Consider a relation schema 𝑅=(𝐴,𝐵,𝐶,𝐷,𝐸,𝐹) on which the following…

Solution:

Key idea: F must be included in every candidate key because F does not appear on the right-hand side of any functional dependency, so no attribute set can determine F unless F is included.

1. Compute closure of AEF:

   AEF+ starts with {A,E,F}. From A->B add B, from E->C add C, and then from B,C->D add D. So AEF+ = {A,B,C,D,E,F} which is all attributes.

2. Compute closure of BEF:

   BEF+ starts with {B,E,F}. From E->C add C, from B,C->D add D, and from D->A add A. So BEF+ = {A,B,C,D,E,F}.

3. Compute closure of DEF:

   DEF+ starts with {D,E,F}. From D->A add A, from A->B add B, and from E->C add C. So DEF+ = {A,B,C,D,E,F}.

4. Minimality check:

   Removing F from any of these sets leaves an attribute (F) that is not produced by the FDs, so the result cannot be a key. Removing other attributes from each set also causes the closure to miss some attributes (for example, EF+ = {C,E,F} misses A and B). Therefore each of AEF, BEF, and DEF is minimal.

Answer: The candidate keys are AEF, BEF, and DEF.