# 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.