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Consider
Statement $-1 :$$\left( {p \wedge \sim q} \right) \wedge \left( { \sim p \wedge q} \right)$ is a fallacy.
Statement $-2 :$$(p \rightarrow q) \leftrightarrow ( \sim q \rightarrow \sim p )$ is a tautology.
Statement $-1$ is false, Statement $-2$ is true
Statement $-1$ is true, Statement $-2$ is false
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$
Solution
Statement-ll: $\quad(p \rightarrow q) \leftrightarrow(\sim q \rightarrow-p)$
$\equiv(p \rightarrow q) \leftrightarrow(p \rightarrow q)$
which is always true
so statement- -II is true
$ \text { Statement-l: } (p \wedge \sim q) \wedge(\sim p \wedge q)$
$=p \wedge \sim q \wedge \sim p \wedge q $
$=p \wedge \sim p \wedge \sim q \wedge q $
$=f \wedge f $
$=f $
so statement- – is true
Alternate
Statement-ll: $\quad(p \rightarrow q) \leftrightarrow(\sim q \rightarrow \sim p)$
$\sim \mathrm{q} \rightarrow \sim \mathrm{p}$ is contrapositive
of $p \rightarrow q$ hence $(p \rightarrow q) \leftrightarrow(p \rightarrow q)$
will be a tautology
statement-ll $\quad(p \wedge \sim q) \wedge(\sim p \wedge q)$
