...
Digital-Logic-Design
October 14, 2023
NVS PGT CS 2017 Part-B
October 14, 2023
Digital-Logic-Design
October 14, 2023
NVS PGT CS 2017 Part-B
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 :

A
True and False
B
False and True
C
False and False
D
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.
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.
0 0 votes
Article Rating
Subscribe
Notify of
0 Comments
Inline Feedbacks
View all comments
0
Would love your thoughts, please comment.x
()
x
error: Alert: Content selection is disabled!!