If the Boolean expression $(\mathrm{p} \wedge \mathrm{q}) \circledast(\mathrm{p} \otimes \mathrm{q})$ is a tautology, then $\circledast$ and $\otimes$ are respectively given by
Correct Option: 1
Option (1)
$(\mathrm{p} \wedge \mathrm{q}) \longrightarrow(\mathrm{p} \rightarrow \mathrm{q})$
$=\sim(\mathrm{p} \wedge \mathrm{q}) \vee(\sim \mathrm{p} \vee \mathrm{q})$
$=(\sim \mathrm{p} \vee \sim \mathrm{q}) \vee(\sim \mathrm{p} \vee \mathrm{q})$
$=\sim \mathrm{p} \vee(\sim \mathrm{q} \vee \mathrm{q})$
$=\sim \mathrm{p} \vee \mathrm{t}$
$=\mathrm{t}$
Option (2)
$(\mathrm{p} \wedge \mathrm{q}) \wedge(\mathrm{p} \vee \mathrm{q})=(\mathrm{p} \wedge \mathrm{q})($ Not a tautology $)$
Option (3) $(\mathrm{p} \wedge \mathrm{q}) \vee(\mathrm{p} \rightarrow \mathrm{q})$
$=(\mathrm{p} \wedge \mathrm{q}) \vee(\sim \mathrm{p} \vee \mathrm{q})$
$=\sim p \vee q$
(Not a tautology)
Option (4) $(\mathrm{p} \wedge \mathrm{q}) \wedge(\mathrm{p} \rightarrow \mathrm{q})$
$=(\mathrm{p} \wedge \mathrm{q}) \wedge(\sim \mathrm{p} \vee \mathrm{q})$
$=\mathrm{p} \wedge \mathrm{q}$
(Not a tautology)
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