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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. en.wikipedia.org/wiki/Karnaugh_maps –  Brandon Carter Nov 5 '10 at 5:27

2 Answers 2

The function is majority (with tie-breaking): the expression is true iff at least two of $A,B,C,D$ are true. This follows from your last term by opening up the parentheses, and can also be checked directly.

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To go along with my comment above, the minimal form is what you arrived at. Using a Karnaugh map, you will get $AB+CD+AC+AD+BC+BD$, which factors to your answer.

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