Let R (ABCDEFGH) be a relation schema and F be the set of dependencies 𝐹 =…

Compute the minimal cover for F = {A→B, ABCD→E, EF→G, EF→H, ACDF→EG}.

Step 1: Split right-hand sides so each dependency has a single attribute on the right.

• A→B

• ABCD→E

• EF→G

• EF→H

• ACDF→E

• ACDF→G

Step 2: Minimize left-hand sides by removing any extraneous attributes.

1. Check ABCD→E: test if B is extraneous. Compute closure of {A,C,D}: A gives B (A→B), so {A,C,D}+ contains B, then ABCD→E yields E. Hence B is extraneous and ABCD→E reduces to ACD→E.

Step 3: Remove redundant dependencies.

• ACDF→E is implied by ACD→E because ACDF contains ACD, so ACDF→E is redundant.

• ACDF→G is implied as follows: ACDF gives E (via ACD→E) and with F present we have EF, and EF→G gives G. Thus ACDF→G is redundant.

Final minimal cover:

• A→B

• ACD→E

• EF→G

• EF→H

This matches the minimal cover provided in the correct option.