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in Databases by Boss (48.8k points)
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Option D is right.

by Boss (35.7k points)
0
why R1(BCD) is in BCNF and not only in 3NF
+1
Only trivial fd is in relation R1(BCD).SO IT IS IN BCNF
0

BCD has f.d BC->CD  (via AB->CD and C->A)  which is not  trivial     and what will be ans of next 

0
Option c is right.
0
how lossy their intersection is a superkey in second relation
0
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
Sorry wrote mistake.

Option d is right.we can not derive functionl dependency AB-->CD.
0
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

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