Given a relation scheme R, with attributes A, B, C, D, E, and a set of functional dependency F = { {A->B}, {C->D}, {B,D->A}, {D->B,E}}. The key concept here is to be able to find the closure of the different attributes. From there, find the super key, candidate key and the minimal cover for the relation.
{A}+ = {A,B,}
{B}+ = {B}
{C}+ = {C,D,B,E,A}
{D}+ = {D,B,E,A}
{E}+ = {E}
Therefore, the super keys will be {C}, and anything that contains C and {C} is a candidate key.
To find minimal cover,
step 1: Break the RHS in F to get F'.
F' = { {A->B}, {C->D}, {B.D->A}, {D->B}, {D->E} }
step 2: Break the LHS in F' to get F'', refer to the closure for each of the attributes to see which attribute on the LHS should be kept.
F''= { {A->B}, {C->D}, {D->A}, {D->B}, {D->E} }
step 3: Remove redundant implications from F'' to get F'''
F'''= { {A->B}, {C->D}, {D->A}, {D->E} } in here {D->B} is removed because it can be formed by transitivity from { {A->B}, {D->A} }
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