આપેલ પૈકી ક્યૂ વિધાન નિત્ય સત્ય છે ?

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Question

$\mathrm{P} \wedge(\mathrm{P} \vee \mathrm{Q}) \equiv \mathrm{P}$

$\mathrm{P} \vee(\mathrm{P} \wedge \mathrm{Q}) \equiv \mathrm{P}$

$\mathrm{Q}

Vedclass · Accepted Answer

$(\mathrm{P} \wedge(\mathrm{P} \rightarrow \mathrm{Q})) \rightarrow \mathrm{Q}$

Vedclass · Answer

$\mathrm{P} \wedge(\mathrm{P} \vee \mathrm{Q}) \equiv \mathrm{P}$

$\mathrm{P} \vee(\mathrm{P} \wedge \mathrm{Q}) \equiv \mathrm{P}$

$\mathrm{Q} \rightarrow(\mathrm{P} \wedge(\mathrm{P} \rightarrow \mathrm{Q}))$

$\equiv \mathrm{Q} \rightarrow(\mathrm{P} \wedge(\sim \mathrm{P} \vee \mathrm{Q})) \equiv \mathrm{Q} \rightarrow(\mathrm{P} \wedge \mathrm{Q})$

$\equiv(\sim Q) \vee(P \wedge Q) \equiv(P \vee(\sim Q))$

$(\mathrm{P} \wedge(\mathrm{P} \rightarrow \mathrm{Q})) \rightarrow \mathrm{Q}$

$\equiv(\mathrm{P} \wedge(\sim \mathrm{P} \vee \mathrm{Q})) \rightarrow \mathrm{Q} \equiv(\mathrm{P} \wedge \mathrm{Q}) \rightarrow \mathrm{Q}$

$\equiv((\sim P) \vee(\sim Q)) \vee Q \equiv(\sim P) \vee t \equiv t$

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