The negation of contrapositive of the statement $\mathrm{p} \rightarrow(\sim \mathrm{q} \wedge \mathrm{r})$ is

The negation of contrapositive of the statement $\mathrm{p} \rightarrow(\sim \mathrm{q} \wedge \mathrm{r})$ is
  1. $(\sim q \vee \sim r) \wedge \sim p$
  2. $(q \vee \sim r) \wedge p$
  3. $(q \wedge \sim r) \vee p$
  4. $\quad(\sim q \wedge \sim r) \vee \sim p$

Solution

Contrapositive of the statement $\mathrm{p} \rightarrow(\sim \mathrm{q} \wedge \mathrm{r})$ is $\begin{aligned} & \sim(\sim q \wedge r) \rightarrow \sim p \\ & \equiv(q \vee \sim r) \rightarrow \sim p \end{aligned}$ ...[De Morgan's law] Negation of contrapositive of $p \rightarrow(\sim q \wedge r)$ is $\begin{aligned} & \sim[(q \vee \sim r) \rightarrow \sim p] \\ & \equiv(q \vee \sim r) \wedge \sim(\sim p) \quad \ldots[\because \sim(p \rightarrow q) \equiv p \wedge \sim q] \\ & \equiv(q \vee \sim r) \wedge p \end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 2)

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