Does minimality of superkey guarantee it to be the candidate key?

Question

Determine the candidate keys and superkeys of the relation R(ABCDEF) with FD's: AEF → C, BF → C, EF → D, and ACDE → F

This is a problem from my book. The book claims that the candidate keys are ABCDE and ABEF. From what I understand, a candidate key is the minimal superkey, and closure test on ABEF captures the relation R perfectly. Since ABEF is more "minimal" than ABCDE, I would argue the only candidate key is in fact, ABEF only. I will grant that ABCDE is a superkey, but not a candidate key. Can somebody explain why I am in the wrong here? Or is it possible that the book is wrong?

Answer 1

“Minimal superkey” does not mean the superkey with the minimum number of attributes of all the other (super)keys, but a (super)key such that, removing any attribute from it, loses the property of being a key, that is of determining all the attributes of the relation. For example, in your case,

ABCDE+ = {ABCDEF}

but:

ABCD+ = ABCD
ABCE+ = ABCE
ABDE+ = ABDE
ACDE+ = ACDEF
BCDE+ = BCDE

so no proper subset of ABCDE determines all the attributes, and for this reason it is a minimal superkey, that is a candidate key.