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2 Answers

5 votes
5 votes

The answer is TRUE. 

A 3NF table that does not have multiple overlapping candidate keys is guaranteed to be in BCNF.

In other words, if the candidate keys in a 3NF relation are

  • all atomic, or
  • non-atomic but non-overlapping,

it is guaranteed that the relation is in BCNF.

Given question is a contradiction to above. So, it is not in BCNF, we can say.

I referred the concept  from stackoverflow. I think it is clear now.

0 votes
0 votes

We'll my reasoning goes like this

consider R(A,B,C,D) and the dependencies : A,B -> C,D and C->B. R is in 3nf with these fds . Now we can write second dependency as A,C -> A,B and $AC^+$ = A,B,C,D so (A,C) will be a candidate key and you can see it overlaps with A,B.

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Somebody please verify the answer