Solve this following

Question:
If $P$ and $Q$ are two statements, then which of the following compound statement is a tautology ?

$((\mathrm{P} \Rightarrow \mathrm{Q}) \wedge \sim \mathrm{Q}) \Rightarrow \mathrm{Q}$
$((\mathrm{P} \Rightarrow \mathrm{Q}) \wedge \sim \mathrm{Q}) \Rightarrow \sim \mathrm{P}$
$((\mathrm{P} \Rightarrow \mathrm{Q}) \wedge \sim \mathrm{Q}) \Rightarrow \mathrm{P}$
$((P \Rightarrow Q) \wedge \sim Q) \Rightarrow(P \wedge Q)$

Correct Option: , 2

Solution:

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) $(\sim \mathrm{P} \wedge \sim \mathrm{Q}) \rightarrow \sim \mathrm{P}$

$\equiv \sim(\sim \mathrm{P} \wedge \sim \mathrm{Q}) \vee \sim \mathrm{P}$

$\equiv(\mathrm{P} \vee \mathrm{Q}) \vee \sim \mathrm{P}$

$\equiv(\mathrm{P} \vee \mathrm{Q}) \vee \mathrm{P} \neq$ 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 :