हिंदी

Using the rules in logic, write the negation of the following: (p ∨ q) ∧ (q ∨ ∼r) - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

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

(p ∨ q) ∧ (q ∨ ∼r)

योग
Advertisements

उत्तर

The negation of (p ∨ q) ∧ (q ∨ ∼r) is
∼ [(p ∨ q) ∧ (q ∨ ∼r)]

≡ ∼(p ∨ q) ∨ ∼(q ∨ ∼r) .....(Negation of conjunction)

≡ (∼p ∧ ∼q) ∨ [∼q ∧ ∼(∼r)] ...............(Negation of disjunction

≡ (∼p ∧ ∼q) ∨ (∼q ∧ r) ...........(Negation of negation)

≡ (∼q ∧ ∼p) ∨ (∼q ∧ r) ..........(Commutative law)

≡ (∼q) ∧ (∼p ∨ r) ..........(Distributive Law)

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Mathematical Logic - Miscellaneous Exercise 1 [पृष्ठ ३४]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Mathematical Logic
Miscellaneous Exercise 1 | Q 11.1 | पृष्ठ ३४

संबंधित प्रश्न

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


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


Without using the truth table show that P ↔ q ≡ (p ∧ q) ∨ (~ p ∧ ~ q)


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.

If a man is a judge then he is honest.


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) ∧ r


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

(∼p ∧ q) ∨ (p ∧ ∼q)


Without using truth table, show that

p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)


Without using truth table, show that

p ∧ [(~ p ∨ q) ∨ ~ q] ≡ p


Without using truth table, show that

~ [(p ∧ q) → ~ q] ≡ p ∧ q


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 [∼r ∧ (p ∨ q) ∧ (p ∨ q) ∧ (∼p ∧ q)] is equivalent to ______ 


(p ∧ ∼q) ∧ (∼p ∧ q) is a ______.


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 ∨ r)] ∨ [∼r ∧ ∼q ∧ p] ≡ p


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


The statement p → (q → p) is equivalent to ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×