Product of Sums simplification

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

  1. frazzman

    Thread Starter New Member

    Apr 8, 2008
    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.

  2. Mark44

    Well-Known Member

    Nov 26, 2007
    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.
  3. Mark44

    Well-Known Member

    Nov 26, 2007
    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
    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.