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Relation: R(ABCDE):

FDs: A->BC, CD->E, B->D, E->A

Find Candidate Keys: {A, E, BC, CD}

All FDs left side are superkey except B->D

So we will decompose at B:

Find B's Closure: {B,D}

BD will be one side and other side ABCE,

Now BD has become a Relation having SuperKey as B (B->D)

Check ABCE:

FDs: A->BC, E->A {We are not talking about D because D is not here}

Find Candidate Keys : {E}

The FD  A->BC, haven't superkey at left side so decompose at A

Find closure of A : {A,B,C}

Now ABC has become a relation having superkey as A ( A->BC )

ABC will be one side and other side EA (EA also have a superkey {E->A} )

So three relations: ABC, EA, BD

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Candidate Keys: {A, E, CD, BC }

decomposition to bcnf:

R(ABC)

R(CDE)

R(BD)

R(EA)

above decomposition is not only in bcnf but also in 3nf.

Hence, it is dependency preserving and lossless.

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