Question:
Consider the two statements:
$(\mathrm{S} 1):(\mathrm{p} \rightarrow \mathrm{q}) \vee(\sim \mathrm{q} \rightarrow \mathrm{p})$ is a tautology.
$(S 2):(p \wedge \sim q) \wedge(\sim p \vee q)$ is a fallacy.
Then :
Correct Option: , 3
Solution:
$S_{1}:(\sim p \vee q) \vee(q \vee p)=(q \vee \sim p) \vee(q \vee p)$
$S_{1}=q \vee(\sim p \vee p)=q \vee t=t=$ tautology
$S_{2}:(p \wedge \sim q) \wedge(\sim p \vee q)=(p \wedge \sim q) \wedge \sim(p \wedge \sim q)=C$
$=$ fallacy
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