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I got this question

Consider The Following Relation R(A,B,C,D,E,G) And The Set Of Functional Dependencies

F = { A → BCD 
      BC → DE 
      B → D 
      D → A}

Prove that AG is a superkey using Armstrong’s axioms.

In the FD sets there is no attribute related to G. So, is it possible to derive anything with those functional dependencies which is related to G? If yes, how can I derive that?

philipxy
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Encipher
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  • What is your textbook name & edition & how does it say to derive FDs that hold in a relation/schema & where are you stuck in it? Please don't ask us to (re)write one. It uses the attributes of the relation/schema. So you get G->G & other FDs via the axioms. PS. Please before considering posting read the manual & google any error message & many clear, concise & precise phrasings of your question/problem/goal, with & without your particular names/strings/numbers, 'site:stackoverflow.com' & tags; read many answers. Reflect your research. [ask] [Help] – philipxy Nov 19 '21 at 00:41
  • Your "I have these FDs" doesn't make sense. "These are all the FDs that hold"?--Not possible. "These are all the non-trivial FDs that hold"?--Not possible. "These are some FDs that hold"?--Question can't be answered. Find out what a *cover* is & what the exact conditions are to apply a particular definition/rule/algorithm. To determine CKs & NFs we must be given FDs that form a cover. Sometimes a minimal/irreducible cover. And the set of all attributes must be given. [See this answer.](https://stackoverflow.com/a/53386492/3404097) – philipxy Nov 19 '21 at 00:44

1 Answers1

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To check that a set of attributes is a superkey, we should see if from those attributes we can derive (through the Armstrong’s axioms) all the attributes of the relation.

So, assuming that F is a cover of the functional dependencies of R, we can check that AG is a superkey if we can derive AG → ABCDEG. Let’s proceed through the following steps:

1. A → BCD (given)
2. BC → DE (given)
3. BCD → BC (by reflexivity)
4. A → BC (by transitivity from 1 and 3)
5. A → DE (by transitivity from 4 and 2)
6. A → BCDE (by union with 1 and 5)
7. A → ABCDE (by augmentation with A from 6)
8. AG → ABCDEG (by augmentation with G from 7)

As final note, each attribute which is not present in the right part of some dependency of a cover of F must be present in any candidate key (otherwise it could be not derived by other attributes).

Renzo
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