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FD1={A->DF, A->B, D->F, DE->F} , FD2={A->BD, D->EF}

from FD1 : A->DF turns into A->D and A->F . Also A->F is not needed cause we can get there with A->D and D->F . So A->BD ,because A->B and A->D

How to prove that D->EF in FD1?

Mitsos
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  • Please read & act on How to Ask, the downvote arrow mouseover text & hits googling 'stackexchange homework'. Show what you tried. PS Given your title we can guess what you're trying to accomplish in the midddle paragraph but you should make yourself clear. Also vague everyday phrases like "turns into", "not needed" & "get there with" are an impediment clarity. – philipxy Jan 19 '18 at 05:04

1 Answers1

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It is not difficult to prove that a functional dependency can be derived from a set of dependencies F. What you have to do is simply to compute the closure of the left part of the dependency with respect to F, and the see if the right part is contained in the closure.

For instance, if you want to discover if D -> EF is implied by FD1 ot not, you can compute the closure of D with respect to FD1 and see if EF is contained in it:

D+ = D
D+ = DF (because of the dependency D -> F in FD1)

no other depenencies in FD1 have the left part inside DF and the right part outside it, so the closure of D, D+ is DF, which does not contain E, so that the dependency D -> EF cannot be derived from FD1 and the two sets of dependencies are not equivalent.

Renzo
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  • I had the same idea, but the textbook that I got this question from, was saying to prove that they are equivalent, so i took it for granted that they are. – Mitsos Jan 17 '18 at 22:06