Which of the following is a tautology?

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Question

Option (1) is

$\sim \mathrm{p} \wedge(\mathrm{p} \vee \mathrm{q}) \rightarrow \mathrm{q}$

$\equiv(\sim \mathrm{p} \wedge \mathrm{p}) \vee(\sim \

Vedclass · Accepted Answer

$(\sim \mathrm{p}) \wedge(\mathrm{p} \vee \mathrm{q}) \rightarrow \mathrm{q}$

Vedclass · Answer

Option (1) is

$\sim \mathrm{p} \wedge(\mathrm{p} \vee \mathrm{q}) \rightarrow \mathrm{q}$

$\equiv(\sim \mathrm{p} \wedge \mathrm{p}) \vee(\sim \mathrm{p} \wedge \mathrm{q}) \rightarrow \mathrm{q}$

$\equiv \mathrm{C} \vee(\sim \mathrm{p} \wedge \mathrm{q}) \rightarrow \mathrm{q}$

$\equiv(\sim p \wedge q) \rightarrow q$

$\equiv \sim(\sim \mathrm{p} \wedge \mathrm{q}) \vee \mathrm{q}$

$\equiv(\mathrm{p} \vee \sim \mathrm{q}) \vee \mathrm{q}$

$\equiv(\mathrm{p} \vee \mathrm{q}) \vee(\sim \mathrm{q} \vee \mathrm{q})$

$\equiv(\mathrm{p} \vee \mathrm{q}) \vee \mathrm{t}$

so $\sim \mathrm{p} \wedge(\mathrm{p} \vee \mathrm{q}) \rightarrow \mathrm{q}$ is a tautology

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