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# 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.

Last updated date: 13th Jun 2024
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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 false
The proposition $p$ and $q$ denoted by $p \vee q$ is false when $p$ and $q$ are false, otherwise true
The 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.

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.
 $p$ $q$ $\sim p$ $\sim q$ $p \to q$ $\sim q \wedge \sim p$ $(p \to q) \leftrightarrow \left( { \sim q \wedge \sim p} \right)$ $T$ $T$ $F$ $F$ $T$ $T$ $T$ $T$ $F$ $F$ $T$ $F$ $F$ $T$ $F$ $T$ $T$ $F$ $T$ $T$ $T$ $F$ $F$ $T$ $T$ $T$ $T$ $T$

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.