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Prove that a relation which is 4NF must be BCNF.
Ans Let R be in 4NF. Suppose it is not in BCNF. Hence, there exists an FD X→Y in R such that x is not a super key. Although by the rule M1 X→Y|=x→→Y. Once again x here is not a super key. This contradicts the fact that R is in 4NF. Hence, our assumption is false. Each R in 4NF should be in BCNF.
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What is candidate key? Candidate Key - A candidate key of an entity set is a minimal superkey, which uniquely identifies each row in the relation.
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