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The statement ( ∼ ( p ⇔ ∼ q ) ) ∧ q     i s :  

Option 1 -

a tautology

Option 2 -

a contradiction

Option 3 -

equivalent to ( p ⇒ q ) ∧ q            

Option 4 -

equivalent to ( p ⇒ q ) ∧ p

0 2 Views | Posted 4 weeks ago
Asked by Shiksha User

  • 1 Answer

  • R

    Answered by

    Raj Pandey | Contributor-Level 9

    4 weeks ago
    Correct Option - 4


    Detailed Solution:

    ∼ ( p ⇔ ∼ q ) ∧ q = ( p ⇔ ∼ q ) ∧ q     i s :

    ∴ ( ∼ ( P ⇔ ∼ Q ) ) ∧ q is equivalent to ( p ⇒ q ) ∧

Similar Questions for you

Let Δ , ∇ ∈ { ∧ , ∨ } be such that p ∇ q ⇒ ( ( p Δ q ) ∇ r ) is a tautology. Then ( p ∇ q ) Δ   r  is logically equivalent to:
R
Raj Pandey

Case – I Δ ≡ ∇ ≡ ∧  

              ( p ∧ q ) → ( ( p ∧ q ) ∧ r )  

              it can be false if r is false,

              so not a tautology

              Case – II If Δ ≡ ∇ ≡ ∨  

              ( p ∨ q ) ?     → ( ( p ∨ q ) ∨ r ) ≡ tautology

              then ( p ∨ q ) ∨ r ≡ ( p Δ r ) ∨ q   

              Case – I

...more

Negation of the Boolean statement ( p ∨ q ) ⇒ ( ( ∼ r ) ∨ p )  is equivalent to
V
Vishal Baghel

pva ( ∼ r   v   p )

∼ ( p ∨ q ) ∨ ( ∼ r v p )

its negation as asked in question

( p ∨ q ) ∧ ( ∼ p ∧ r )

= ( p ∧ ∼ p ∧ r ) ∨ ( q ∧ r ∧ ∼ p )

= ( p ∧ r ∧ ∼ p )                   [ a s     p ∧ ∼ p     i s     f a l s e ]

The remainder when (2021)2023 is divided by 7 is:
R
Raj Pandey

2 0 2 1 ≡ − 2 mod (7)

⇒ ( 2 0 2 1 ) 2 0 2 3 ≡ ( − 2 ) 2 0 2 3 m o d ( 7 )   …… (i)

Now,   ( − 2 ) 3 ≡ − 1 m o d ( 7 )  

⇒ ( − 2 ) 2 0 2 3 ≡ ( − 2 ) m o d ( 7 ) ≡ 5 m o d ( 7 )       ……. (ii)

(i) & (ii)

⇒ ( 2 0 2 1 ) 2 0 2 3 ≡ 5 m o d ( 7 )  

∴  Remainder = 5

Let p, q, r be three logical statements. Consider the compound statements

  S 1 : ( ( ∼ p ) ∨ q ) ∨ ( ( ∼ p ) ∨ r )     a n d  

  S 2 : p → ( q ∨ r )  

 Then, which of the following is NOT true?
R
Raj Pandey

kindly consider the following Image 

 

The negation of the statement “If a quadrilateral is a square then it is a rhombus”.
V
Vishal Baghel

  ∼ ( p → q ) = p ∧ ∼ q

A quadrilateral is a square and it is not a rhombus

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