# Find the Super Key for the Relation R ( A, B, C, D, E, F, G, H )

Consider a relation R with the schema R = (A, B, C, D, E, F, G, H) and a set of functional dependencies F as:

{ABC, ADE, B F, FGH}

Find the super key for this relation.

## A: Find the Super Key for the relation R ( A, B, C, D, E, F, G, H ) : 49

Given Relation:

R (A, B, C, D, E, F, G, H)

Given Functional Dependencies (F):

AB C
A DE
B F
F GH

Attributes A and B are not present on the right side of any Functional Dependency(FD). Hence, AB must be present in the candidate key and super key.

Let's find the closure of AB.

{AB}+

= {AB} (axiom of reflexivity)

= {ABC} (AB C, given FD)

= {ABCDE} (A DE, given FD)

= {ABCDEF} (B F, given FD)

= {ABCDEFGH} (F GH, given FD)

The closure of AB is:

{AB}+ = {ABCDEFGH}

This means AB can determine all other attributes of the relation R. Hence, {AB} is a Super Key for the relation R.

Besides {AB}, all the possible combinations of {AB} with other attributes are also Super Keys.

For example,

`{ABCD}`

`{ABF}`

`{ABE}`

`{ABFG}`

etc. are also Super Keys.