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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. – Brandon Carter Nov 5 '10 at 5:27

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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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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