4
$\begingroup$

I know that

$$\vdash_{\mathcal G}\exists x(\exists yP(x,y) → \forall z \exists wP(z,w))$$

(I have read and done a syntactic proof of this.)
And therefore also

$$\models \exists x(\exists yP(x,y) → \forall z \exists wP(z,w))$$

By soundness of FOL.

But I can't make sense of how this formula can be valid. It looks wrong to me. The way I read it is: if there exists some $x$ and $y$ such that $P(x,y)$, then forall $z$ and some $w$, $P(z,w)$. But how can that be valid?

The order of quantifiers is significant in FOL, so it might be simpler to consider a logically equivalent formula on a simpler form. This formula can be rewritten to a formula in Prenex Normal Form, for example like this:

$$\exists x \forall y \forall z \exists w: P(x,y) → P(z,w)$$

But this formula does not look like it is valid, either. Here is an attempt at a counter-example in the form of an interpretation of the formula (a structure for it): Let $P = \{0\} \times \mathbb{N}$. Then $P(0,x)$ for all $x \in \mathbb{N}$, but there is no $w \in \mathbb N$ s.t. $P(z,w)$ for all $z \in \mathbb N$. In fact, there is only one $z \in \mathbb N$ s.t. $P(z,w)$ holds for some $w$, namely $z = 0$.

$\endgroup$
  • $\begingroup$ I believe it should be read as "There exists an $x$ such that if there also exists a $y$ such that $P(x,y)$ is true, then for every $z$ there is $x$ such that $P(z,x)$." $\endgroup$ – Hayden May 30 '14 at 17:26
  • $\begingroup$ It would help if you could express the formula without using $x$ twice as a bound variable. If not please check if the formula is correctly written $\endgroup$ – Rene Schipperus May 30 '14 at 17:41
  • $\begingroup$ @ReneSchipperus good point, I've edited $\endgroup$ – Guildenstern May 30 '14 at 17:44
  • $\begingroup$ Also see Why is this true? (∃x)(P(x)⇒(∀y)P(y)). $\endgroup$ – Joshua Taylor Jun 17 '14 at 19:56
4
$\begingroup$

This is just the Drinker Paradox in disguise.

Note the following:

$$\exists x(\exists yP(x,y) → \forall z \exists xP(z,x))\iff \exists x(\exists yP(x,y) → \forall y \exists xP(y,x)).$$

Denoting the predicate $\exists x_2P(x_1,x_2)$ by $D(x_1)$ turns the subject statement into

$$\exists x(D(x)\to \forall yD(y)).$$

If you allow me to rewrite this as $\exists x(\neg D(x)\lor \forall yD(y))$, then I think it is intuitive to think of this is as saying "either someone doesn't drink, or everyone does".

Note that this isn't what this formula conveys immediately. The statement "either someone doesn't drink, or everyone does" should be formalized as $\exists x\neg D(x)\lor \forall yD(y)$, but they are equivalent and since this is a matter of intuition, I think this close enough.

If it is any help, see my comment here.

$\endgroup$
3
$\begingroup$

Assuming the universe is non-empty, consider two cases:

  1. $\forall x\ \exists y\ P(x,y)$.

    Here your formula is true, because the conclusion of implication is trivially true.

  2. $\exists x\ \forall y\ \neg P(x,y)$.

    Here your formula is true, because for that $x$ we have $\neg P(x,x)$ and so the premise of implication is false.

In fact your formula is very similar to reordering quantifiers, namely

$$\exists x\ \forall y\ P(x,y)\ \to\ \forall y\ \exists x\ P(x,y).$$

I hope this helps $\ddot\smile$

$\endgroup$
  • $\begingroup$ The universe is always non empty. This is one of the assumptions of mathematical logic. $\endgroup$ – Rene Schipperus May 31 '14 at 13:08
  • 2
    $\begingroup$ @ReneSchipperus No, this is only a convention. Some consider only non-empty models, others consider the possibility of empty model (e.g. see free logic). Besides, if you already have this assumption, asserting it a second time won't make the statement invalid. $\endgroup$ – dtldarek May 31 '14 at 13:16
  • $\begingroup$ It's a convention of classical first-order logic that should be assumed unless otherwise stated. The interested case where it's relaxed, as mentioned, is free logic. There's some discussion of this in the comments to the answer to Why is this true? (∃x)(P(x)⇒(∀y)P(y)). $\endgroup$ – Joshua Taylor Jun 17 '14 at 19:58
3
$\begingroup$

But the reading :

if there exists some $x$ and $y$ such that $P(x,y)$, then for all $z$ and some $w$, $P(z,w)$

is not correct; the first $\exists$ is "outside" the parentheses.

Try with an "informal" argument, starting rewriting the formula as :

$∃x(\lnot ∃yP(x,y) \lor ∀z∃wP(z,w))$

you immediately see that you cannot "move inside" the first $\exists$ without changing the meaning of the formula.

If we remember that we can "distribute" $\exists$ over $\lor$, because $\exists$ "is like" an "inifinite" disjunction", we have that our formula is "equivalent" to :

$∃x(\lnot ∃yP(x,y)) \lor ∃x∀z∃wP(z,w)$

which is :

$\lnot ∀x∃yP(x,y) \lor ∃x∀z∃wP(z,w)$.

In the second disjunct, the quantifier $\exists x$ does not do any job, because $x$ in not free in it; thus, forget it.

Finally, we have :

$\lnot ∀x∃yP(x,y) \lor ∀z∃wP(z,w)$

which is "clearly" valid.

$\endgroup$
1
$\begingroup$

The intuition is clear, take $x$ to be an element with no corresponding $y$ such that $P(x,y)$, if that is possible. Then the hypothesis will be false, and the statement true. On the other have if for every $x$ there is a $y$, $P(x,y)$ then the conclusion is correct and you can take any element for $x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.