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$\begingroup$
AB+CD+A’BD+A’BC+AB’D+AB’C AB+CD+A’B(D+C)+AB’(D+C) CD+(AB+A'B(D+C)+AB’(D+C)) CD+(AB+B(D+C)+A(D+C)) CD+AB+(B(D+C)+A(D+C)) CD+AB+(A+B)(D+C) 

This is what I got, but I'm not sure if it's correct. It's as good as I can get it algebraically. Can it be simplified further?

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    Karnaugh maps are your best friend for this type of thing. http://en.wikipedia.org/wiki/Karnaugh_maps2010-11-05

2 Answers 2