Given the following two statements: S1: Every table with two single-valued attributes is in 1NF, 2NF, 3NF and BCNF. S2: $AB\rightarrow C, D\rightarrow E, E\rightarrow C$ is a minimal cover for the set of functional dependencies $AB\rightarrow C, D\rightarrow E, AB\rightarrow E, E\rightarrow C$. Which one of the following is CORRECT?

Evaluating Statement S1:

• Any relational schema with two single-valued attributes $R(A, B)$ is inherently in 1NF.

• The possible non-trivial functional dependencies are:

  1. $A \rightarrow B$: Then $A$ is a candidate key, so the LHS is a superkey.

  2. $B \rightarrow A$: Then $B$ is a candidate key, so the LHS is a superkey.

  3. Both $A \rightarrow B$ and $B \rightarrow A$: Both $A$ and $B$ are candidate keys (superkeys).

  4. No non-trivial dependencies: The candidate key is $AB$.

• In all cases, for every non-trivial functional dependency $X \rightarrow Y$, $X$ is a superkey. Therefore, every two-attribute relation is always in BCNF (and consequently in 1NF, 2NF, and 3NF).

• Thus, S1 is TRUE.

Evaluating Statement S2:

• Given original set of FDs $F = \{AB \rightarrow C,\ D \rightarrow E,\ AB \rightarrow E,\ E \rightarrow C\}$.

• To find a minimal (canonical) cover:

  1. Check for redundant dependencies:

     • For $AB \rightarrow C$: Using the other FDs $\{D \rightarrow E, AB \rightarrow E, E \rightarrow C\}$, compute $(AB)^+$:

       $$(AB)^+ = \{A, B, E, C\}$$

       Since $C \in (AB)^+$, $AB \rightarrow C$ is redundant and can be removed.

  2. The remaining set is $F^{\prime} = \{D \rightarrow E,\ AB \rightarrow E,\ E \rightarrow C\}$.

  3. Testing for extraneous attributes on the LHS of $AB \rightarrow E$ shows neither $A$ nor $B$ can be dropped.

• Thus, a minimal cover for $F$ is $\{D \rightarrow E,\ AB \rightarrow E,\ E \rightarrow C\}$.

• The set given in S2, $\{AB \rightarrow C,\ D \rightarrow E,\ E \rightarrow C\}$, is missing $AB \rightarrow E$ (which cannot be derived from this set because $(AB)^+ = \{A, B, C\}$ does not contain $E$). Hence it is not equivalent to $F$.

• Thus, S2 is FALSE.

Therefore, S1 is TRUE and S2 is FALSE (Option A).