Is the relation R(ABCDE) with set of FD's {AB->CD, AC->BED, D->A} in 3NF?

Question

Is the relation R(ABCDE) with set of FD's {AB->CD, AC->BED, D->A} in 3NF? I doubt it but the notes i am reading say so. Could someone please explain how is this correct? My understanding is that assuming we take AB as key then the attributes B,E & D are dependent on only a part of the key (i.e. A) thereby violating 2NF property.

Answer 1

Yes, This Relation is in 3NF Because, For the given FD set there are total 4 candidate keys which are AB,AC,DB,DC. And let us suppose we have key AB, then for the given relation, there is no Partial FD and Transitive FD. So the given relation is in 3NF but is is not in BCNF because of D-->A violates the BCNF rule.