Consider the two statements : ( S _ 1 ): ( p → q ) ∨ ( ~ q → p ) is a tautology. ( S _ 2 ): ( p ∧ ~ q ) ∧ ( ~ p ∨ q ) is a fallacy. Then :

JEE Main 202126 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 26 August
Shift: Evening Shift
Subject: Mathematics
Chapter: Mathematical Reasoning

JEE Main 2021 · 26 August · Evening Shift · Mathematics · Mathematical Reasoning

Consider the two statements : ( S _ 1 ): ( p → q ) ∨ ( ~ q → p ) is...

Consider the two statements :

\left({S}_{1}\right): ( p \to q ) ∨ ( ~ q \to p )

is a tautology.
\left({S}_{2}\right): ( p ∧ ~ q ) ∧ ( ~ p ∨ q ) is a fallacy.
Then :

Options

Aonly $( {S}_{1} )$ is true.
Bboth $( {S}_{1} )$ and $( {S}_{2} )$ are false.
Conly $( {S}_{2} )$ is true.
Dboth $( {S}_{1} )$ and $( {S}_{2} )$ are true.

Correct answer

D. both $( {S}_{1} )$ and $( {S}_{2} )$ are true.

Step-by-step solution

$\begin{cases}p & q & ~ p & ~ q & p \to q & ~ q \to p & ( p \to q ) ∨ ( ~ q \to p ) & p ∧ ~ q & ~ p ∨ q & ( p ∧ ~ q ) ∧ ( ~ p ∨ q ) \\ T & T & F & F & T & T & T & F & T & F \\ T & F & F & T & F & T & T & T & F & F \\ F & T & T & F & T & T & T & F & T & F \\ F & F & T & T & T & F & T & F & T & F\end{cases}$

From table
$( p \to q ) ∨ ( ~ q \to p )$ is tautology and $( p ∧ ~ q ) ∧ ( ~ p ∨ q )$ is fallacy.

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