Product of Sums simplification

Discussion in 'Homework Help' started by frazzman, Apr 8, 2008.

  1. frazzman

    Thread Starter New Member

    Apr 8, 2008
    1
    0
    Hi guys,
    Just looking for some clarification on product of sums simplification. I have tried to simplifiy this a few times now and my truth tables just dont seem to match up. If you could kindly show what it should be, and how to get there, i would be very appreciative.

    (!A + B + C) (A + B + C)
    where ! represents a compliment.

    The actual sum has a third set of brackets with three more values, but I would like to save some of the learning for me to figure out.

    Thanks,
    Frazz
     
  2. Mark44

    Well-Known Member

    Nov 26, 2007
    626
    1
    Frazz,
    You need to multiply each term in the 2nd sum by each one in the first sum, which gives you 9 products. Then group terms to get the simplified final expression.

    Carrying out the multiplication, I get this:
    !AA + !AB + !AC + BA + BB + BC + CA + CB + CC

    You can get rid of !AA (not A AND A) = the empty set (or 0 as far as the calculation is concerned).
    CC and BB are the same as C and B, respectively.
    !AB + !AC = !A(B + C), and
    AB + AC = A (B + C)
    The two expressions on the right sides immediately above can be combined to make
    (!A + A)(B + C), which is 1(B + C) or more simply, B + C.

    After simplifying as much as possible, I end up with B + BC + C.
    Mark
     
  3. Mark44

    Well-Known Member

    Nov 26, 2007
    626
    1
    From your post it sounds like you have the answer. Whether you do or no, you can check your work by comparing the truth values for (!A + B + C)(A + B + C) and whatever you end up with. Your truth table will need 8 rows for the 3 variables (2^3 = 8), and 5 columns (one each for A, B, and C, one for the product, and one for your simplified result. If the last two columns agree in all 8 rows, you know that what you started with and what you ended up with are equivalent.
     
  4. Mark44

    Well-Known Member

    Nov 26, 2007
    626
    1
    Just for my own reassurance, I made a table as described in my previous post and compared the truth values of (!A + B + C)(A + B + C) and B + BC + C for all possible combinations of the truth values of A, B, and C. Since all eight truth values for (!A + B + C)(A + B + C) and B + BC + C agreed, I was convinced that my answer was correct.
     
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