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Is there some understandable rationale why $\forall x\, \exists y\, P(x,y) \not \equiv \exists y\, \forall x\, P(x,y)$?

I'm looking for a sentence I can explain to students, but I am failing every time I try to come up with one.

Example

Let $P(x,y)$ mean that $x$ is greater than $y$.

  • $\forall x\, \exists y\, P(x,y)$ means that for all $x$, there is a number $y$, such that $x$ is greater than $y$.
  • $\exists y\, \forall x\, P(x,y)$ means that there is some $y$, that every number $x$ is greater than.

These don't seem to mean different things to me. Is this perhaps an example where they do mean the same thing or am I just translating to English incorrectly?

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    $\begingroup$ "There is an MSE question that everyone can answer" vs. "For each person, there is at least one MSE question they know the answer to" $\endgroup$
    – Xodarap
    Commented Sep 12, 2013 at 16:36
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    $\begingroup$ Your first statement says that for any choice of $x$, you can find a $y$ such that $y<x$. This is easily seen to be true for any $x$ (choose $y=x-1$, for example). However, the second statement states that for some $y$, then all values of $x$ are greater than $y$. This is clearly false, as choosing $x=y$ shows. $\endgroup$
    – copper.hat
    Commented Sep 12, 2013 at 16:59

7 Answers 7

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Take $P(x, y)$ to mean $y$ is a parent of $x$.

Then $\forall x \exists y P(x, y)$ means everybody has a parent, while $\exists y \forall x P(x, y)$ means there is someone who is the parent of every son and daughter.

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    $\begingroup$ Nice. So $\forall x \exists y$ reads "everyone has a [whatever]," whereas $\exists x \forall y$ reads "some person has everyone for a [whatever]." $\endgroup$ Commented Sep 12, 2013 at 16:38
  • $\begingroup$ Very nice. Could you give me any insight into why the example I came up with doesn't work? $\endgroup$
    – danmcardle
    Commented Sep 12, 2013 at 16:46
  • $\begingroup$ Note that the $\exists\forall$ statement in the question was $\exists y\,\forall x\,P(x,y)$, not $\exists x\,\forall y\,P(x,y)$ as in this answer. (The parent-child relation still give a good example, though, even for the originally intended sentence.) $\endgroup$ Commented Sep 12, 2013 at 18:26
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    $\begingroup$ Why nobody notice that question asks about ∀x∃y ≢ ∃y∀x whereas you answer about ∀x∃y ≢ ∃x∀y. How do you all synchronize to answer, accept and upvote the answer to a different question? It is also incredable that your duplicate question has the same problem, math.stackexchange.com/questions/201051/…. $\endgroup$
    – Val
    Commented Jan 9, 2014 at 23:31
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    $\begingroup$ @Val Yes exactly! It baffles me too. I asked the question once again, hopefully now without ambiguity.. math.stackexchange.com/questions/1600798/… $\endgroup$ Commented Jan 5, 2016 at 13:38
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Let $P(O,C)$ mean car $C$ is owned by owner $O$.

Then $\forall C \ \exists O \ P(O,C)$ means every car has an owner.

However, $\exists O \ \forall C \ P(O,C)$ means that some owner owns all cars.

Clearly these mean different things.

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    $\begingroup$ Very similar to the answer by Rob Arthan, but I really like this answer because I find it easier to think about cars and owners than family relationships (particularly because they're not both people). $\endgroup$
    – danmcardle
    Commented Sep 12, 2013 at 16:45
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    $\begingroup$ Families are much more complicated than cars :-). $\endgroup$
    – copper.hat
    Commented Sep 12, 2013 at 16:51
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Take $P(x,y)$ to be "$x$ is friends with $y$".

One statement says 'everyone has a friend', the other says 'someone is friends with everyone'.

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A well-formed formula in prenex normal form can be interpreted as a game. The classical logical connectives can also be interpreted as a game: $\lor$ is a choice that the player owns and $\lnot$ exchanges the roles of the players. I think it's less confusing though to just consider statements in prenex normal forms and not assign a game interpretation to connectives, though.

So, we have a game. We're trying to make the statement surrounded by the quantifiers true and the opponent is trying to make it false.

$ \exists x \cdots $ can be read as we choose $x$. Using I choose conveys the same intuition, but is confusing when speaking to people.

$ \forall x \cdots $ can be read as they choose $x$. Using you choose likewise conveys the same intuition, but is confusing when speaking.

$ \exists x \forall y \mathop. P(x, y)$ is intuitively a more difficult game to win than $\forall y \exists x \mathop. P(x, y)$. In the first game, we have to move first and the opponent can react to whatever we did. In the second game, the opponent moves first, so we only need to make a move that beats the move that the opponent happened to pick and we don't need a single strategy that always works.

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  • $\begingroup$ +1 Interesting take on an often asked question (this one is almost a decade old). $\endgroup$ Commented Jul 30, 2021 at 15:04
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Oops! My first answer was to the question I thought you were asking. Here is the answer to the question you actually asked: if $P(x, y)$ means $x > y$, then $\forall x \exists y P(x, y)$ is true iff there is no greatest number and $\exists y \forall x (P(x, y)$ is true if there is a smallest number. Your translations into natural language are OK, but these statements are not the same: the second one is false for any non-empty set of numbers, since for any $y$, $\lnot y < y$. The first sentence is true over the natural numbers but not over any non-empty finite set of natural numbers.

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Order tells you the valid dependencies. For example, if we say $$\forall x \exists y P(x,y),$$

what we're really saying is, for any $x$, we can find a $y$ dependent on that $x$ such that $P(x,y)$.

It may help to Skolemize. The formula becomes: there exists a functions $y(*)$ such that for all $x$ we have $P(x,y(x))$.

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I stumbled on this page asking the same question, but I think I understand now why $\forall x\exists yP(x,y) \not\equiv \exists y\forall xP(x,y)$. It seems that the quantifiers for x and y are essentially nested in each other, such that the second quantifier is dependent on the first.

Let's use the parent example where $P(x,y)$ is a predicate function meaning that y is the parent of x, which can also be stated that x is the child of y. Then, $\forall x\exists yP(x,y)$ can be interpreted as "For all children, there exists a parent." Switching the order of the quantifiers gives $\exists y\forall xP(x,y)$, which can be interpreted as "For some parent, there exists all their children." It seems that the second interpretation has the domain of children limited to the children of the parent, whereas in the first it was all children.

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