Exams Joint Entrance Examination Mathematics Mathematical Reasoning Question 88 Question 88 of 103 Medium MCQ
Among the two statements
\[(\mathrm{S} 1):(p \Rightarrow q) \wedge(p \wedge(\sim q))\] is a contradiction and
\[(\mathrm{S} 2):(p \wedge q) \vee((\sim p) \wedge q) \vee(p \wedge(\sim q)) \vee((\sim p) \wedge(\sim q))\] is a tautology
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\[
S 1 : ( p ⇒ q ) ∧ ( p ∧ ∼ q ) ≡ ( ∼ p ∨ q ) ∧ ( p ∧ ∼ q ) ≡ ( ∼ p ∧ p ∧ ∼ q ) ∨ ( q ∧ p ∧ ∼ q ) ≡ ( f ∧ ∼ q ) ∨ ( f ∧ p ) ≡ f ∨ f ≡ f S 2 : ( p ∧ q ) ∨ ( ∼ p ∧ q ) ∨ ( p ∧ ∼ q ) ∨ ( ∼ p ∧ ∼ q ) ≡ ( ( p ∨ ∼ p ) ∧ q ) ∨ ( ( p ∨ ∼ p ) ∧ ∼ q ) ≡ ( t ∧ q ) ∨ ( t ∧ ∼ q ) ≡ q ∨ ∼ q ≡ t \begin{aligned}
S_1: & (p \Rightarrow q) \wedge(p \wedge \sim q) \\\\
& \equiv(\sim p \vee q) \wedge(p \wedge \sim q) \\\\
& \equiv(\sim p \wedge p \wedge \sim q) \vee(q \wedge p \wedge \sim q) \\\\
& \equiv(f \wedge \sim q) \vee(f \wedge p) \\\\
& \equiv f \vee f \equiv f \\\\
S_2: & (p \wedge q) \vee(\sim p \wedge q) \vee(p \wedge \sim q) \vee(\sim p \wedge \sim q) \\\\
& \equiv((p \vee \sim p) \wedge q) \vee((p \vee \sim p) \wedge \sim q) \\\\
& \equiv(t \wedge q) \vee(t \wedge \sim q) \equiv q \vee \sim q \equiv t
\end{aligned} S 1 : S 2 : ( p ⇒ q ) ∧ ( p ∧ ∼ q ) ≡ ( ∼ p ∨ q ) ∧ ( p ∧ ∼ q ) ≡ ( ∼ p ∧ p ∧ ∼ q ) ∨ ( q ∧ p ∧ ∼ q ) ≡ ( f ∧ ∼ q ) ∨ ( f ∧ p ) ≡ f ∨ f ≡ f ( p ∧ q ) ∨ ( ∼ p ∧ q ) ∨ ( p ∧ ∼ q ) ∨ ( ∼ p ∧ ∼ q ) ≡ (( p ∨ ∼ p ) ∧ q ) ∨ (( p ∨ ∼ p ) ∧ ∼ q ) ≡ ( t ∧ q ) ∨ ( t ∧ ∼ q ) ≡ q ∨ ∼ q ≡ t
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