Question

Consider the schema R = (A, B, C, D, E) and let the following set F of functional dependencies holdforR: F = {A -> BC, CD -> E, C -> A, B -> D,}

1) Prove or disprove ADE is in the closure of F. A proof can be made by using inference rules IR1 through IR3. A disproof should be done by showing a relational instance (counter example) that refutes the rule.

2) What are the candidate keys of R = (A, B, C, D, E)? Explain your answer.

Answer 1

Given Relation , R = (A, B, C, D, E)

         Functional dependencies F = {A -> BC, CD -> E, C -> A, B -> D}

1)

         we have A -> BC from Augmentation rule we can add attribute D to this FD,

                   AD -> BCD

                   AD -> BE         ( as CD -> E )

             from decomposition rule we can write X -> YZ as X -> Y and X -> Z

   so AD -> BE implies AD -> B and AD -> E

   Hence AD -> E dervied.

2)

        To find candidate keys we can use closure set of attributes

                  A⁺ = A

                       = ABC            ( from A -> BC )

                       = ABCD          ( from B -> D )

   = ABCDE    ( from CD -> E )

Hence A⁺ = ABCDE , here all the attributes of the table are derived so A is a candidate key.

                    

                  C⁺ = C

                       = CA            ( from C -> A)

                       = ABC          ( from A -> BC )

   = ABCD    ( from B -> D )

                       = ABCDE      ( from CD -> E )

Hence C⁺ = ABCDE , here all the attributes of the table are derived so C is a candidate key.

So the candidate keys in the table are A and C.