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
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?
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.
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
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.
Thanks for your words of wisdome Georacer. I found the hard way and now have reason to believe what you said. K-map is the method to use.