Prove that if a schema, S =(S, F), is in 3NF, then every FD in F + (not only those that are in F) satis?es the 3NF requirements.
What will be an ideal response?
Suppose to the contrary, that there is an FD X ? A ? (F+? F) such that X is not a superkey, A does not belong to any key of S, and the FD is non-trivial, i.e., A ? X . As we know, it must be the case that A ? X+F and X+F should be computable by the attribute closure algorithm. The main step in that algorithm hinges on being able to ?nd an FD Z ? V ? F such that Z ? closure. Since closure ? X+F and X is not a superkey, none of the FDs in F whose left-hand side is a superkey can be used in that algorithm. Therefore, Z ? V must be one of those FDs where the attributes of V belong to the various keys of S. But in order to add A to closure, it must belong to V for some such FD, i.e., A itself must be an attribute of one of the keys in S —a contradiction.
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