F1(A,B,C)=(A∧∼B)∨[∼C∧(A∨B)]∨∼A ≡{(A∧∼B)∨[∼C∧(A∨B)]‌‌ (Associative law) ≡{(A∨∼A)∧(∼B∨∼A)}∨[(A∨B)∧C]‌‌ (distributive law) ≡(∼B∨∼A)∨[(A∨B)∧∼C]‌‌(∵A∨∼A≡T,T∧p≡p) ≡[(∼A∼B)∨(A∨B)]∧[(∼A∨∼B)∨∼C] ≡T∧(∼A∨∼B)∨∼C ≡(∼A∨∼B)∨∼C Which is not a tautology. Now, F2(A,B)‌‌=(A∨B)∨(B→∼A) ‌≡(A∨B)∨(∼B∨∼A) Hence, F2 is a tautology.