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Given the set of dependencies AB->C, BD->EF, AD->GH, A->I, H->J. How would you find a minimal cover? By applying a process described in a book I get: AB->C, A->I, BD->EF, AD->GH, H->J instead of AB->CI, BD->EF, AD->GHIJ. Is it possible to combine AB->C and A->I into AB->CI and get rid of A->I?

Mike Sherrill 'Cat Recall'
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Drimades Boy
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    See [Armstron's axioms - Composition](https://en.wikipedia.org/wiki/Armstrong%27s_axioms#Composition) – reaanb Feb 25 '18 at 06:32
  • I see. But can I say that AB->CI infers A->I? – Drimades Boy Feb 25 '18 at 19:16
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    No, given AB->CI we can't derive A->I. – reaanb Feb 25 '18 at 20:03
  • Hi. Tell us the book & show how you followed it. How else can we point out your errors? Also why should others have to rederive a solution instead of just checking yours? Google 'stackexchange homework'. PS To "get rid of" a FD, it would need to be implied by the FDs you keep. That's not so for A->I given AB->CI. But you can probably show that for yourself by making up some simple ABCI data. But since dropping that FD is not part of any step in the process you were trying to follow, why are you asking? – philipxy Feb 27 '18 at 10:56
  • PS The set you say you got is just the input list. Also the set you say the book gives is not a cover of the input, since it doesn't imply A->I. So you seem very confused. Please confirm the input, your output and the book output. Also "combine into" is not useful phrasing. AB->C and A->I *imply* AB->CI. But it's the things that are *implied* by others that we can drop. Having derived AB->CI, we can only drop A->I if all the other FDs imply it. Follow the reference process. – philipxy Mar 02 '18 at 04:47

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The functional dependencies of a minimal cover of a set of dependencies F must satisfy four conditions:

  1. They must be a cover of F (of course!)
  2. Each right part has only one attribute
  3. Each left part must not have extraneous attributes (that is attributes such that original dependency can be derived even if we remove them)
  4. No dependency of the cover is redundant (i.e. can be derived from the remaining dependencies).

So this means that a minimal cover of the example is (note that there can be more then one minimal cover, that is set satisfying the above conditions):

{ AB → C
  AD → G
  AD → H
  A → I
  BD → E
  BD → F
  H → J }

Of course to this set you can apply the Armstrong’s axioms to derive many other dependencies (for instance AD → GH, AB → CI, AD → GHIJ, ABD → EJ, etc.) but these are not part of any minimal cover of F (that is, they do not satisfy the above definition).

Renzo
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  • I bothered to examine the FD sets & the input is already just a trivial rearrangement of your one-attribute-on-the-RHS minimal form in a multiple-attributes-on-the-RHS minimal form. That & other odd things in the question (see my comment there) suggest that maybe the alleged input is actually the book's output. Anyway something's off up there. PS Your answer although useful doesn't actually answer either of questions in the question. – philipxy Mar 02 '18 at 05:45