JEE Main202330 Jan 2023Morning ShiftMathematicsMathematical ReasoningActual
Among the statements: S 1 : p ∨ q ⇒ r ⇔ p ⇒ r S 2 : p ∨ q ⇒ r ⇔ p ⇒ r ∨ q ⇒ r
Options
- AOnly ( S 1 ) is a tautology
- BNeither ( S 1 ) nor ( S 2 ) is a tautology
- COnly ( S 2 ) is a tautology
- DBoth ( S 1 ) and ( S 2 ) are tautologies
Correct answer
B. Neither ( S 1 ) nor ( S 2 ) is a tautology
Step-by-step solution
We have, S 1 :   p ∨ q ⇒ r ⇔ p ⇒ r For S 1 :     If   p   is   F ,   q   is   T ,   r   is   F   , then p ∨ q ⇒ r ≡ F p ⇒ r ≡ T So, p ∨ q ⇒ r ⇔ p ⇒ r ≡ F then F ⇒ F ≡ F i.e. S 1 is not a tautology S 2 :   p ∨ q ⇒ r ⇔ p ⇒ r ∨ q ⇒ r S 2 :   If   p   is   T ,  q   is   F ,  r