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The given circuit is equivalent to
Options:
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Solution:
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Verified Answer
The correct answer is:
The symbolic form of the given circuit is
$\begin{aligned}
& (\mathrm{p} \vee \sim \mathrm{q} \vee \sim \mathrm{r}) \wedge(\mathrm{p} \vee(\mathrm{q} \wedge \mathrm{r})) \\
& \equiv \mathrm{p} \vee[(\sim \mathrm{q} \vee \sim \mathrm{r}) \wedge(\mathrm{q} \wedge \mathrm{r})]
\end{aligned}$
...[Distributive law]
$\equiv p \vee[\sim(q \wedge r) \wedge(q \wedge r)]$
...[De Morgan's law]
$\equiv \mathrm{p} \vee \mathrm{F}$
...[Complement law]
$\equiv \mathrm{p}$
[Identity law]
$\begin{aligned}
& (\mathrm{p} \vee \sim \mathrm{q} \vee \sim \mathrm{r}) \wedge(\mathrm{p} \vee(\mathrm{q} \wedge \mathrm{r})) \\
& \equiv \mathrm{p} \vee[(\sim \mathrm{q} \vee \sim \mathrm{r}) \wedge(\mathrm{q} \wedge \mathrm{r})]
\end{aligned}$
...[Distributive law]
$\equiv p \vee[\sim(q \wedge r) \wedge(q \wedge r)]$
...[De Morgan's law]
$\equiv \mathrm{p} \vee \mathrm{F}$
...[Complement law]
$\equiv \mathrm{p}$
[Identity law]
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