Negation of contrapositive of statement pattern $(p \vee \sim q) \rightarrow(p \wedge \sim q)$ is

Negation of contrapositive of statement pattern $(p \vee \sim q) \rightarrow(p \wedge \sim q)$ is
  1. $(\sim p \wedge q) \vee(p \wedge \sim q)$
  2. $(\sim \mathrm{p} \vee \mathrm{q}) \wedge(\mathrm{p} \vee \sim \mathrm{q})$
  3. $(p \wedge \sim q) \vee(\sim p \wedge \sim q)$
  4. $(\sim \mathrm{p} \vee \sim \mathrm{q}) \wedge(\mathrm{p} \vee \mathrm{q})$

Solution

Contrapositive of $(p \vee \sim q) \rightarrow(p \wedge \sim q)$ is $\sim(p \wedge \sim q) \rightarrow \sim(p \vee \sim q)$ $\begin{aligned} & \equiv \sim[\sim(p \wedge \sim q)] \vee \sim(p \vee \sim q) \ldots[p \rightarrow q \equiv \sim p \vee q] \\ & \equiv(p \wedge \sim q) \vee(\sim p \wedge q) \quad \ldots[\text { De Morgan's law] }\end{aligned}$ Negation of contrapositive of $(p \vee \sim q) \rightarrow(p \wedge \sim q)$ is $\sim[(p \wedge \sim q) \vee(\sim p \wedge q)]$ $\begin{array}{ll}\equiv \sim(p \wedge \sim q) \wedge \sim(\sim p \wedge q) & \ldots[\text { [De Morgan's law] } \\ \equiv(\sim p \vee q) \wedge(p \vee \sim q) & \ldots[\text { De Morgan's law] }\end{array}$

Asked in: MHT CET 2023 (13 May Shift 2)

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