Let \(F_1(A,B,C)=(A\wedge\sim B)\vee[\sim C\wedge(A\vee B)]\vee\sim A\) and
\(F_2(A,B)=(A\vee B)\vee(B\to\sim A)\) be two logical expressions. Then :
\(F_2(A,B)=(A\vee B)\vee(B\to\sim A)\) be two logical expressions. Then :
Show Hint
- Both F1 and F2 are not tautologies
- F1 and F2 both are tautologies
- F1 is not a tautology but F2 is a tautology
- F1 is a tautology but F2 is not a tautology
The Correct Option is C
Solution and Explanation
\(F_1\): \((A\wedge\sim B)\vee\sim A=\sim A\vee\sim B\) (absorption), so \(F_1=\sim A\vee\sim B\vee[\sim C\wedge(A\vee B)]\).
Take \(A=T,B=T,C=T\): \(\sim A\vee\sim B=F\) and \(\sim C\wedge(A\vee B)=F\wedge T=F\), so \(F_1=F\). Hence \(F_1\) is not a tautology.
Answer: (3) F1 is not a tautology but F2 is a tautology
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