Q139 / 200 · Manipal (MET) 2013 · Mathematics

$(p \wedge \sim q) \wedge(\sim p \wedge q)$

  • A
    A tautology
  • Correct answer
    B
    A contradiction
  • C
    Both a tautology and a contradiction
  • D
    Neither a tautology nor a contradiction

Solution

$\begin{aligned}(p \wedge \sim q)(\sim p \wedge q) & =(p \wedge \sim p) \wedge(\sim q \wedge q) \\ & =f \wedge f=f\end{aligned}$ (By using associative laws and commutative laws.) $\therefore \quad(p \wedge \sim q) \wedge(\sim p \wedge q)$ is a contradiction.

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

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