Sơ đồ suy luận nào đúng?

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

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

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

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