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+21 votes

Given the following two statements:

**S1:** Every table with two single-valued attributes is in 1NF, 2NF, 3NF and 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**?

- S1 is TRUE and S2 is FALSE.
- Both S1 and S2 are TRUE.
- S1 is FALSE and S2 is TRUE.
- Both S1 and S2 are FALSE.

+25 votes

Best answer

+13 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.

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.

+8 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**

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