Consider the relation R(A, B, C, D, E, P, G) with the following functional…

Concept

Second Normal Form (2NF) requires that the relation is in 1NF and that no non-prime attribute is fully functionally dependent on only a part of any candidate key. A non-prime attribute is one that belongs to no candidate key. A partial dependency occurs when a non-prime attribute is determined by a proper subset of a candidate key. The normal forms are nested: a relation must clear 2NF before it can be considered for 3NF, and 3NF before BCNF, so a 2NF violation rules out all higher forms.

Application to this relation

Step 1 - Find the candidate key. Attributes A and B never appear on the right-hand side of any FD, so they must belong to every candidate key. Compute the closure of {A, B}:

1. Start: {A, B}.

2. Apply AB → CD to add C and D, giving {A, B, C, D}.

3. Apply C → E to add E, giving {A, B, C, D, E}.

4. Apply DE → P to add P, giving {A, B, C, D, E, P}.

5. Apply B → G to add G, giving {A, B, C, D, E, P, G}.

The closure {A, B}+ equals all seven attributes, and neither A nor B alone closes to the whole relation, so {A, B} is the unique, minimal candidate key. The prime attributes are A and B; the non-prime attributes are C, D, E, P, and G.

Step 2 - Test for a partial dependency. The proper subsets of the key {A, B} are {A} and {B}. The dependency B → G uses only {B}, a proper subset of the key, to determine G, and G is non-prime. That is exactly a partial dependency of a non-prime attribute on part of the candidate key.

Cross-check and contrast

Because a partial dependency exists, the relation fails 2NF. Since the normal forms are nested, failing 2NF automatically means it is neither in 3NF nor in BCNF, so the strongest correct statement is that the relation is not in 2NF. The remaining FDs (AB → CD, DE → P, C → E, P → C) do not repair this: the single FD B → G is sufficient to break 2NF.