Relation \(R\) has eight attributes \(ABCDEFGH\). Fields of \(R\) contain only…

Answer: R is in 1NF but not in 2NF.

• 1NF: Satisfied because every field contains only atomic values.

• Find candidate keys: D does not appear on the right side of any functional dependency, so every candidate key must include D.

  1. Closure of B,D: Start with {B,D}. B → C,F,H adds C,F,H. F → E,G adds E,G. E → A adds A. A → B,C adds B,C. Closure yields all attributes A,B,C,D,E,F,G,H, so {B,D} is a candidate key.

  2. Closure of A,D: A → B,C adds B,C. B → C,F,H adds F,H. F → E,G adds E,G. All attributes are obtained, so {A,D} is a candidate key.

  3. Similarly, {E,D} and {F,D} are candidate keys because E → A and F → E,G let them derive all attributes when combined with D.

• Prime and non-prime attributes: Attributes that appear in some candidate key are prime: A, B, E, F, D. Non-prime attributes are C, G, H.

• Check 2NF: There is a functional dependency A → B,C. Attribute C is non-prime and is functionally dependent on A, which is a proper subset of the candidate key {A,D}. This is a partial dependency, so the relation violates 2NF.

• Conclusion: Since 1NF holds but 2NF is violated, the relation is in 1NF but not in 2NF. Consequently it is not in 3NF or BCNF.