Why is this relation not in Boyce-Codd Normal Form (BCNF)

You dropped "F" somewhere.

Since you're given {ABC} is a candidate key, then you know that ABC->DEF. (That just follows from the definition of "candidate key".) And ABC->D, ABC->E, and ABC->F, by Armstrong's axiom of decomposition.

The important part as far as BCNF is concerned is E->A.

So the final result should be: E1(ABC), E2(ABD), E3(CDE), E4(EA), E5(ABCF)

I think that's right, except for E1. You don't need a separate relation for a candidate key. I'm not sure how you derived that one given what's in the problem.

Usually, your textbook will have at least one algorithm for normalizing relations. In the real world, there's often more than one 5NF schema for a given set of FDs.

"Armstrong's axioms are a set of axioms (or, more precisely, inference rules) used to infer all the functional dependencies on a relational database"(link).

• R: ABCDE
• F: C->AB, D->A, BE->CE, E->B

BE->CE can be split in BE->C and BE->E. The trivial functional dependency BE->E can be skipped. BE->C can be replaced by E->C because from BE->C and E->B one can deduce E->C. Therefore the set of functional dependencies can be reduced to:

• R: ABCDE
• F: C->AB, D->A, E->C, E->B

A and B cannot be member of a key because they cannot be found on the left site of a functional dependency. So it is sufficient to search all keys of the relational system (remove the Attributes A B from R and all functional dependencies containing A and B)

• R: CDE
• F: E->C

C is not on the left side of a functional dependcy so again one can reduce the system to

• R: DE
• F:

So the list F of functional identities is empty. The only key of this relational system is DE which is also the only key of the unreduced systems. So your answer of your first question is true.

You can use this page to find your candidate keys

To check if the system is in BCNF it is not necessary to find all candidate keys. It is sufficient to find one functional dependency which has a left side that is no a key.

C->AB is such a functional dependency: C is not a key because the closure of C is C. This means that no further attributes can be generated by applying functional dependencies on C.

You can use this page to find check if a relation is in normal form.