Let R ={A, B, C, D, E, F} be a relation schema with the following dependencies…

Solution: Find the candidate key(s) by computing attribute closures.

Compute the closure of EC:

• Start with {E, C}.

• E → A, so add A. Now have {E, C, A}.

• C → F, so add F. Now have {E, C, A, F}.

• A → B, so add B. Now have {A, B, C, E, F}.

• EC → D (we already have E and C), so add D. Now have {A, B, C, D, E, F}.

Since EC+ = {A, B, C, D, E, F}, EC determines all attributes and is therefore a key.

Check minimality:

• E+ = {E, A, B} (from E → A and A → B) — missing C, D, F, so E alone is not a key.

• C+ = {C, F} (from C → F) — missing A, B, D, E, so C alone is not a key.

Thus EC is minimal and is a candidate key.

Briefly check the other choices:

• CD+: start {C, D}; C → F gives F, so CD+ = {C, D, F}. Cannot derive A, B, or E, so CD is not a key.

• AE+: start {A, E}; E → A and A → B give {A, B, E}. Cannot derive C, D, or F, so AE is not a key.

• AC+: start {A, C}; C → F and A → B give {A, B, C, F}. Cannot derive E or D, so AC is not a key.

Final answer: EC is a candidate key for R.