Question 40

Evaluate Statement S1:
Let $R(A, B)$ be any relation schema with two atomic attributes.

• If no non-trivial FDs hold, the candidate key is $\{A, B\}$, and all dependencies are trivial, so $R$ is in BCNF.
• If $A \to B$ holds, the candidate key is $A$. The LHS of the only non-trivial FD is a superkey, so $R$ is in BCNF.
• Similarly if $B \to A$ holds, or both hold.
  Since every relation in BCNF is also in 3NF, 2NF, and 1NF, any binary relation with single-valued attributes is automatically in 1NF, 2NF, 3NF, and BCNF.
  Therefore, S1 is TRUE.

Evaluate Statement S2:
Original FD set $F = \{AB \to C, D \to E, AB \to E, E \to C\}$.
Notice that from $AB \to E$ and $E \to C$, we can infer $AB \to C$ by transitivity. Thus $AB \to C$ is redundant.
A minimal cover of $F$ would remove $AB \to C$, resulting in $\{D \to E, AB \to E, E \to C\}$.
The set given in S2 is $\{AB \to C, D \to E, E \to C\}$. In this set, we cannot derive $AB \to E$ (closure $(AB)^+ = \{A, B, C\}$). Thus it is not even an equivalent cover for $F$.
Therefore, S2 is FALSE.

Hence, S1 is TRUE and S2 is FALSE.