# Boolean Algebra: Can this be simplified further?

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?

-
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

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.
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.