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Given the following two statements: 

S1: Every table with two single-valued attributes is in $\text{1NF, 2NF, 3NF}$ and $\text{BCNF}.$ 

S2: $AB \to C, D \to E, E \to C$ is a minimal cover for the set of functional dependencies $AB \to C, D \to E, AB \to E, E \to C$.

Which one of the following is CORRECT?

  1. S1 is TRUE and S2 is FALSE.
  2. Both S1 and S2 are TRUE.
  3. S1 is FALSE and S2 is TRUE.
  4. Both S1 and S2 are FALSE.
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Best answer
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48 votes

(A) S1 is TRUE and S2 is FALSE.

A relation with $2$ attributes is always in BCNF

The two sets of functional dependencies are not the same. We can not derive $AB \to E$ from the $1^{\text{st}}$ set.

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31 votes

S1 is true bcoz if there are only 2 attributes then relation is always in bcnf.

let R(A B) possible cases are:

(1) {A->B } here A is cand key so BCNF

(2) {B->A}  here B is cand key so BCNF

(3) { A->B  B->A} here A and B both are cand key so BCNF

​​​​​​​(4) no non-trivial FD's here AB is cand key so BCNF

S2 is false bcoz minimal cover should be { D->E , AB->E , E->C }

so ans is A

18 votes
18 votes
Table with 2 single valued attributes will be in 1NF,

2NF because there can not be a partial key dependencies. Key can be 1 attribute or 2 attributes. If one is key then other can be dependent, if 2 attributes make key then it is trivial.

3NF, there is no question of Transitive dependencies.

BCNF, because if at all a dependency exists, the determinant will be a Key.
Answer:

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