We get CK = {AB,BD,BE} so, Prime Attributes(PA)= A,B,D,E
now if we check the given fd’s E→ A, D→ E are in 3NF but not in BCNF. now we decompose them into BCNF.
Decomposition:
D+ = DEA, E+ = EA . 1st decompose into DEA , then E→ A, D→ E goes to this table and CK = {D}
but E→ A is not in 3NF. so we again decompose DEA into DE and EA .
if we remove EA from the given R(ABCDE) , we get BCD which will be the another relation.
So, we get total 3 relations(DE, EA and BCD)
Checking lossless : D is common and is a key from DE, EA. it forms DEA.
D is common and is a key from DEA, BCD
Checking FD preserving: AB→ CD can be preserved directly or indirectly from these 3 relations.
Answer is 3 BCNF Relations which is lossless but not fd preserving.