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

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प्रश्न

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

p ∧ (q ∨ r)

बेरीज
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उत्तर

The negation of p ∧ (q ∨ r) is
∼ [p ∧ (q ∨ r)]
≡ ∼p ∨ ∼(q ∨ r) ..........(Negation of conjunction)
≡ ∼p ∨ (∼q ∧ ∼r) ..........(Negation of disjunction)

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पाठ 1: Mathematical Logic - Miscellaneous Exercise 1 [पृष्ठ ३४]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 1 Mathematical Logic
Miscellaneous Exercise 1 | Q 11.2 | पृष्ठ ३४

संबंधित प्रश्‍न

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


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


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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


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

(∼p ∧ q) ∨ (p ∧ ∼q)


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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)


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 ______.


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Negation of the Boolean expression `p Leftrightarrow (q \implies p)` is ______. 


∼ ((∼ p) ∧ q) is equal to ______.


Without using truth table, prove that:

[p ∧ (q ∨ r)] ∨ [∼r ∧ ∼q ∧ p] ≡ p


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Without using truth table prove that

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