If $P$ and $Q$ are two statements, then which of the following compound statement is a tautology ?
Correct Option: , 2
LHS of all the options are some i.e.
$((\mathrm{P} \rightarrow \mathrm{Q}) \wedge \sim \mathrm{Q})$
$\equiv(\sim \mathrm{P} \vee \mathrm{Q}) \wedge \sim \mathrm{Q}$
$\equiv(\sim \mathrm{P} \wedge \sim \mathrm{Q}) \vee(\mathrm{Q} \wedge \sim \mathrm{Q})$
$\equiv \sim \mathrm{P} \wedge \sim \mathrm{Q}$
(A) $(\sim \mathrm{P} \wedge \sim \mathrm{Q}) \rightarrow \mathrm{Q}$
$\equiv \sim(\sim \mathrm{P} \wedge \sim \mathrm{Q}) \vee \mathrm{Q}$
$\equiv(\mathrm{P} \vee \mathrm{Q}) \vee \mathrm{Q} \neq$ tautolog $\mathrm{y}$
(B) $(\sim \mathrm{P} \wedge \sim \mathrm{Q}) \rightarrow \sim \mathrm{P}$
$\equiv \sim(\sim \mathrm{P} \wedge \sim \mathrm{Q}) \vee \sim \mathrm{P}$
$\equiv(\mathrm{P} \vee \mathrm{Q}) \vee \sim \mathrm{P}$
(C) $(\sim \mathrm{P} \wedge \sim \mathrm{Q}) \rightarrow \mathrm{P}$
$\equiv(\mathrm{P} \vee \mathrm{Q}) \vee \mathrm{P} \neq$ Tautology
(D) $(\sim \mathrm{P} \wedge \sim \mathrm{Q}) \rightarrow(\mathrm{P} \wedge \mathrm{Q})$
$\equiv(P \vee Q) \vee(P \wedge Q) \neq$ Tautology
Aliter :
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