Consider: S 1 : p ⇒ q ∨ p ∧ ~ q is a tautology. S 2 : ~ p ⇒ ~ q ∧ ~ p ∨ q is a contradiction. Then

Question

Consider: S 1 : p ⇒ q ∨ p ∧ ~ q is a tautology. S 2 : ~ p ⇒ ~ q ∧ ~ p ∨ q is a contradiction. Then, only S 2 is correct, both S 1 and S 2 are correct, both S 1 and S 2 are wrong, only S 1 is correct

MARKS App · Accepted Answer

p q p ⇒ q ~ q p ∧ ~ q p ⇒ q ∨ p ∧ ~ q T T T F F T T F F T T T F T T F F T F F T T F T So, p ⇒ q ∨ p ∧ ~ q is a tautology. ~ p ~ q ~ p ⇒ ~ q ~ p ∨ q ~ p ⇒ ~ q ∧ ~ p ∨ q F F T T T F T T F F T F F T F T T T T T

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

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

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