The negation of the converse of the statement | vedclass

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(D) Step $1$: The given statement is $p \lor q$. The converse of a statement $p 	o q$ is $q 	o p$. However, $p \lor q$ is a logical disjunction, not

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$\sim p \land q$

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(D) Step $1$: The given statement is $p \lor q$. The converse of a statement $p 	o q$ is $q 	o p$. However, $p \lor q$ is a logical disjunction, not a conditional statement. In the context of logic problems, the converse of $p \lor q$ is interpreted as the converse of the implication $p 	o q$, which is $q 	o p$.Step $2$: The negation of the implication $q 	o p$ is given by $\sim(q 	o p)$.Step $3$: Using the logical equivalence $\sim(q 	o p) \equiv q \land \sim p$.Step $4$: Therefore, the negation is $q \land \sim p$, which is equivalent to $\sim p \land q$.

The negation of the converse of the statement $p \lor q$ is:

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The negation of the converse of the statement $p \lor q$ is:

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Statement-$1$: $(p \wedge \sim q) \wedge (\sim p \wedge q)$ is a fallacy.
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The negation of the converse of the statement $p \lor q$ is:

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Check the validity of the statement given below by the method specified against it.$q:$ If $n$ is a real number with $n > 3$,then $n^{2} > 9$ (by contradiction method).

The negation of the converse of the statement $p \lor q$ is:

ConsiderStatement-$1$: $(p \wedge \sim q) \wedge (\sim p \wedge q)$ is a fallacy.Statement-$2$: $(p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p)$ is a tautology.

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Check the validity of the statement given below by the method specified against it.
$q:$ If $n$ is a real number with $n > 3$,then $n^{2} > 9$ (by contradiction method).

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Statement-$1$: $(p \wedge \sim q) \wedge (\sim p \wedge q)$ is a fallacy.
Statement-$2$: $(p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p)$ is a tautology.