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The notation $\exists ! x P(x)$ denotes the proposition “there exists a unique $x$ such that $P(x)$ is true”. Give the truth values of the following statements : 

I.$\exists ! x P(x) \rightarrow \exists x P(x)$

II.$\exists ! x \neg P(x) \rightarrow \neg \forall x P(x)$

  1. Both $I$ & $II$ are true. 
  2. Both $I$ & $II$ are false. 
  3. $I$ - false, $II$ - true 
  4. $I$ - true, $II$ - false
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2 Answers

Best answer
5 votes
5 votes
If there exists a particular x such that p(x) is

true..then obviously there exists some x for which

p (x) is ture . .

Hence 1 is correct

Now ∽∀p(x) means ∃∽p(x)

Therefore 2 is also correct
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8 votes
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I: If there exists a unique x with P(x) true, then there exist an x with P(x) true. This is TRUE as exactly 1 is a subset of at least one.

II: If there exists a unique x with P(x) false, then there does not exist an x with P(x) true. This is FALSE (contradiction) as all except one x can are having P(x) true.
Answer:

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