Let F1(A,B,C) =(A∧~B)∨{−C∧(A∨B)∨~A and F2(A,B)=(A∨B)∨(B→∼A)be two logical expressions. Then,

F1(A,B,C)=(~A∨B)∨(~A)∨(~c∧(A∨B))F2(A,B)=A∨B∨B→~AF1 not tautology; F2 is tautology ABC~A~CA∨B~A∨B~C∧A∨B~A∨B∨~C∧A∨B∨~ATTTFFTTFTTTFFTTTTTTFTFFTFFFTFFFTTFTTFTTTFTTFTFTFTTTTTTFFTTFFTFTFFFTTFTFT ABA∨B~AB→~AA∨B∨B→~ATTTFFTTFTFTTFTTTTTFFFTTT

Let F1(A,B,C) =(A∧~B)∨{−C∧(A∨B)∨~A and F2(A,B)=(A∨B)∨(B→∼A)be two logical expressions. Then,

F1(A,B,C)=(~A∨B)∨(~A)∨(~c∧(A∨B))F2(A,B)=A∨B∨B→~AF1 not tautology; F2 is tautology ABC~A~CA∨B~A∨B~C∧A∨B~A∨B∨~C∧A∨B∨~ATTTFFTTFTTTFFTTTTTTFTFFTFFFTFFFTTFTTFTTTFTTFTFTFTTTTTTFFTTFFTFTFFFTTFTFT ABA∨B~AB→~AA∨B∨B→~ATTTFFTTFTFTTFTTTTTFFFTTT

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Let F₁(A,B,C) $= (A \land ~ B) \lor {- C \land (A \lor B) \lor ~ A$ and $F_{2} (A, B) = (A \lor B) \lor (B \to\sim A)$ be two logical expressions. Then,

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F₁ is not a tautology but F₂ is a tautology

F₁ and F₂ both are tautologies

F₁ is a tautology but F₂ is not a tautology

Both F₁ and F₂ are not tautologies

answer is D.

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Detailed Solution

$\begin{array}{l} F_{1} (A, B, C) = (~ A \lor B) \lor (~ A) \lor (~ c \land (A \lor B)) \\ F_{2} (A, B) = (A \lor B) \lor (B \to ~ A) \\ F_{1} not tautology; F_{2} is tautology \end{array}$

A	B	C	$~ A$	$~ C$	$A \lor B$	$~ A \lor B$	$~ C \land (A \lor B)$	$(~ A \lor B) \lor (~ C \land (A \lor B)) \lor (~ A)$
T	T	T	F	F	T	T	F	T
T	T	F	F	T	T	T	T	T
T	F	T	F	F	T	F	F	F
T	F	F	F	T	T	F	T	T
F	T	T	T	F	T	T	F	T
F	T	F	T	T	T	T	T	T
F	F	T	T	F	F	T	F	T
F	F	F	T	T	F	T	F	T