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.
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.
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.
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.
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