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Boolean Algebra Trouble

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data

    1.
    Simplify the following problem using boolean algebra:
    (p+q'r')(p'q'+r)

    2.
    Use algebra or karnaugh maps to simplify the following circuit:
    (z+xy')+yx+xzy'


    2. Relevant equations
    Boolean algebra rules and demorgan's law


    3. The attempt at a solution

    1. (p+q'r')(p'q'+r)
    =pq'p'+pr+q'r'p'q'+q'r'r
    =0q'+pr+'q'r'p+0q
    =pr+q'r'p'
    =0r+q'r'
    =00+q'
    =q'
    Not sure if correct

    2.(z+xy')+yx+x(z+y')
    =zxy'+yx+xzy'
    =zxy'+1x+xz
    =x(zy'+z)
    =xz
    Not sure about this one either
     
    Last edited: Sep 29, 2011
  2. jcsd
  3. Sep 29, 2011 #2

    LCKurtz

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    OK to there. Then next equation is wrong. Use the identity a + a'b = a + b.
    What rule did you use to get that step?
     
  4. Sep 29, 2011 #3
    So i get
    pr+q' for the first one
    and i used demorgan's law to get to the second step in the second question
     
  5. Sep 29, 2011 #4

    LCKurtz

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    Actually, after looking at it a bit more leisurely than I did this morning, I think you should have just stopped at pr + q'r'p' which is correct. I don't think my suggestion was good.

    For the second one, use the associative law to group the two center terms first to get started.
     
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