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