Question

Let F₁(A,B,C) = (A∧~B) ∨ [~C ∧ (A ∨ B)] ∨~A and F₂(A, B) = (A ∨ B) ∨ (B →~A) be two logical expressions. Then :

(1) F₁ and F₂ both are tautologies

(2) F₁ is a tautology but F₂ is not a tautology

(3) F₁ is not tautology but F₂ is a tautology

(4) Both F₁ and F₂ are not tautologie

Answer 1

Correct option is (3) (3) F₁ is not tautology but F₂ is a tautology

F₁ : (A ∧~B) ∨[~C ∧ (A ∨ B)] ∨~A

F₂ : (A ∨ B) ∨ (B →~A)

F₁ : {(A ∧~B) ∨~A} ∨ [(A ∨ B) ∧~C]

: {(A ∨ ~A) ∧ (~A ∨ ~B)} ∨ [(A ∨ B) ∧ ~C]

: {t ∧ (~A ∨ ~B)} ∨ [(A ∨ B) ∧ ~C]

: (~A ∨ ~B) ∨ [(A ∨ B) ∧ ~C]

: \(\underbrace{[(\text ~A∨\text~B∨(A∨B))]}_t\)\([(\text ~A∨\text~B)∧\text~C]\)

F₁ : (~A ∨ ~B) ∧ ~ C ≠ t (tautology)

F₂ : (A ∨ B) ∨ (~B ∨ ~A) = t (tautology)