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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

in MATHEMATICAL REASONING by Expert (11.5k points) | 49 views

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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)

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