I tried to solve the following question, but I don't know if my solution is correct!
Q: Transform the following proposition into conjunction normal form (CNF):
( ∧ ) ∨ ( ∧ ) ∨ ( → )
A: Let P ≡ The Whole Proposition, 1 ≡ ( ∧ ), 2 ≡ ( ∧ ), and 3 ≡ ( → )
≡ ¬¬( ( ∧ ) ∨ ( ∧ ) ∨ ( → ) ) [Negation Law (P)]
≡ ¬¬( ( ∧ ) ∨ ( ∧ ) ∨ (¬ ∨ ) ) [Material Implication Law (3)]
≡ ¬ ( (¬ ∨ ¬) ∧ (¬ ∨ ¬) ∧ ( ∧ ¬) ) [De Morgan Law (P)]
≡ ¬ ( (¬ ∨ ¬) ∧ (¬ ∨ ¬) ∧ ¬¬( ∧ ¬) ) [Negation Law (3)]
≡ ¬ ( (¬ ∨ ¬) ∧ (¬ ∨ ¬) ∧ ¬(¬ ∨ ) ) [De Morgan Law (3)]
≡ CNF
UPDATE
The correct answer is as follow:
( ∧ ) ∨ ( ∧ ) ∨ ( → )
≡ ( ∧ ) ∨ ( ∧ ) ∨ (¬ ∨ ) [Material Implication Law]
≡ [( ∧ ) ∨ ] ∧ [( ∧ ) ∨ ] ∨ (¬ ∨ ) [Distribution Law]
≡ ( ∨ ) ∧ ( ∨ ) ∧ (( ∨ ) ∧ ( ∨ ) ∨ (¬ ∨ ) [Distribution Law]
≡ ( ∨ ) ∧ ( ∨ ) ∧ ( ∨ ) ∧ ( ∨ ∨ ¬ ∨ ) [Associative Law]
≡ CNF
Thanks to (@Bram28).
HINT
The equivalence rule you want to use is
Distribution
$P \lor (Q \land R) \Leftrightarrow (P \lor Q) \land (P \lor R)$
Using this you can turn disjunctions of conjunctions into conjunctions of disjunctions, which is what you want to do to get things into CNF.