Let r∈P,q,~p,~q be such that the logical statement r∨~p⇒p∧q∨r is a tautology. Then r is equal to

Question

MARKS Content Team · Accepted Answer

Correct answer: (C) ~p. Solving all option one by one we get following truth table, (1) p=r q ~p r∨~p p∧q p∧q∨r r∨~p⇒p∧q∨r T F F T F T T F T T T F F F T T F T T T T F F T T F F T (2) p ~p r∨~p q=r p∧q p∧q∨r r∨~p⇒p∧q∨r T F T T T T T F T T T F T T T F F F F F T F T T F F F F (3) p q r=~p r∨~p p∧q p∧q∨r r∨~p⇒p∧q∨r T T F F T T T F T T T F T T T F F F F F T F F T T F T T (4) ~p p q r∨~p r=~q p∧q p∧q∨r r∨~p⇒p∧q∨r F T T F F T T T F T F T T F T T T F T T F F F F T F F T T F T T Now final answer is option C.

$p = r$	$q$	$~ p$	$r \lor (~ p)$	$(p \land q)$	$(p \land q) \lor r$	$r \lor (~ p) \Rightarrow (p \land q) \lor r$
$T$	$F$	$F$	$T$	$F$	$T$	$T$
$F$	$T$	$T$	$T$	$F$	$F$	$F$
$T$	$T$	$F$	$T$	$T$	$T$	$T$
$F$	$F$	$T$	$T$	$F$	$F$	$T$

$p$	$~ p$	$r \lor (~ p)$	$q = r$	$(p \land q)$	$(p \land q) \lor r$	$r \lor (~ p) \Rightarrow (p \land q) \lor r$
$T$	$F$	$T$	$T$	$T$	$T$	$T$
$F$	$T$	$T$	$T$	$F$	$T$	$T$
$T$	$F$	$F$	$F$	$F$	$F$	$T$
$F$	$T$	$T$	$F$	$F$	$F$	$F$

$p$	$q$	$r = ~ p$	$r \lor (~ p)$	$(p \land q)$	$(p \land q) \lor r$	$r \lor (~ p) \Rightarrow (p \land q) \lor r$
$T$	$T$	$F$	$F$	$T$	$T$	$T$
$F$	$T$	$T$	$T$	$F$	$T$	$T$
$T$	$F$	$F$	$F$	$F$	$F$	$T$
$F$	$F$	$T$	$T$	$F$	$T$	$T$

$~ p$	$p$	$q$	$r \lor (~ p)$	$r = ~ q$	$(p \land q)$	$(p \land q) \lor r$	$r \lor (~ p) \Rightarrow (p \land q) \lor r$
$F$	$T$	$T$	$F$	$F$	$T$	$T$	$T$
$F$	$T$	$F$	$T$	$T$	$F$	$T$	$T$
$T$	$F$	$T$	$T$	$F$	$F$	$F$	$F$
$T$	$F$	$F$	$T$	$T$	$F$	$T$	$T$

Let $r \in (P, q, ~ p, ~ q)$ be such that the logical statement $r \lor (~ p) \Rightarrow (p \land q) \lor r$ is a tautology. Then $r$ is equal to

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