Here is my solution: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.
Thread starter | Similar threads | Forum | Replies | Date |
---|---|---|---|---|
Boolean Algebra question | FPGAs (Field Programmable Gate Array) | 14 | ||
Boolean Algebra | Off-Topic | 0 | ||
R | Boolean algebra simplification | Homework Help | 22 | |
V | Boolean algebra excercises | Homework Help | 2 | |
Digital Electronics | Homework Help | 1 |