Given the following two statements: S1: Every table with two single-valued…

Answer: S1 is true and S2 is false.

• Why S1 is true: With only two atomic attributes, consider the possible FD scenarios:

  1. No nontrivial FD: the key is the pair of attributes, so both attributes are prime and 2NF, 3NF and BCNF hold (no partial or transitive violations).

  2. One attribute determines the other (A -> B or B -> A): the determinant is a single-attribute key, so the determinant is a superkey and 3NF/BCNF hold.

  3. Both directions (A -> B and B -> A): either attribute is a key, so normal-form conditions are satisfied.

• Why S2 is false: Start from the given FDs: AB -> C, D -> E, AB -> E, E -> C. To form a minimal cover:

  • Ensure each FD has a single attribute on the right: they already do.

  • Check for extraneous attributes on left sides: AB -> E has no extraneous attribute (neither A nor B alone implies E from the given set), and the other left sides are single attributes.

  • Check for redundant FDs: AB -> C is redundant because AB -> E and E -> C imply AB -> C by transitivity. Concretely, AB+ contains E (from AB -> E) and then contains C (from E -> C), so AB -> C follows from the other FDs.

  • Remove the redundant FD and the remaining set is a minimal cover: AB -> E, D -> E, E -> C.