GATE CS 2014 Set 1 Q30 — Normal Forms

Statement S1 is TRUE:

Consider a relation $R(A, B)$ with two single-valued attributes. Since the attributes are single-valued, it is in 1NF. For any non-trivial functional dependency (FD) $X \to Y$ to exist, the only possibilities are:

• $A \to B$: Here, $A$ is the candidate key, making the left-hand side (LHS) a superkey.
• $B \to A$: Here, $B$ is the candidate key, making the LHS a superkey.
• Both $A \to B$ and $B \to A$: Both $A$ and $B$ are candidate keys, so the LHS of both FDs are superkeys.
• No non-trivial FDs exist: BCNF is vacuously satisfied.

In all cases, for every non-trivial FD $X \to Y$, $X$ is a superkey. Thus, $R$ is always in BCNF, which implies it is also in 3NF, 2NF, and 1NF.

Statement S2 is FALSE:

For a set of FDs $G$ to be a minimal cover of $F$, they must be equivalent ($F^+ = G^+$). Let's test the proposed cover $G = \{AB \to C, D \to E, E \to C\}$ against the original set $F = \{AB \to C, D \to E, AB \to E, E \to C\}$:

• Under $G$, the attribute closure of $AB$ is: $(AB)^+_G = \{A, B, C\}$.
• Under $F$, the attribute closure of $AB$ is: $(AB)^+_F = \{A, B, C, E\}$.

Since $AB \to E$ cannot be derived from $G$, the sets are not equivalent. Thus, $G$ is not a minimal cover.