For R=(,B,C,D,E,F,G,H,I,J) and functional dependencies
H,D->A
D->E,F,G
H,D,B->C,J
H->I,J
A->BC
What will be all the candidate keys?
(I am getting DH as the candidate key)
Yes, you are getting the correct answer:
DH -> A,B,C,D,E,F,G,H,I,J
Thus, the relation is in the Second Normal form, because NOT-key attributes are dependent on other NOT-key attributes.
P.S. You have forgotten the A in the brackets ;)
