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Relation R which satisfy 3NF and atmost one compound candidate key is also in BCNF

(pls explain give counter examples for both true and false)

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the reason for a relation to be 3nf but not bnf is because of { subset of ck(i) ----> subset of ck(j)}

for make it true you must need at least two compound keys.

hence the relation with only one compound key is in bcnf also.

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Consider the example:R(A,B,C,D,E,F)

ABC-->DEF

D-->A

This is in 3NF(key is ABC,there is no Transitive or partial dependency)

But,here there exists non key to key FD(that is LHS is not super key[D-->A])

Also here we have one compound candidate key -ABC

But it is not in BCNF

So the statement{Relation R which satisfies 3nf and atmost one compound candidate key is also bcnf} should be false

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