Are these relations in BCNF

"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.

The only key is {BC}. You have a partial key dependency: C->D. Since you have a partial key dependency, R is not in 2NF. So R is also not in BCNF.

An informal way of expressing BCNF is that every arrow in every FD is an arrow that comes out of a candidate key. In R, the arrow in "C->D" doesn't come out of a candidate key.