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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 $p \vee q$

  4. a contradiction

  5. none of the above

in Mathematical Logic by Veteran (52.2k points)
edited by | 1.1k views
0
why not option a? it is modus ponens clearly and hence it is a tautology... pls pls clear my doubt...
0
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.

4 Answers

+15 votes
Best answer

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

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

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

$\equiv (p \wedge q)$


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

by Boss (41k points)
edited by
0
why not option a? it is modus ponens clearly and hence it is a tautology... pls pls clear my doubt...
+13 votes
OPTION (B)
by Active (2.6k points)
+11 votes
p ^ (~p v q)

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

= False V (p^q)

= (p^q)
by (297 points)
+1 vote
$P \wedge (\neg P \vee Q ) \equiv  ( P \wedge \neg P ) \vee ( P \wedge Q )$

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

                        $\equiv ( P \wedge Q )$
by Veteran (54.9k points)
edited ago by
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