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Negation of the Boolean expression $\mathrm{p} \Leftrightarrow(\mathrm{q} \Rightarrow \mathrm{p})$ is
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Verified Answer
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$(\sim p) \wedge(\sim q)$
Given expression is $p \Leftrightarrow(q \Rightarrow p)$
$\sim(\mathrm{p} \leftrightarrow(\mathrm{q} \rightarrow \mathrm{p}))$
$\sim(\mathrm{p} \leftrightarrow \mathrm{q})=(\mathrm{p} \wedge \sim \mathrm{q}) \vee(\mathrm{q} \wedge \sim \mathrm{p})$
$\sim(\mathrm{p} \leftrightarrow(\mathrm{q} \rightarrow \mathrm{p}))$ $=(\mathrm{p} \wedge \sim(\mathrm{q} \rightarrow \mathrm{p})) \vee((\mathrm{q} \rightarrow \mathrm{p}) \wedge \sim \mathrm{p})$
$(\mathrm{p} \wedge \sim(\mathrm{q} \rightarrow \mathrm{p}))=\mathrm{p} \wedge(\mathrm{q} \wedge \sim \mathrm{p})$ $=(p \wedge \sim p) \wedge q=c$
$(q \rightarrow p) \wedge \sim p=(\sim q \vee p) \wedge \sim p$ $=\sim p \wedge(\sim q \vee p)$
$=(\sim \mathrm{p} \wedge \sim \mathrm{q}) \vee(\sim \mathrm{p} \wedge \mathrm{p})=\sim \mathrm{p} \wedge \sim \mathrm{q}$
$\sim(p \leftrightarrow(\dot{q} \rightarrow p))=c \vee(\sim p \wedge \sim q)=\sim p \wedge \sim q$
$\sim(\mathrm{p} \leftrightarrow(\mathrm{q} \rightarrow \mathrm{p}))$
$\sim(\mathrm{p} \leftrightarrow \mathrm{q})=(\mathrm{p} \wedge \sim \mathrm{q}) \vee(\mathrm{q} \wedge \sim \mathrm{p})$
$\sim(\mathrm{p} \leftrightarrow(\mathrm{q} \rightarrow \mathrm{p}))$ $=(\mathrm{p} \wedge \sim(\mathrm{q} \rightarrow \mathrm{p})) \vee((\mathrm{q} \rightarrow \mathrm{p}) \wedge \sim \mathrm{p})$
$(\mathrm{p} \wedge \sim(\mathrm{q} \rightarrow \mathrm{p}))=\mathrm{p} \wedge(\mathrm{q} \wedge \sim \mathrm{p})$ $=(p \wedge \sim p) \wedge q=c$
$(q \rightarrow p) \wedge \sim p=(\sim q \vee p) \wedge \sim p$ $=\sim p \wedge(\sim q \vee p)$
$=(\sim \mathrm{p} \wedge \sim \mathrm{q}) \vee(\sim \mathrm{p} \wedge \mathrm{p})=\sim \mathrm{p} \wedge \sim \mathrm{q}$
$\sim(p \leftrightarrow(\dot{q} \rightarrow p))=c \vee(\sim p \wedge \sim q)=\sim p \wedge \sim q$
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