0 votes 0 votes 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 Aboveallplayer asked Nov 25, 2016 Aboveallplayer 567 views answer comment Share Follow See 1 comment See all 1 1 comment reply Hradesh patel commented Nov 25, 2016 reply Follow Share is it C??? 1 votes 1 votes Please log in or register to add a comment.
Best answer 2 votes 2 votes 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 . Amit Pal answered Nov 25, 2016 • selected Nov 25, 2016 by Aboveallplayer Amit Pal comment Share Follow See all 0 reply Please log in or register to add a comment.