For a relational schema R(A, B, C, D, E, F) consider the following Functional…

A set of attributes K is a superkey of R if its closure K+ — found by repeatedly applying the given functional dependencies — contains every attribute of R. K is a candidate key if it is additionally minimal, meaning no proper subset of K is already a superkey. Any attribute that never appears on the right-hand side of any FD must belong to every candidate key, because no other attribute can ever derive it.

1. Scan the right-hand side of every given FD (A → B, C → DF, AC → E, D → F): B, D, E and F each appear as some FD's right-hand side, but A and C never do.

2. Since A and C never appear on any right-hand side, both must belong to every candidate key, so start by testing the closure of {A, C}.

3. Apply A → B: B is added, giving {A, B, C}.

4. Apply C → DF: D and F are added, giving {A, B, C, D, F}.

5. Apply AC → E: E is added, giving {A, B, C, D, E, F}, which is every attribute of R.

6. Because (AC)+ = R, the set {A, C} is a superkey.

• Test A alone: only A → B applies, so A+ = {A, B}, which does not reach R — A alone is not a superkey.

• Test C alone: only C → DF applies (D → F adds nothing new), so C+ = {C, D, F}, which also does not reach R — C alone is not a superkey either.

• Since neither proper subset of {A, C} is a superkey, {A, C} is minimal, and since A and C must appear in every candidate key, it is the only candidate key of R.

The candidate key of R is AC.