Digital-Logic-Design
October 14, 2023एनवीएस पीजीटी सीएस 2017 पार्ट-बी
October 14, 2023Propositional-Logic
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Question 153
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The truth value of the statements :
∃!xP(x) → ∃xP(x) and ∃!x ⌐P(x) →⌐∀xP(x), (where the notation ∃!xP(x)
denotes the proposition “There exists a unique x such that P(x) is true”) are :
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True and False
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False and True
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False and False
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True and True
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Question 153 Explanation:
From the given question,
→ The symbol ∃ is call the existential quantifier and represents the phrase “there exists” or “for some”. The existential quantification of P(x) is the statement “P(x) for some values x in the universe”, or equivalently, “There exists a value for x such that P(x) is true”, which is written ∃xP(x). So the statement ∃!xP(x) → ∃xP(x) is true.
→ If P(x) is true for at least one element in the domain, then ∃xP(x) is true. Otherwise it is false.
→ Accordingly DeMorgan’s laws for quantifiers:the following statements are true.
¬∀xP(x) ≡ ∃x¬P(x)
¬∃xP(x) ≡ ∀x¬P(x) then the statement is ∃!x ⌐P(x) →⌐∀xP(x) is true.
→ The symbol ∃ is call the existential quantifier and represents the phrase “there exists” or “for some”. The existential quantification of P(x) is the statement “P(x) for some values x in the universe”, or equivalently, “There exists a value for x such that P(x) is true”, which is written ∃xP(x). So the statement ∃!xP(x) → ∃xP(x) is true.
→ If P(x) is true for at least one element in the domain, then ∃xP(x) is true. Otherwise it is false.
→ Accordingly DeMorgan’s laws for quantifiers:the following statements are true.
¬∀xP(x) ≡ ∃x¬P(x)
¬∃xP(x) ≡ ∀x¬P(x) then the statement is ∃!x ⌐P(x) →⌐∀xP(x) is true.
Correct Answer: D
Question 153 Explanation:
From the given question,
→ The symbol ∃ is call the existential quantifier and represents the phrase “there exists” or “for some”. The existential quantification of P(x) is the statement “P(x) for some values x in the universe”, or equivalently, “There exists a value for x such that P(x) is true”, which is written ∃xP(x). So the statement ∃!xP(x) → ∃xP(x) is true.
→ If P(x) is true for at least one element in the domain, then ∃xP(x) is true. Otherwise it is false.
→ Accordingly DeMorgan’s laws for quantifiers:the following statements are true.
¬∀xP(x) ≡ ∃x¬P(x)
¬∃xP(x) ≡ ∀x¬P(x) then the statement is ∃!x ⌐P(x) →⌐∀xP(x) is true.
→ The symbol ∃ is call the existential quantifier and represents the phrase “there exists” or “for some”. The existential quantification of P(x) is the statement “P(x) for some values x in the universe”, or equivalently, “There exists a value for x such that P(x) is true”, which is written ∃xP(x). So the statement ∃!xP(x) → ∃xP(x) is true.
→ If P(x) is true for at least one element in the domain, then ∃xP(x) is true. Otherwise it is false.
→ Accordingly DeMorgan’s laws for quantifiers:the following statements are true.
¬∀xP(x) ≡ ∃x¬P(x)
¬∃xP(x) ≡ ∀x¬P(x) then the statement is ∃!x ⌐P(x) →⌐∀xP(x) is true.
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