# GATE | GATE-CS-2014-(Set-1) | Question 40

Given the following two statements:

S1: Every table with two single-valued attributes is in 1NF, 2NF, 3NF and BCNF. S2: AB->C, D->E, E->C is a minimal cover for the set of functional dependencies AB->C, D->E, AB->E, E->C.

Which one of the following is CORRECT?

**(A)** S1 is TRUE and S2 is FALSE.

**(B)** Both S1 and S2 are TRUE.

**(C)** S1 is FALSE and S2 is TRUE.

**(D)** Both S1 and S2 are FALSE.

**Answer:** **(A)** **Explanation:**

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

A relational schema R is in BCNF iff in Every non-trivial Functional Dependency X->Y, X is Super Key. If we can prove the relation is in BCNF then by default it would be in 1NF, 2NF, 3NF also.

Let R(AB) be a two attribute relation, then

- If {A->B} exists then BCNF since {A}+ = AB = R
- If {B->A} exists then BCNF since {B}+ = AB = R
- If {A->B,B->A} exists then BCNF since A and B both are Super Key now.
- If {No non trivial Functional Dependency} then default BCNF.

Hence it’s proved that a Relation with two single – valued attributes is in BCNF hence its also in 1NF, 2NF, 3NF.

**Hence S1 is true.**

S2: AB->C, D->E, E->C is a minimal cover for the set of functional dependencies AB->C, D->E, AB->E, E->C.

As we know Minimal Cover is the process of eliminating redundant Functional Dependencies and Extraneous attributes in Functional Dependency Set.

So each dependency of F = {AB->C, D->E, AB->E, E->C} should be implied in minimal cover.

As we can see AB->E is not covered in minimal cover since {AB}+ = ABC in the given cover {AB->C, D->E, E->C}

**Hence, S2 is false.**

This explanation has been contributed by** Manish Rai.**

**Learn more about Normal forms here:**

Database Normalization | Introduction

Database Normalization | Normal Forms

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