Q.
R(ABCD)
F = {A → B, B → C, C → D}
D = {AB, BC, CD}
So, here all FDs of F present at F1 ∪ F2. So, it is dependency preserving.
Q.
R (ABCD)
F = {A → B, B → C, C → D, D → A, A → C A → D}
D = {AD, BC, CD}
Only left A → B, which can easily derive from F1 ∪ F2.
Or
So, F1 ∪ F2 ⊇ F is true and F1 ∪ F2 ⊆ F always true then F1 ∪ F2 = F (true), it is dependency preserving.
Don’t think that at F1 ∪ F2 number of FDs is larger compare to the given F, but actually, both are the same F1 ∪ F2 contain some redundant FDs,
Both are contained the same number of FDs.
Q.
R (ABCDE)
F = {AB → CD, C → D, D → E}
D = {ABC, CD, DE}
It is dependency preserving.
Q.
R (ABCDEF)
F = {AB → CD, C → D, D → E, E → F} and D = {AB, CDE, EF}
No Functional Dependencies derive or possible.
But it better to write minimize the way
Here AB → CD FD not present at F1 ∪ F2 so, it is not dependency preserving.
Q.
R (ABCDEG)
F = {AB → C, AC → B, AD → E, B → D, BC → A, E → G}
D = {ABC, ACDE, ADG}
Here B → D and E → G FDs are lost so, it is not dependency preserving.
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