# Boolean Algebra Problems

Thread Starter

#### molder22

Joined Oct 7, 2006
3
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.

#### cat

Joined Sep 25, 2006
8
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

#### mik3ca

Joined Feb 11, 2007
189
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.
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)

#### fxbit

Joined Mar 15, 2007
4

#### Blue_turnip

Joined Mar 5, 2007
8
lol throwing in a xor doesn't really help simplify it, mate. Similar threads