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Which one of the following is a tautology ?
$\mathrm{P} \wedge(\mathrm{P} \vee \mathrm{Q})$
$\mathrm{P} \vee(\mathrm{P} \wedge \mathrm{Q})$
$\mathrm{Q} \rightarrow(\mathrm{P} \wedge(\mathrm{P} \rightarrow \mathrm{Q}))$
$(\mathrm{P} \wedge(\mathrm{P} \rightarrow \mathrm{Q})) \rightarrow \mathrm{Q}$
Solution
$\mathrm{P} \wedge(\mathrm{P} \vee \mathrm{Q}) \equiv \mathrm{P}$
$\mathrm{P} \vee(\mathrm{P} \wedge \mathrm{Q}) \equiv \mathrm{P}$
$\mathrm{Q} \rightarrow(\mathrm{P} \wedge(\mathrm{P} \rightarrow \mathrm{Q}))$
$\equiv \mathrm{Q} \rightarrow(\mathrm{P} \wedge(\sim \mathrm{P} \vee \mathrm{Q})) \equiv \mathrm{Q} \rightarrow(\mathrm{P} \wedge \mathrm{Q})$
$\equiv(\sim Q) \vee(P \wedge Q) \equiv(P \vee(\sim Q))$
$(\mathrm{P} \wedge(\mathrm{P} \rightarrow \mathrm{Q})) \rightarrow \mathrm{Q}$
$\equiv(\mathrm{P} \wedge(\sim \mathrm{P} \vee \mathrm{Q})) \rightarrow \mathrm{Q} \equiv(\mathrm{P} \wedge \mathrm{Q}) \rightarrow \mathrm{Q}$
$\equiv((\sim P) \vee(\sim Q)) \vee Q \equiv(\sim P) \vee t \equiv t$
