##
**Set of solved exercises in
Normalization / Normalization Solved Examples / How to find candidate keys, and
primary keys in database? / Sets of examples to find the keys of a tables /
Process of Key finding in a database - Examples**

__Question:__
Consider a relation R(A, B, C, D, E) with FD's AB → C, CD → E, C → A, C → D, D → B.

Determine all the keys of
relation R. Do not list superkeys that are not a candidate key.

__Solution:__

__Key 1:__
From above points, we would know
that the attributes AB together can identify all the other attributes of R uniquely.
Hence,

**AB**is one of the keys.

__Key 2:__
From AB, we can obtain the rest
of the attributes. (See the discussion in Key 1 above). Hence,

**C**is one of the keys.

__Key 3:__
From AB, we can obtain the rest
of the attributes. (See the discussion in Key 1 above). Hence

**AD**is one of the keys.
The keys are

**AB**,**AD**, and**C**.**Go back to Normalization - Solved Exercise page.**

Hello Admin, Why can't CD be a key as well? We know that , with reflexivity rule, CD -> E: CDE, and with C ->AD, and D->B. Isn't CD be a key?

ReplyDeleteHi Sherman, CD is a super key because we found that C is a key. CD is not a candidate key.

DeleteCandidate key is a minimal super key for which no proper subset is a key. Proper subset for CD is {C} and {D}. But already C is a key. Hence, CD cannot be a key (candidate).