Find canonical cover for the given set of functional dependencies.

Question

Consider the following set of functional dependency on the scheme (A, B,C)
A→BC, B→C, A→B, AB→C

The canonical cover for this set is:
A. A→BC and B→C
B. A→BC and AB→C
C. A→BC and A→B
D. A→B and B→C

Answer 1

In the given FDs,

1) Convert elements on LHS as singleton

           Thus, A -> BC can be written as A-> B and A->C

        

2) Remove composite attributes from LHS

           Here, AB -> C can be written as A -> C, because we have an FD {A -> B}.

3) Remove redundant attributes

        We have FD = {A -> B, A -> C , B -> C}

       this is a transitive depenency A -> B -> C. After remove redundancy, we get A->B, B->C.

 

Option D is the required answer

Comments

it is correct

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if ans is d

by using above conical set not producing the fd AB----->C

The canonical cover for this set is: 
  A→B and B→C

it covers only A-->B

            A--->C

A--->BC 

BUT NOT COVERING AB---->C

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Hi
A-->C         means C can uniquely be identified by A
AB---> C    means C can uniquely be identified  by AB
Both are same.
So if you know 'A' u can uniquely find C and their is no need of using AB-----> C. And if you don't know 'A', then you can't even use AB--->C as it will be needing 'A'

Answer 2

The answer must be D as A->B and B->C must be the minimum FDs that follow the rules.

Answer 3

D. A→B and B→C