Question

Consider the following definition of equivalent sets of functional dependencies on a

relation:
“Two sets of functional dependencies F and F’ on a relation R are equivalent if all FD’s

in F’ follow from the ones in F, and all the FD’s in F follow from the ones in F’.” Given a relation R(A, B, C) with the following sets of functional dependencies:

F1 = {A  B, B  C},

F2 = {A  B, A  C}, and

F3 = {A  B, AB  C}:

[a] Are F1 and F3 equivalent? If so, prove that all FD’s in F1 follow from the ones in F3, and vice-versa. If not, give a counterexample by showing an instance (set of tuples) of R such that the FD’s in one of the sets all hold, while some FD in the other set does not hold.

[b] Are F2 and F3 equivalent? If so, prove that all FD’s in F2 follow from the ones in F3, and vice-versa. If not, give a counterexample by showing an instance (set of tuples) of R such that the FD’s in one of the sets all hold, while some FD in the other set does not hold.

[c] Does the given definition of functional dependency equivalence imply that two sets of functional dependencies on a relation are equivalent if and only if the keys of the relation due to each set are identical? Explain your answer.

Answer 1

Given sets of functional dependencies are:
F1 = {A  B, B  C},
F2 = {A  B, A  C}, and
F3 = {A  B, AB  C}:

[a]

Solution :
Step 1 : Take Set F1 and Check F3 is covered from F1 or not.
(A)+ = {B}
(A)+ = {BC}
A → B covered. But AB-->C not coverd.
⇒ F3 is not derived from F1. Hence F3 is not covered by F1.
  
Step 2 : Take Set F3 and Check F1 is covered from F3 or not.
(A)+ = {B}
(AC)+ = {BC}
(E)+ = {ABC}
F1 = {A  B, B  C} is covered.
⇒ F1 is derived from F3. Hence F1 is covered from F3.

Since F3 is not derived from F1, F1 and F3 are not equivalent.

[b]Solution :
Step 1 : Take Set F2 and Check F3 is covered from F1 or not.
(A)+ = {B}
(A)+ = {AC}
A → B covered. But AB-->C not coverd.
⇒ F3 is not derived from F1. Hence F3is not covered by F1.
  
Step 2 : Take Set F3 and Check F2 is covered from F3 or not.
(A)+ = {B}
(AC)+ = {BC}
(E)+ = {ABC}
F1 = {A  B, B  C} is covered.
⇒ F1 is derived from F3. Hence F1 is covered from F3.

Since F3 is not derived from F1, F1 and F3 are not equivalent.