Statement $-1 :$ $\sim (p \leftrightarrow \sim q)$ is equivalent to $p\leftrightarrow q $
Statement $-2 :$ $\sim (p \leftrightarrow \sim q)$ s a tautology
Statement $-1 :$ $\sim (p \leftrightarrow \sim q)$ is equivalent to $p\leftrightarrow q $
Statement $-2 :$ $\sim (p \leftrightarrow \sim q)$ s a tautology
- [AIEEE 2009]
- A
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$
- B
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$
- C
Statement $-1$ is false, Statement $-2$ is true
- D
Statement $-1$ is true, Statement $-2$ is false
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- [JEE MAIN 2019]
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- [JEE MAIN 2022]
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Consider
Statement $-1 :$$\left( {p \wedge \sim q} \right) \wedge \left( { \sim p \wedge q} \right)$ is a fallacy.
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Statement $-1 :$$\left( {p \wedge \sim q} \right) \wedge \left( { \sim p \wedge q} \right)$ is a fallacy.
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- [JEE MAIN 2013]