Normalization

Question

For R(ABCDE), given candidate keys are ABC, ACD.
Now if given FD is ABD->E then what is the highest normal form?

ABD is not a super key/ candidate key or E is prime so it's not in 3NF. And because it's a part of CK/ set of prime attributes on LHS then it is not in 2NF also. Is it the right logic?

From a different pov where we say this is not partial dependecy and is in 2NF:
Now by other simple logic this has no transitive depedency (because set prime determines non-prime) and no partial dependency then why does it fail 3NF test??

Comments

some FD's are missing, give relevant example for doubt :)

---

Assume that other dependencies are in 2NF but this is the determining FD
Other eg can be: CK is BE for R(ABCDE) and FDs are AB->C and C->D.

Answer 1

On the given information, you cannot say that it is not in 2NF as for violation of 2NF, proper subset of CK should not determine any non-prime attribute. Here ABD is not a proper subset of CK so cannot comment that it is not in 2NF. Violation of 2NF will be there if A,B,C,D,AB,AC,BC,AD,CD determine any non-prime attribute

Comments

So even if the ABD is a subset of prime attributes but is not subset of any candidate key then it has no partial dependency... the partial dependency states that non-prime is dependent on part of CK but not about part of CK+ some non-prime and hence we can't comment...
What about the example where CK is BE for R(ABCDE) and FDs are AB->C and C->D?
Again for AB->C I cannot comment about the 2NF?

---

Yes, B alone or E alone should not determine any non-prime attribute

Source: NAVATHE.
I hope this helps you.