# For w y w is not a super key but y is part of a

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Unformatted text preview: r W Y, W is not a super key but Y is part of a candidate key. (c) R is not in BCNF. Z X causes violation, because Z is not a super key. Therefore, decompose R into XZ and YZW. YZW is not in BCNF. W Y causes violation, because W is not a super key. Therefore, decompose YZW into YW and ZW. As a result, R is replaced by XZ, YW, and ZW that are all in BCNF. Q.7 [15 pts] Given the relation schema R(A, B, C, D) with the functional dependency set F = {AB C, B D, A B, DA C, A DC}. Find a canonical cover of F. Answer: Combine A B and A DC into A BDC, the set becomes {AB C, B D, A BDC, DA C}. B is extraneous in AB C, since A+ using F contains C. D is extraneous in DA C, since A+ using F contains C. The set now becomes {A C, B D, A BDC, A C} = {B D, A BDC}. D is extraneous in A BDC, since A+ using {B D, A BC} contains D. There is no other extraneous attributes. Therefore, the canonical cover is {B D, A BC} Q.8 [15 pts] Consider a relation schema R(A,B,C,D,E,F) and a set of functional dependencies {AB C, C FA, F E} that hold on R. Using only Armstrong’s axioms show that the functional dependency AB E also holds on R. (You should not use rules other than the Armstrong axioms, such as union or decomposition.) An...
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