Consider, Statement-I $$(p \wedge \sim q) \wedge \left( { \sim p \wedge q} \right)$$ is a fallacyStatement-II $$\left( {p \to q} \right) \leftrightarrow \left( { \sim q \to \sim p} \right)$$ is a tautologyA) Statement-I is false, Statement-II is trueB) Statement-I is true, Statement-II is correct explanation for Statement-IC) Statement-I is true, Statement-II is true and Statement-II is not correct explanation for Statement-ID) Statement-I is true, Statement-II is false.

Question

Consider, Statement-I $$(p \wedge  \sim q) \wedge \left( { \sim p \wedge q} \right)$$ is a fallacyStatement-II $$\left( {p \to q} \right) \leftrightarrow \left( { \sim q \to  \sim p} \right)$$ is a tautologyA) Statement-I is false, Statement-II is trueB) Statement-I is true, Statement-II is correct explanation for Statement-IC) Statement-I is true, Statement-II is true and Statement-II is not correct explanation for Statement-ID) Statement-I is true, Statement-II is false.

Vedantu Content Team · Accepted Answer

Hint: Here, the concept is solve by using truth table in the manner,The proposition $$p$$ and $$q$$ denoted by $$p \wedge q$$ is true when $$p$$ and $$q$$are true, otherwise falseThe proposition $$p$$ and $$q$$ denoted by $$p \vee q$$ is false when $$p$$ and $$q$$ are false, otherwise trueThe proposition$$p$$, $$ \sim p$$ is called negation of $$p$$The implication $$p \to q$$ is the proposition that is false when $$p$$ is true and $$q$$ is false and true otherwise.The bi-conditional $$p \leftrightarrow q$$ is the proposition that is true when $$p$$ and $$q$$ have the same truth values as false otherwise.We use the above concept to find whether the statements are true or false.Complete step-by-step answer:Every statement is either true or false.$$p$$$$q$$$$ \sim p$$$$ \sim q$$$$p \wedge  \sim q$$$$ \sim p \wedge q$$$$\left( {p \wedge  \sim q} \right) \wedge ( \sim p \wedge q)$$$$T$$$$T$$$$F$$$$F$$$$F$$$$F$$$$F$$$$T$$$$F$$$$F$$$$T$$$$T$$$$F$$$$F$$$$F$$$$T$$$$T$$$$F$$$$F$$$$T$$$$F$$$$F$$$$F$$$$T$$$$T$$$$F$$$$F$$$$F$$A compound proposition that is always false, no matter what the truth values of the propositions that occur is called a fallacy.Hence the statement-I is true,   $$(p \wedge  \sim q) \wedge \left( { \sim p \wedge q} \right)$$ is fallacy.$$p$$$$q$$$$ \sim p$$$$ \sim q$$$$p \to q$$$$ \sim q \wedge  \sim p$$\t$$(p \to q) \leftrightarrow \left( { \sim q \wedge  \sim p} \right)$$$$T$$$$T$$$$F$$$$F$$$$T$$$$T$$$$T$$$$T$$$$F$$$$F$$$$T$$$$F$$$$F$$$$T$$$$F$$$$T$$$$T$$$$F$$$$T$$$$T$$$$T$$$$F$$$$F$$$$T$$$$T$$$$T$$$$T$$$$T$$A compound proposition that is always true, no matter what the truth values of the propositions that occur is called a tautology.Hence the statement-II is true, $$(p \to q) \leftrightarrow \left( { \sim q \wedge  \sim p} \right)$$is tautology.Hence the option (c) is correct, the statement-I is true, statement-II is true and the statement-II is not the correct explanation for statement-I.Note: The statements are either true or false. This is called the law of the excluded middle.A truth table shows how the truth or falsity of a compound statement depends on the truth or falsity of the simple statements from which it’s constructed.

\[p\]	\[q\]	\[ \sim p\]	\[ \sim q\]	\[p \wedge \sim q\]	\[ \sim p \wedge q\]	\[\left( {p \wedge \sim q} \right) \wedge ( \sim p \wedge q)\]
\[T\]	\[T\]	\[F\]	\[F\]	\[F\]	\[F\]	\[F\]
\[T\]	\[F\]	\[F\]	\[T\]	\[T\]	\[F\]	\[F\]
\[F\]	\[T\]	\[T\]	\[F\]	\[F\]	\[T\]	\[F\]
\[F\]	\[F\]	\[T\]	\[T\]	\[F\]	\[F\]	\[F\]

\[p\]	\[q\]	\[ \sim p\]	\[ \sim q\]	\[p \to q\]	\[ \sim q \wedge \sim p\]	\[(p \to q) \leftrightarrow \left( { \sim q \wedge \sim p} \right)\]
\[T\]	\[T\]	\[F\]	\[F\]	\[T\]	\[T\]	\[T\]
\[T\]	\[F\]	\[F\]	\[T\]	\[F\]	\[F\]	\[T\]
\[F\]	\[T\]	\[T\]	\[F\]	\[T\]	\[T\]	\[T\]
\[F\]	\[F\]	\[T\]	\[T\]	\[T\]	\[T\]	\[T\]