BCNF decomposition excercise

"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.

2NF: Remove Partial Dependencies

R = {A, B, C, D, E, F, G, H, I, J} includes partial dependencies.

D and E depend only on A, F depends only on B, G, H, I and J don't depend on the key (directly) at all.

R0 = {A, B, C}

R1 = {A, D, E, I, J}

R2 = {B, F, G, H}

R0, R1, and R2 contain no partial dependencies (or repeating groups) so they are 2NF. However R1 and R2 are still an issue, because they contain transitive dependencies.

3NF: Remove Transitive Dependencies

I and J depend on D, not on the key of R1. Therefore you need to further normalize R1 as follows:

R1 = {A, D, E, I, J}

R1a = {A, D, E}

R1b = {D, I, J}

Similarly, G and H depend only on F so R2 must be decomposed as follows:

R2 = {B, F, G, H}

R2a = {B, F}

R2b = {F, G, H}

Now all of your remaining relations (R0, R1a, R1b, R2a, R2b) are devoid of repeating groups, partial dependencies and transitive dependencies. That means your relations are in 3NF.

When you are looking at an relation that hasn't been normalized and a series of dependencies, you can often normalize by inspection just by recognizing what your primary keys are going to be. Any attribute or combination of attributes that functionally determine other attributes are going to end up as primary keys. Once you've got your primary keys defined, you just need to figure out which non-key attributes go with each key. This is obvious from the statement of what your functional dependencies are.