i am trying it out:
Lets make it simple-
A-> B E
So according to the correct FD: A,C is not getting derived in RHS, so you can consider AC to be the Candidate key.So, combine them,you can derive the rest data in the relation.So, u can break the relation as :
Its BCNF as def: x->a , x should be a super key. So in both the FD , LHS is a super key/candidate key.
Now, your case
A-> B C D E
you can think its in BCNF,
but the basic criteria of FD is canceled.
A B C D E
A1 SAM WATSON C1 PETER PARKER C1
A1 SAM WATSON C2 BRUCE WAYNE C1
a Student can have exactly 2 different advisers.
ATD: x-> a = x is functionally dependent to a if the value of a can uniquely identify x.But in the above example its violated,A1 derives 2 different combination .
Now if break it in my way:
A B E
A1 SAM WATSON C1
C1 PETER PARKER
C2 BRUCE WAYNE
but the decomposition is not Loss-less(lossy), as there is no common attribute to unite the tables.
if i try to add :
so the FD definition is again violated.
I've ried my best, i am still the beginner so if i've made any mistake please point it out, i would love to learn from it.