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

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):

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.