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

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प्रश्न

Using the algebra of statement, prove that

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

योग
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उत्तर

L.H.S.

= (p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q)

≡ (p ∧ q) ∨ [(p ∧ ~ q) ∨ (~ p ∧ ~ q)]    ....[Associative Law]

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

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

≡ (p ∧ q) ∨ (~q ∧ t)      .....[Complement Law]

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

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

≡ (p ∨ ~ q) ∧ t              ....[Complement Law]

≡ p ∨ ~ q                   .....[Identity Law]

= R.H.S.

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अध्याय 1: Mathematical Logic - Exercise 1.9 [पृष्ठ २२]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 1 Mathematical Logic
Exercise 1.9 | Q 2.2 | पृष्ठ २२

संबंधित प्रश्न

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


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


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(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


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


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p ↔ q ≡ ∼(p ∧ ∼q) ∧ ∼(q ∧ ∼p)


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∼ (p ∨ q) ∨ (∼p ∧ q) ≡ ∼p


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(p ∨ q) ∧ (q ∨ ∼r)


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p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)


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


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