1. Quiz 2 Let R(A,B,C,D) be a relation schema with
the FDs ABCD and BCD and
the MVD A B.
Answer the following:
Find all possible keys of R.
Is the relation in 4NF? If not, propose a decomposition of R into 4NF relations.
(Answer only what is asked - Remember to justify your answer.)

2. Answer
There is no FD that has A in the right hand side, so any key of R must contain A. Since {A}+={A,B,C,D} we conclude that A (or {A}) is the only key of R.
To figure out whether the relation is in 4NF or not, we need to look for MVDs that will potentially violate the 4NF condition. Since A is a key of R, we only need to consider MVDs whose left hand side is not A. In this case, BCD is a candidate (because each FD is a MVD). This is a nontrivial MVD. Furthermore B is not a superkey. Therefore, BCD violates the 4NF condition. This implies that R is not in 4NF.
This suggests a decomposition: {B,C,D}, {B,A} (using BCD). The FDs of these relations are B CD (and its derived FDs: BC, BD) and A B, respectively. These imply that the newly defined relations only have trivial MVDs or MVDs with the left hand side is a key (BC, BD for the first one) . In other words, the decomposition yields 4NF relations. Hence, we can stop here.