हिंदी

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

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

L.H.S. = (p ∨ q) ∧ (~ p ∨ ~ q)

≡ [(p ∨ q)] ∧ [(p ∨ q) ∧ ~ q]      ...[Distributive law]

≡ [(p ∧ ~ p) ∨ (q ∧ ~ p)] ∨ [(p ∧ ~ q) ∨ (q ∧ ~ q)]        ...[Distributive law]

≡ [F ∨ (q ∧ ~ p)] ∨ [(p ∧ ~ q) ∨ F]    ...[Complement law]

≡ (q ∧ ~ p) ∨ (p ∧ ~ q)   ...[Identity law]

≡ (p ∧ ~ q) ∨ (~ p ∧ q)    ...[Commutative law]

≡ R.H.S.

shaalaa.com

Notes

The question is modified.

  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Mathematical Logic - Exercise 1.9 [पृष्ठ २२]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 1 Mathematical Logic
Exercise 1.9 | Q 2.3 | पृष्ठ २२

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


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

The Sun sets in the East. 


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

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


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 rules in logic, prove the following:

∼p ∧ q ≡ (p ∨ q) ∧ ∼p


Using rules in logic, prove the following:

∼ (p ∨ q) ∨ (∼p ∧ q) ≡ ∼p


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

(p ∨ q) ∧ (q ∨ ∼r)


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


Without using truth table, show that

p ↔ q ≡ (p ∧ 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) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q) ≡ (p ∨ ~ q)


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


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


(p ∧ ∼q) ∧ (∼p ∧ q) is a ______.


The negation of the Boolean expression (r ∧ ∼s) ∨ s 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 ______.


Without using truth table, prove that:

[p ∧ (q ∨ r)] ∨ [∼r ∧ ∼q ∧ p] ≡ p


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


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×