Relation: R(ABCDE):
FDs: A->BC, CD->E, B->D, E->A
Find Candidate Keys: {A, E, BC, CD}
All FDs left side are superkey except B->D
So we will decompose at B:
Find B's Closure: {B,D}
BD will be one side and other side ABCE,
Now BD has become a Relation having SuperKey as B (B->D)
Check ABCE:
FDs: A->BC, E->A {We are not talking about D because D is not here}
Find Candidate Keys : {E}
The FD A->BC, haven't superkey at left side so decompose at A
Find closure of A : {A,B,C}
Now ABC has become a relation having superkey as A ( A->BC )
ABC will be one side and other side EA (EA also have a superkey {E->A} )
So three relations: ABC, EA, BD