# simplify pos by hand

#### lbond

Joined Aug 19, 2011
4
I am in need of assistance. I am following a prof on u-tube, he gave a digital simplification problem, f=(a+b+c)(a+b+!c)(!a+!b+!c) [ != compliment]. He gave the final answer as f=(a+b)(a+!c). When I multiply I get the following(hang on to your hat, there are 18 terms) (aa+ab+a!c+a!a+a!b+a!c+ba+bb+b!c+b!a+b!b+b!c+ca+cb+c!c+c!a+c!b+c!c). Im not trying to use the karnaugh, but im doing the simplification by hand using identities. Any help will be greatly appreciated. thanks

#### Zazoo

Joined Jul 27, 2011
114
Try applying theorems before expanding. For example, using:

(x + y!)(x + y) = x

will help simplify the original function.

Edit: Also, I get a different simplified PoS form when minimizing the original function. Are you sure the original and the answer are the same as shown in the video?

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#### lbond

Joined Aug 19, 2011
4
Thank you very much for your input. I will check the problem that's given, and the answer again to be sure. In the mean time your approach hadn't occured to me yet. I have spent two days so far trying to get the answer. I know we are supposed to show our work on here, i have gone thru three sheets of paper front and back. I will show a couple of answers (a+c): (a+b+(b+!c)+c): (!c(a+b)+c), and couple more, but I threw those papers away. Again, thank you very much. I will keep at it, and try the approach you suggest.

#### lbond

Joined Aug 19, 2011
4
I tried to simplify using karnaugh map, and I got the answer. However this is not the way I wanted to solve it. I will keep trying to do it by hand, because I want to increase my skill of using logic identities. I am still having problems combining multi-variable pos terms. ex. (a+b+c)(a+b+!c)(a+!b+!c)[! is complement]. Can anyone point me in the right direction? Thanks

#### Georacer

Joined Nov 25, 2009
5,182
On the contrary, I would discourage you even further. A K-map is a fool-proof way to nail down the simplest (not shortest) expression, especially in multi-variable expressions.