Let R(ABCDEFGH) be a relation schema and F = {A → B, ABCD → E, EF → G, EF → H, ACDF → EG}. The minimal cover of this set of dependencies is

1. AA → B, ACD → E, EF → G and EF → H
2. BA → B, ACD → E, EF → G, EF → H and ACDF → G
3. CA → B, ACD → E, EF → G, EF → H and ACDF → E
4. DA → B, ABCD → E, EF → H and EF → G

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Correct answer: A. A → B, ACD → E, EF → G and EF → H

Explanation

A → B makes B extraneous in ABCD → E, giving ACD → E. ACDF → E follows from ACD → E, and ACDF → G follows from ACD → E and EF → G, so ACDF → EG is redundant. The minimal cover is A → B, ACD → E, EF → G, EF → H.