So, as part of my assignment, I have to prove that any relation with two attributes is in BCNF.
As per my understanding, if for a relation we have 3rd normal form and one non key attribute functionally determine key attribute, it violates the BCNF.
Say my relation consists of two attributes A1,A2
Scenario1(only one functional dependency)
A1 -> A2 (so A1 is the key, and A2 does not FD A1 : so no violation)
same applies for
A2 -> A1
But what if
A1->A2 and A2->A1
Here key can be either A1, A2. And the other non key attribute functionally determines the key.