मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

Let p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r). Then, this law is known as ______. - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Let p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r). Then, this law is known as ______.

पर्याय

  • Commutative law

  • Associative law

  • De-Morgan's law

  • Distributive law

MCQ
रिकाम्या जागा भरा

उत्तर

Let p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r). Then, this law is known as distributive law.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1: Mathematical Logic - Miscellaneous Exercise 1 [पृष्ठ २९]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] 12 Standard HSC Maharashtra State Board
पाठ 1 Mathematical Logic
Miscellaneous Exercise 1 | Q 1.03 | पृष्ठ २९

संबंधित प्रश्‍न

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


If A = {2, 3, 4, 5, 6}, then which of the following is not true?

(A) ∃ x ∈ A such that x + 3 = 8

(B) ∃ x ∈ A such that x + 2 < 5

(C) ∃ x ∈ A such that x + 2 < 9

(D) ∀ x ∈ A such that x + 6 ≥ 9


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

(a) p ∧ (q → r)

(b)  ~P ∨ ~q


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) ∧ 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


Without using truth table, show that

~r → ~ (p ∧ q) ≡ [~ (q → r)] → ~ p


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 negation of p → (~p ∨ q) is ______ 


(p ∧ ∼q) ∧ (∼p ∧ q) is a ______.


The negation of the Boolean expression (r ∧ ∼s) ∨ s is equivalent to: ______ 


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×