Consider the following set of functional dependencies on the schema (A, B, C).…
2022
Consider the following set of functional dependencies on the schema (A, B, C). A → BC, B → C, A → B, AB → C. Then the canonical cover for this set is —
- A.
A → BC and B → C
- B.
A → BC and A → B
- C.
A → BC and AB → C
- D.
A → B and B → C
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Correct answer: D
Split A → BC into two dependencies: A → B and A → C .
We now have: A → B, A → C, B → C, AB → C.
Check AB → C : it is implied by B → C (so AB → C is redundant ) — remove it.
Check A → C : since A → B and B → C hold, A → C is implied by transitivity (A→B and B→C ⇒ A→C). So A → C is redundant — remove it.
Remaining minimal set (canonical cover) is { A → B, B → C } , which is equivalent to the original set.