We get 3 BCNF Relations(DBE, CD, AC) which is lossless but not fd preserving due to AB→ C
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From given R(ABCDE) and FD’s : { AB->C, C->D, D>B, D->E} => CK={ AB,AC,AD} and Prime attributes = A,B,C,D .
So, D→ E is not in 2nf and C→ D, D→ B are not in BCNF.
find closure of all these C+ = CDBE and D+ = DBE
1st decompose into CDBE and AC.
in R1(CDBE) only C->D, D->B,D->E holds. we get CK ={C} and now again D->B,D->E are not in 3nf. so decompose into DBE(D->BE) and CD(C->D)
Therefore we get total 3 Relations(DBE, CD, AC) which are in BCNF.