----for first relation,candidate key is AB and C.
so,all the FDs except D->E are in BCNF
now we will make a separate relation for D->E
so,now we got two relation R1(A,B,C,D) R2(D,E)
so,it is both lossless as well as dependency preserving.
---for second relation,candidate key is AB aand CB
so,here,except AB -> C,all are not in BCNF
so,we need to make separate relation for C->A ,C->D , D->E
now,we get 4 relations,R1(B,C) ,R2(C,A) ,R3(C,D) ,R4(D,E)
now we wont be able to derive id AB->C
and hence it is not a dependency preserving BCNF decomposition.
am i corect??