## Membership Test for Functional Dependency or Identification of additional Functional Dependencies by Closure Set –

#### What is the Membership Test for Functional Dependency ??

Let F be the Functional Dependency set. X → Y be any non trivial functional dependency.

X → Y implied in Functional Dependency Set only if X^{+} determines Y.

##### Method :

- To determine, whether a particular Functional Dependency can be derived from the original set or not, compute the closure from the original functional dependencies. If the closure contains the RHS of the original functional dependency or the determined attributes, then it will become additional functional dependency. For example, AB→DE is a functional dependency. Compute the closure of AB=AB+.

If AB^{+}includes ‘DE’, then AB→DE is true and will become additional functional dependency. - Repeatedly check all additional functional dependencies to find out new functional dependency set.

Question : Given the following FDs A → BC CD → E E → C D → AEH ABH → BD DH → BC

```
a) Find out, whether BCD → H or not from the following Functional Dependencies?
```

**Solution :
**Compute the closure of BCD = BCD^{+}
BCD^{+} = BCD
= ABCDEH {as D → AEH}
Hence BCD → H is true as the closure contains H.

**b)** Find out, whether AED→C or not from the above given Functional Dependencies?

**Solution:
**Compute the closure of AED.
ADE^{+} = ADE
= ABCDE {as A → BC}
Hence AED → C is true as the closure contains C

Question 2:Following dependencies are given: AB → CD AF → D DE → F C → G F → E G → A Which of the following is false ? (a) (CF)^{+}= {ACDEFG} (b) (BG)^{+}= {ABCDG} (c) (AF)^{+}= {ACDEFG} (d) (AB)^{+}= {ABCDG}

**Solution : option (c) is false . as (AF)**^{+} will give {AFED} but not {ACDEFG}

```
Question 2:
Consider a relational scheme R with attributes A,B,C,D,F and the FDs
A → BC
B → E
CD → EF
Prove that functional dependency AD → F holds in R.
```

```
Solution :
Compute the closure of AD.
(AD)
```^{+} = ADBCEF
Hence AD → E is true as the closure contains E.

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