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The proposition $p \wedge (\sim p \vee q)$ is:

1. a tautology

2. logically equivalent to $p \wedge q$

3. logically equivalent to $p \vee q$

5. none of the above

why not option a? it is modus ponens clearly and hence it is a tautology... pls pls clear my doubt...
modus ponens is an inference rule to conclude something.. It is not used to show tautology. Here, put 'P'  as False. It is not a tautology.

$p \wedge (\sim p \vee q)$

$\equiv (p \wedge \sim p) \vee (p \wedge q)$

$\equiv F \vee (p \wedge q)$

$\equiv (p \wedge q)$

Hence, Option(B) logically equivalent to $(p \wedge q)$.

why not option a? it is modus ponens clearly and hence it is a tautology... pls pls clear my doubt...
Just substitute values debasree , it’s not ALWAYS TRUE, so not tautology.
p ^ (~p v q)

= (p ^ ~p) v (p ^ q)

= False V (p^q)

= (p^q)
$P \wedge (\neg P \vee Q ) \equiv ( P \wedge \neg P ) \vee ( P \wedge Q )$

$\equiv F \vee ( P \wedge Q )\:\:\: \bigg[\because (P \wedge \neg P) \: \text{ is always False and} \: (P \vee \neg P)\:\text{ is always True} \bigg]$

$\equiv ( P \wedge Q )$