In a schema R(A, B, C, D, E, F, G, H), each field of R contains only atomic values. F = {CH→G, A→BC, B→CFH, E→A, F→EG} is a set of functional dependencies so that F+ is exactly the set of FDs that holds R. The relation R is :
In 1NF, but not in 2NF
2NF forbids partial dependencies: no non-prime attribute may depend on a proper subset of a candidate key. To test it, find all candidate keys, mark the prime attributes and look for a dependency from part of a key to a non-prime attribute.
Step 1 — D never appears on any right-hand side, so D is in every key. A+ = A → BC → (B → CFH) F, H → (F → EG) E, G = {A, B, C, E, F, G, H}. So AD+ covers all eight attributes and AD is a candidate key.
Step 2 — E → A gives ED as a key, F → E gives FD as a key and B → F gives BD as a key. C and H do not reach A, so the candidate keys are AD, BD, ED and FD.
Step 3 — Prime attributes: A, B, D, E, F. Non-prime attributes: C, G, H.
Step 4 — A → BC makes the non-prime attribute C depend on A, which is a proper subset of the key AD. This is a partial dependency (B → CFH is another one), so R violates 2NF.
All fields are atomic, so R is in 1NF. Because 2NF already fails, R cannot be in 3NF or BCNF either, which rules out the higher forms.
In 1NF, but not in 2NF option is correct.
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