Q46 / 120 · VITEEE 2013 · Mathematics

The proposition $\sim(p \Leftrightarrow q)$ is equivalent to

  • A
    $(\mathrm{p} \vee \sim \mathrm{q}) \wedge(\mathrm{q} \wedge \sim \mathrm{p})$
  • Correct answer
    B
    $(\mathrm{p} \wedge \sim \mathrm{q}) \vee(\mathrm{q} \wedge \sim \mathrm{p})$
  • C
    $(\mathrm{p} \wedge \sim \mathrm{q}) \wedge(\mathrm{q} \wedge \sim \mathrm{p})$
  • D
    None of the above

Solution

$\sim(p \Leftrightarrow q) \equiv \sim[(p \Rightarrow q) \wedge(q \Rightarrow p)]$ $\equiv \sim(p \Rightarrow q) \vee(\sim(q \Rightarrow p))$ $(\because$ De-Morgan's law $)$ $\equiv(p \wedge \sim q) \vee(q \wedge \sim p)$

Hence, the correct answer is option: 'B'.

Relevant Questions