Prove that the algorithm for synthesizing 3NF decompositions yields schemas that satisfy the conditions of 3NF.
Use the proof-by-contradiction technique. Assume that some FD violates 3NF and then show that this contradicts the fact that the algorithm synthesized schemas out of a minimal cover.
Suppose, to the contrary, that one of the schemas, say (XA; {X ? A,...}), which results from the synthesis using the FD set F, is not in 3NF. Then there must be Y ? B ? F+, where YB? XA, which violates the 3NF conditions.
If A = B ,then Y ? X . However, this means that the FD X ? A cannot be in the minimal cover of F, because Y ? X and so Y ? A holds and has a smaller left-hand side.
If A = B ,thenB ? X and B but it cannot be part of a key (because otherwise Y ? B would not have violated 3NF). Thus, there is a key, W ,of XA such that W ? X ?{A}. In particular, it implies that W ? A ? F+. Again, this means that X ? A cannot be in a minimal cover of F.
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A) Row labels B) Banded columns C) Summary statistics D) Column labels
Using the class Pet from Listing 6.1, write a program to read data for five pets and display the following data: name of smallest pet, name of largest pet, name of oldest pet, name of youngest pet, average weight of the five pets, and average age of the five pets.
This project is a little more challenging if it is written to find and display all the names if more than one pet weighs the most or least, or is the oldest or youngest, which is how the solution in this manual was written.