Which of the following/s is/are FALSE statement? A. An all key relation is always in BCNF since it has no FDs. B. A relation that is not in 4NF due to nontrivial MVD must be decomposed to convert it into a set of relations in 4NF. C. The decomposition removes the redundancy by the MVD. D. 3NF is stronger than BCNF. Choose the correct answer from the options given below:
D only
The normal forms form a chain 1NF ⊃ 2NF ⊃ 3NF ⊃ BCNF ⊃ 4NF, each later form being stricter than the earlier one. Each statement is checked against this.
A — In an all-key relation the only key is the whole set of attributes, so it has no nontrivial functional dependencies; BCNF can never be violated, and the relation is in BCNF (Navathe states this directly). TRUE.
B — If a nontrivial MVD X ↠ Y holds with X not a superkey, the relation is not in 4NF and must be decomposed (into XY and X(R − Y)) to reach a set of 4NF relations. TRUE.
C — That decomposition removes the redundancy (repeated combinations of independent multivalued facts) caused by the MVD. TRUE.
D — BCNF requires every determinant of a nontrivial FD to be a superkey, while 3NF also allows the right side to be a prime attribute. So BCNF is stronger than 3NF, not the reverse. FALSE.
Only D is false, so the choices that include A or B, or that name C, are wrong.
D only option is correct.
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