English

The negation of p ∧ (q → r) is - Mathematics and Statistics

Advertisements
Advertisements

Question

The negation of p ∧ (q → r) is ______________.

Options

  • p ∨ ( ~q ∨ r )

  • ~p ∧ ( q → r )

  • ~p ∧ ( ~q → ~r )

  • ~p ∨ ( q ∧ ~r )

MCQ
Advertisements

Solution

~ [P ∧ (q → r)

=~[( P)] ∨ [~ (q → r)]                             ...(By De Morgan's law)
=~[( P)] ∨ [~ (~q ∨ r )]                           ...(By Conditional Law)
=~[( P)] ∨ [( q ∧ ~r )]                             ...(By De Morgan's law)

~ [ P ∧ ( q → r )] = ~ P ∨ ( q ∧ ~r ) 

shaalaa.com
  Is there an error in this question or solution?
2015-2016 (March)

APPEARS IN

RELATED QUESTIONS

Without using truth tabic show that ~(p v q)v(~p ∧ q) = ~p


Using the rules of negation, write the negatlon of the following: 

(a) p ∧ (q → r)

(b)  ~P ∨ ~q


Write the Truth Value of the Negation of the Following Statement :

The Sun sets in the East. 


Write the truth value of the negation of the following statement : 

cos2 θ + sin2 θ = 1, for all θ ∈ R 


Rewrite the following statement without using if ...... then.

It 2 is a rational number then `sqrt2` is irrational number.


Rewrite the following statement without using if ...... then.

It f(2) = 0 then f(x) is divisible by (x – 2).


Without using truth table prove that:

(p ∨ q) ∧ (p ∨ ∼ q) ≡ p


Without using truth table prove that:

(p ∧ q) ∨ (∼ p ∧ q) ∨ (p ∧ ∼ q) ≡ p ∨ q


Using rules in logic, prove the following:

p ↔ q ≡ ∼(p ∧ ∼q) ∧ ∼(q ∧ ∼p)


Using the rules in logic, write the negation of the following:

(p ∨ q) ∧ (q ∨ ∼r)


Using the rules in logic, write the negation of the following:

p ∧ (q ∨ r)


Without using truth table, show that

p ∧ [(~ p ∨ q) ∨ ~ q] ≡ p


Without using truth table, show that

~ [(p ∧ q) → ~ q] ≡ p ∧ q


Without using truth table, show that

~r → ~ (p ∧ q) ≡ [~ (q → r)] → ~ p


Without using truth table, show that

(p ∨ q) → r ≡ (p → r) ∧ (q → r)


Using the algebra of statement, prove that

[p ∧ (q ∨ r)] ∨ [~ r ∧ ~ q ∧ p] ≡ p


Using the algebra of statement, prove that (p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∧ ~ q) ∨ (~ p ∧ q).


The statement pattern p ∧ ( q v ~ p) is equivalent to ______.


For any two statements p and q, the negation of the expression (p ∧ ∼q) ∧ ∼p is ______ 


(p → q) ∨ p is logically equivalent to ______ 


The logically equivalent statement of (p ∨ q) ∧ (p ∨ r) is ______ 


The negation of p → (~p ∨ q) is ______ 


The statement pattern p ∧ (∼p ∧ q) is ______.


(p ∧ ∼q) ∧ (∼p ∧ q) is a ______.


The logical statement [∼(q ∨ ∼r) ∨ (p ∧ r)] ∧ (q ∨ p) is equivalent to: ______ 


Without using truth table prove that (p ∧ q) ∨ (∼ p ∧ q) v (p∧ ∼ q) ≡ p ∨ q


If p ∨ q is true, then the truth value of ∼ p ∧ ∼ q is ______.


Which of the following is not a statement?


∼ ((∼ p) ∧ q) is equal to ______.


Without using truth table, prove that : [(p ∨ q) ∧ ∼p] →q is a tautology.


The simplified form of [(~ p v q) ∧ r] v [(p ∧ ~ q) ∧ r] is ______.


Without using truth table prove that

[(p ∧ q ∧ ∼ p) ∨ (∼ p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r) ≡ (p ∨ q) ∧ r


Show that the simplified form of (p ∧ q ∧ ∼ r) ∨ (r ∧ p ∧ q) ∨ (∼ p ∨ q) is q ∨ ∼ p.


The logically equivalent statement of \[\left(\sim p\wedge q\right)\vee\left(\sim p\wedge\sim q\right)\] \[\vee\left(\ p\wedge\sim q\right)\] is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×