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IITM November 2016

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FD s given P1P3->P4, P1->P2, P2->P1

now

a) bcnf and p2p3->p4 holds

b)bcnf and  p2p3->p4 does not hold

c)3nf not in bcnf and p2p3->p4 holds

d)3nf not in bcnf and  p2p3->p4 does not hold
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According to given FD's  candidate keys ={P1P3, P2P3}

so p1p3->p4 is in BCNF

p1->p2 is in 3NF

p2->p1 is in 3NF

so as whole relation is in 3NF not in BCNF. and p2p3 is a candidate key so p2p3->p4 will also hold .

so option (c) is correct .
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