-1

I have a question about counter examples I want to show for example

$\exists x ( Px \rightarrow Qx) $

is not logically equivalent to

$\exists x Px \rightarrow \exists x Qx $

I have the model

Domain=[1,2]

$P=[{1}]$

$Q=[]$

so I choose $x=2$

then

$\exists x Px \rightarrow Qx $

I got $f \rightarrow f $

so this is true

But if chose $x=1$ then $\exists x (Px \rightarrow Qx )$ is false

If $\exists x Px \rightarrow \exists Qx $

if I have x=1, x=2 then it is false

My question is for counter example do $x$ value does the have to be the same because if $x=2$

then both things are true, but if $x=1$ both statement are false.

Or it does not matter what $x$ is as long you can make one true and one false in the model.

Fernando Martinez
  • 6,826
  • 22
  • 76
  • 114
  • It's not clear to me what you ask. Do you want to show a (counter)example to the given statrement? Or you want to learn a way of finding (counter)examples for statements of some kind, using the given one as an example problem? – CiaPan Feb 10 '25 at 19:33
  • I am asking like like for the counter example do you have to use the same domain value like x=1 and the same one has to be true for one statement and false for another statement – Fernando Martinez Feb 10 '25 at 19:36
  • Cause I know they are not logically equivalent but if I chose x=1 I get both are false but x=2 both are true – Fernando Martinez Feb 10 '25 at 19:39
  • But in counter example I know you got to show one is false and one is true. – Fernando Martinez Feb 10 '25 at 19:40
  • 3
    "If I choose $x=1$ then $\exists x (Px \rightarrow Qx )$ is false" doesn't make sense. $\exists x H$ is true as long as there exists a value of $x$ making $H$ true. So you don't "choose" a particular value of $x$ when evaluating it. – Karl Feb 10 '25 at 19:52
  • ∃x(Px→Qx) ≡ (∀xPx)→∃xQx – ryang Feb 11 '25 at 04:27

2 Answers2

2

I think you're confused when you write "if I choose $x=1$ then $\exists x (Px \rightarrow Qx )$ is false." The sentence $\exists x H$ is true as long as there exists a value of $x$ making $H$ true - you don't "choose" a particular value of $x$ when evaluating it.

So $\exists x (Px \rightarrow Qx )$ is true in your model because $2$ exists. $\exists x Px$ is true because $1$ exists. And $\exists x Qx$ is false because there does not exist a value of $x$ making $Qx$ true.

Karl
  • 12,752
1

The trick is to notice the second formula contains two quantifiers: in $$\exists x\, ( Px \rightarrow Qx) $$ both sides of the implication are evaluated on the same element $x$, BUT in $$(\exists x\, Px) \rightarrow (\exists x\, Qx) $$ the $x$ for which $P$ is evaluated and $x$ for which $Q$ is evaluated may be different elements of the domain.

Let $x$ be 'some person', $P$ be 'killed somebody', and $Q$ be 'has been killed'.
Then sentence 1. becomes 'if there is a person who killed somebody, then that person has been killed' (which IRL most probably resolves to 'we have a suicide'),
while sentence 2. resolves to 'if someone killed somebody, then someone has been killed' (most probably IRL: some other one).

;)

The simplest counter-example would be an empty domain. An expression with the existential quantifier is false on such domain: its NOT true, there exists such $x$ that ...anything.
So, the first expression $\exists x\, (\ldots)$ is false.

OTOH, both $\exists x\,Px$ and $\exists x\,Qx$ are false, so the second expression becomes $\text {false} \rightarrow \text {false}$ which is
...$\text {true}$.

CiaPan
  • 13,884