DNF to CNF conversion So i tried to convert this 
(ABCD+AB¬C¬D+¬A¬CD+¬AC¬D+¬B¬CD+¬BC¬D)
into CNF form but im stuck at 
(¬A+¬B+¬C+D).(¬A+¬B+C+¬D).(A+¬B+C+D).(A+¬B+¬C+¬D).(A+C+D).(A+¬C+¬D).(B+C+D).(B+¬C+¬D) 
and the right answer should be (¬A+¬B+¬C+D).(¬A+¬B+C+¬D).(A+C+D).(A+¬C+¬D).(B+C+D).(B+¬C+¬D).
Any help would be great.
 A: Your answer is logically equivalent, but has some redundant terms
Your answer: 
$(\neg A \lor \neg B \lor \neg C \lor D) \land (\neg A \lor \neg B \lor C \lor \neg D) \land (A \lor \neg B \lor C \lor D) \land (A \lor \neg B \lor \neg C \lor \neg D) \land (A \lor C \lor D) \land (A \lor \neg C \lor \neg D) \land (B \lor C \lor D) \land (B \lor \neg C \lor \neg D)$
The book's answer:
$(\neg A \lor \neg B \lor \neg C \lor D) \land (\neg A \lor \neg B \lor C \lor \neg D) \land (A \lor C \lor D) \land (A \lor \neg C \lor \neg D) \land (B \lor C \lor D) \land (B \lor \neg C \lor \neg D)$
Note the only difference between your answer are the two terms below:
$(A \lor \neg B \lor C \lor D) \land (A \lor \neg B \lor \neg C \lor \neg D)$
These terms are made redundant by 
$(A \lor C \lor D)$ and $(A \lor \neg C \lor \neg D)$ respectively
Do you see how $(A \lor C \lor D)$ is equivalent to $(A \lor \neg B \lor C \lor D) \land (A \lor B \lor C \lor D)$? This means it is equivalent and redundant to an extra term you included.
