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.

Similar Questions

Statement$-I :$ $\sim (p\leftrightarrow q)$ is equivalent to $(p\wedge \sim q)\vee \sim (p\vee \sim q) .$
Statement$-II :$ $p\rightarrow (p\rightarrow q)$ is a tautology.

The negative of $q\; \vee \sim (p \wedge r)$ is

Statement $-1$ : The statement $A \to (B \to A)$ is equivalent to $A \to \left( {A \vee B} \right)$.

Statement $-2$ : The statement $ \sim \left[ {\left( {A \wedge B} \right) \to \left( { \sim A \vee B} \right)} \right]$ is a Tautology

Given the following two statements :

$\left( S _{1}\right):( q \vee p ) \rightarrow( p \leftrightarrow \sim q )$ is a tautology.

$\left( S _{2}\right): \sim q \wedge(\sim p \leftrightarrow q )$ is a fallacy.

Then

The statement $(p \wedge(\sim q) \vee((\sim p) \wedge q) \vee((\sim p) \wedge(\sim q))$ is equivalent to

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.