Q92 / 150 · MHT CET 2024 (11 May Shift 2) · Mathematics

The following statement $(\mathrm{p} \rightarrow \mathrm{q}) \rightarrow((\sim \mathrm{p} \rightarrow \mathrm{q}) \rightarrow \mathrm{q})$ is

  • A
    a fallacy.
  • B
    equivalent to $(\sim \mathrm{p}) \rightarrow \mathrm{q}$.
  • C
    equivalent to $\mathrm{p} \rightarrow(\sim \mathrm{q})$.
  • Correct answer
    D
    a tautology.

Solution

$\begin{aligned} & (p \rightarrow q) \rightarrow((\sim p \rightarrow q) \rightarrow q) \\ & \equiv(p \rightarrow q) \rightarrow((p \vee q) \rightarrow q) \\ & \equiv(p \rightarrow q) \rightarrow(\sim(p \vee q) \vee q) \\ & \equiv(p \rightarrow q) \rightarrow((\sim p \wedge \sim q) \vee q) \\ & \equiv(p \rightarrow q) \rightarrow((\sim p \vee q) \wedge(\sim q \vee q)) \\ & \equiv(p \rightarrow q) \rightarrow((\sim p \vee q) \wedge T) \\ & \equiv(p \rightarrow q) \rightarrow(\sim p \vee q) \\ & \equiv(p \rightarrow q) \rightarrow(p \rightarrow q) \\ & \equiv T\end{aligned}$

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

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