I read oodles of examples stating that 3NF is BCNF as long as there are no overlapping candidate keys...
Given:
R(a,b,c,d,e,f)
a,b,c -> d
a,b,c -> e
a,b,c -> f
so, a,b,c is the only candidate key. So far it's BCNF.
Now, I add the FD f-> c , which is non-prime -> prime but I still don't have overlapping candidate keys.
Am I back to 3NF or am I still in BCNF?
If you add a new FD then you have to redetermine the CKs, which attributes are prime & non-prime, whether it's in 3NF, whether there are overlapping CKs & whether it's in BCNF.
We have a new given cover that has one more FD. (If we want all the FDs that now hold then we must reapply Armstrong's axioms. If we want a minimal cover for the new cover then we must calculate it.) Applying a definition of CK and/or an algorithm to find CKs we have CKs {a,b,c} & {a,b,f}. The non-prime attributes are d & e. Applying a definition of 3NF, eg that every non-prime attribute is non-transitively functionally dependent on every CK, we have 3NF. There are overlapping CKs, so we might not have BCNF. Applying a definition of BCNF, eg that every determinant of a non-trivial FD is a superkey, we notice f->c is not out of a superkey, so we don't have BCNF.