Question 42

Given schema $X = (PQRS)$ and $F = \{QR \to S, R \to P, S \to Q\}$.
Candidate keys of $X$:

• $(QR)^+ = QRSP = PQRS \implies QR$ is a candidate key.
• Since $S \to Q$, $(SR)^+ = SRQP = PQRS \implies SR$ is a candidate key.

Checking Statement II (Lossless and Dependency Preserving):

1. Lossless Join: $Y \cap Z = (PR) \cap (QRS) = R$.
   In $F$, we have $R \to P$, which means $R$ functionally determines $Y = (PR)$ ($R$ is a superkey of $Y$). By the lossless join theorem, the decomposition is lossless.
2. Dependency Preservation:
   • $R \to P$ is preserved in $Y(PR)$.
   • $QR \to S$ and $S \to Q$ involve only attributes of $Z(QRS)$ and are preserved in $Z(QRS)$.
     All functional dependencies in $F$ are preserved. Thus the decomposition is dependency preserving.
     Hence, Statement II is TRUE.

Checking Statement I (BCNF for Y and Z):

1. In $Y(PR)$, the candidate key is $R$, and the only dependency is $R \to P$. Since $R$ is a superkey, $Y$ is in BCNF.
2. In $Z(QRS)$, candidate keys are $QR$ and $SR$.
   The dependency $S \to Q$ holds in $Z$. Here, the determinant $S$ is NOT a superkey of $Z$ because $S^+ = SQ \ne QRS$.
   Thus, $Z$ violates BCNF.
   Hence, Statement I is FALSE.

Therefore, only Statement II is correct (Option C).