a mistake as the last term was q'qpr, but it cancelled because of q'q.

    There are no general recipes.
    Use what you notice first, but it helps if you notice when something cancels, for example E+a=E, a+a'=E, aa'=0. Factor out what you can, so a+ab=a(E+b)=a. But it is not a rule; sometimes you need to apply other tricks.

    ehild
     
  6. Oct 24, 2011 #5
    Ahh fair enough, thanks for the help.
     
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