Question

give an example to show it is false or argue why it is true.

∃!xP(x)⇒∃xP(x)

∃xP(x)⇒∃!xP(x)

∃!x¬P(x)⇒¬∀xP(x)

Answer 1

1. ∃!xP(x)⇒∃xP(x)

This is true. Since the premise says that there exists a unique x such that P(x) is true, so this means existence of x is guaranteed for P(x) to be true. So the right side is true. Hence the imlication is true.

2. ∃xP(x)⇒∃!xP(x)

This is false. Since the premise says that there exists an x in the universe such that P(x) is true. But it does not gurantee that there is only one such x for which P(x) is true. This means there is possibility of several such x that P(x) is true.

Let us take an example to establish that it is false.

Let the domain set be irrational numbers, and range be positive is integers. Let for

Then this is true that there exists irrational numbers like

Thus ∃xP(x) is true. However, ∃!xP(x) is false, since we have

So uniqueness is not true.

3. ∃!x¬P(x)⇒¬∀xP(x)

This is true. Since the premise says that there exists a unique x such that P(x) is not true, So this means existence of one such x is guaranteed for P(x) not to be true. So this means for all x P(x) cannot be true. There may be some x for which P(x) may be true, but it is certainly not true for all x. Hence the imlication is true.

For example, suppose P(x) stands for the result of division of 1 by the number x.

Consider x=0, then division of 1 by the number 0 is not possible. That is ∃!x¬P(x).

Because if and

So we cannot say that

That is, ¬∀xP(x) is true.

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