# How to prove that two sets of functional dependencies are equal or not - Proving equality of two sets of functional dependencies - How to find whether two sets of functional dependencies are equal or not? - Equivalent sets of functional dependencies

__Exercise:__
Let F = {A → C, AC →
D, E → AD}, and let G = {A → CD, E → AHE}. Are they
equivalent?

__Solution:__
To check whether
two sets of functional dependency are equivalent, we have to verify every
functional dependency (FD) of F is in G

^{+}and every FD of G is in F^{+}.

__Theorem:__

*If***F**

**⊆**

**G**

^{+}*and*G**⊆**

**F**

^{+}*, then F and G are equivalent.*

__To check for F__

__⊆__

__G__^{+}
Let us take the
first FD of F, A → C and find the closure using FDs from G as follows;

- A
^{+}= ACD through the FD A → CD of G. The result includes the RHS attribute of FD A → C. Hence, A → C is in G^{+}.

Let us do the same
for other FDs of F;

- AC
^{+}= ACD through the FD A → CD of G. The result includes the RHS attribute of FD AC → D. Hence, AC → D is in G^{+}.

- E
^{+}= AHECD through the FDs E → AHE and A → CD of G. The result includes the RHS attribute of FD E → AD. Hence, E → AD is in G^{+}.

As all the FDs of F
can be derived using G,

**F****⊆****G**^{+}is true.

__To check for G__

__⊆__

__F__^{+}- A
^{+}= ACD through the FDs A → C and AC → D of F. The result includes the RHS attribute of FD A → CD. Hence, A → CD is in F^{+}.

- E
^{+}= ACDE through the FDs E → AD, A → C and AC → D of F. The result does not include the RHS attribute of FD E → AHE. Hence, E → AHE is in F^{+}.

As all the FDs of G
cannot be derived using F,

**G****⊆****F**^{+}is false.

__Result:__
We found that

**F****⊆****G**is true but not^{+ }**G****⊆****F**. Hence, the given set of functional dependencies F and G are not equivalent (F ≠ G).^{+}
******************

## How to find closure of set of functional dependencies?

## How to find closure of attributes?

## How to find canonical cover for a set of functional dependencies?

## How to find extraneous attribute?

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