0 0 votes Databases + – Sanjay Sharma 1.2k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
0 0 votes Option D is right. abhishekmehta4u answered May 21, 2018 abhishekmehta4u comment Share Follow See all 8 Comments 8 8 Comments reply Sanjay Sharma commented May 21, 2018 reply Follow flag why R1(BCD) is in BCNF and not only in 3NF 0 0 replyShare abhishekmehta4u commented May 21, 2018 reply Follow flag Only trivial fd is in relation R1(BCD).SO IT IS IN BCNF 1 1 replyShare Sanjay Sharma commented May 26, 2018 reply Follow flag BCD has f.d BC->CD (via AB->CD and C->A) which is not trivial and what will be ans of next 0 0 replyShare abhishekmehta4u commented May 26, 2018 reply Follow flag Option c is right. 0 0 replyShare Sanjay Sharma commented May 26, 2018 reply Follow flag how lossy their intersection is a superkey in second relation 0 0 replyShare Sanjay Sharma commented May 26, 2018 reply Follow flag and dependency also preserved since AB->CD and C->A can give BC->CD which is present in CBD and from BC->CD and C->A we can get AB->CD 0 0 replyShare abhishekmehta4u commented May 26, 2018 reply Follow flag Sorry wrote mistake. Option d is right.we can not derive functionl dependency AB-->CD. 0 0 replyShare Sanjay Sharma commented May 26, 2018 reply Follow flag why dependency is not preserved can't we do like this AB->CD and C->A can give BC->CD which is present in CBD and from BC->CD and C->A we can get AB->CD 0 0 replyShare Please log in or register to add a comment.
