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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