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Consider,
Statement-I \[(p \wedge \sim q) \wedge \left( { \sim p \wedge q} \right)\] is a fallacy
Statement-II \[\left( {p \to q} \right) \leftrightarrow \left( { \sim q \to \sim p} \right)\] is a tautology
A) Statement-I is false, Statement-II is true
B) Statement-I is true, Statement-II is correct explanation for Statement-I
C) Statement-I is true, Statement-II is true and Statement-II is not correct explanation for Statement-I
D) Statement-I is true, Statement-II is false.

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Last updated date: 13th Jun 2024
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Answer
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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.

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

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.