###### Digital-Logic-Design

October 14, 2023###### एनवीएस पीजीटी सीएस 2017 पार्ट-बी

October 14, 2023# Propositional-Logic

Question 153 |

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 :

True and False | |

False and True | |

False and False | |

True and True |

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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