Boolean Algebra Problems

I am having trouble reducing the following probems.
F = AB’CD’ + A’BCD’+AB’C’D+A’BC’D
F = A’BC’+ A’B’C+ AC’D+ ACD’
F = (A′ + BD′ + AD) (B + C′) (A+ B′C)

Any help would be appreciated.
Thanks in advance.

 

You can't simplify that function further. If you've learned karnaugh maps, just draw a karnaugh map for the function and you'll see why.
cat

 

Here is my solution:

using: F = AB’CD’ + A’BCD’+AB’C’D+A’BC’D

F = C(AB'D + A'BD') + C'(AB'D + A'BD)
F = C(AB'D + A'BD') + C'D(AB' + A'B)
F = C(AB'D + A'BD') + C'D(A xor B)

 

No advertising.

 

lol throwing in a xor doesn't really help simplify it, mate.