Equivalent sets of Functional dependencies

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

Question:

4. Let F1 = {A → C, AC → D, E → AD} and F2 = {A → CD, E → AH}. Are F1 and F2 are equivalent?

Solution:

Equivalent sets of functional dependencies: Two sets of FDs F and G over schema R are equivalent, if F+ = G+. That is, if every functional dependency of F is in G+ and every functional dependency of G is in F+, then we would say that the sets of functional dependencies F and G are equivalent. (here, F+ and G+ means the closure of sets of functional dependencies) - Further, refer here.

We first check if F1 ⊆ F2⁺:

If we are able to find the closure for all the determinants of functional dependencies of F2, and if the derived closures are able to show the functional dependencies of F1, then we say that F1 ⊆ F2⁺.

A⁺_F2 = ACD.

That is, if you know A then you know ACD according to the FDs of F2. Hence, the functional dependency A → C of F1 is in F2.

AC⁺_F2 = ACD.

That is, if you know AC then you know ACD according to the FDs of F2. Hence, the functional dependency AC → D of F1 is in F2.

E⁺_F2 = ACDEH. (from E → AH, A → CD)

That is, if you know E then you know ACDEH according to the FDs of F2. Hence, the functional dependency E → AD of F1 is in F2.

From this, we are able to derive all the FDs of F1 from F2+. Hence, F1 ⊆ F2⁺ is TRUE.

We first check if F2 ⊆ F1⁺:

We need to follow the above steps for this too;

A+F1 = ACD.

So, A → CD of F2 is in F1.

E+F1 = ACDE.

So, E → AH of F2 is not in F1.

We are not able to derive all the FDs of F2 from F. Hence, F2 ⊆ F1⁺ is FALSE.

From the above, it is proved that F1 and F2 are not equivalent.

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