Sơ đồ suy luận nào đúng?

(a→b)∧(c→d)∧(b∨d){(a → b) ∧ (c → d)} ∧ (b ∨ d)(a→b)∧(c→d)∧(b∨d) ⇒ (a ∨ c).

(a→ b)∧(c→ d)∧(b∨d){(a → ~b) ∧ (c → ~d)} ∧ (b ∨ d)(a→ b)∧(c→ d)∧(b∨d) ⇒ ~(a ∧ c).

(a→b)∧(c→d)∧( b∨ d){(a → b) ∧ (c → d)} ∧ (~b ∨ ~d)(a→b)∧(c→d)∧( b∨ d) ⇒ (a ∨ c).

(a→b)∧(c→d)∧( b∨ d){(a → b) ∧ (c → d)} ∧ (~b ∨ ~d)(a→b)∧(c→d)∧( b∨ d) ⇒ ~(a ∨ c).