Solution

Correct answer is Option1

Explanation:Use the union rule to replace

EF → G and EF → H 

EF → GH

F = { A → B  ABCD → E  EF → GH  ACDF → EG }

B is extraneous in ABCD → E because B ∈ ABCD and {A → B, ABCD → E, EF → GH, ACDF → EG}

logically implies {A → B, ACD → E, EF → GH, ACDF → EG}.

This is because every  ACD → E.

This FD can be derived using Armstrong’s Axioms from A → B and ABCD → E via transitivity rule

So remove B from ABCD → E.

F = { A → B ACD → E  EF → GH ACDF → EG }

E is extraneous in ACDF → EG because E ∈ EG and {A → B, ACD → E, EF → GH, ACDF → G}

logically implies {A → B, ACD → E, EF → GH, ACDF → EG}

remove E from ACDF → EG

F = { A → B ACD → E EF → GH ACDF → G}

G is extraneous in ACDF → G. Note that ACDF → G is already implied by ACD → E and EF → GH in F

remove ACDF → G from F.

None of the remaining FD's in F have extraneous attributes so the minimal cover is

A → B, ACD → E, EF → G, EF → H.

Hence the correct answer is option 1.