Question

# given a set of functional dependencies

given a set of functional dependencies

F = {AB → C, C → B}, above the relational scheme R (A, B, C).

Prove that {AB → AC, AB → BC, AC → B, AC → AB} F +

Solution:

Given,

=>Set of functional dependency F = {AB -> C, C -> B}

=>Relation = R(A, B, C)

To prove: F+ = {AB -> AC, AB -> BC, AC -> B, AC -> AB}

Explanation:

Deriving AB -> AC:

=>(AB)+ = ABC using functional dependency {AB -> C, C -> B} hence we can write AB -> AC

Deriving AB -> BC:

=>(AB)+ = ABC using functional dependency {AB -> C, C -> B} hence we can write AB -> BC

Deriving AC -> B:

=>(AC)+ = ABC using functional dependency {AB -> C, C -> B} hence we can write AC -> B

Deriving AC -> AB:

=>(AC)+ = ABC using functional dependency {AB -> C, C -> B} hence we can write AC -> AB.

=>As we can see that we can cover all the functional dependency of the F+ set using attribute closure; hence F+ set is correct.

=>Hence based on the above statements, we have proved our result.

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