Consider

Statement -1 :left( {p wedge | vedclass

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Question

Statement-ll: $\quad(p \rightarrow q) \leftrightarrow(\sim q \rightarrow-p)$

$\equiv(p \rightarrow q) \leftrightarrow(p \rightarrow q)$

which is

Vedclass · Accepted Answer

Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$

Vedclass · Answer

Statement-ll: $\quad(p ightarrow q) \leftrightarrow(\sim q ightarrow-p)$

$\equiv(p ightarrow q) \leftrightarrow(p ightarrow q)$

which is always true

so statement- -II is true

$ 	ext { 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: $\quad(p ightarrow q) \leftrightarrow(\sim q ightarrow \sim p)$

$\sim \mathrm{q} ightarrow \sim \mathrm{p}$ is contrapositive

of $p ightarrow q$ hence $(p ightarrow q) \leftrightarrow(p ightarrow q)$

will be a tautology

statement-ll $\quad(p \wedge \sim q) \wedge(\sim p \wedge q)$

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.

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