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I have a concern with nested quantifiers.

I have: $$ \forall x \exists y \forall z(x^2-y+z=0) $$ such that $$ x,y,z \in \Bbb Z^+$$

My first question, can it be read like this:

$$ \forall x \forall z \exists y(x^2-y+z=0) $$

The way I did it, is I started off with $x=1, z=1 $

$$ 2-y = 0 $$ $$ y =2 $$

Is this a good approach?

Dimitri
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3 Answers3

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No, you cannot interchange the $\exists y$ and $\forall z$: doing so changes the meaning of the statement. The original statement,

$$\forall x\exists y\forall z\left(x^2-y+z=0\right)$$

says that no matter what positive integer $x$ you choose, I can find a $y\in\Bbb Z^+$ such that $x^2-y+z=0$ no matter what positive integer $z$ you choose. For instance, if you choose $x=1$, I can find some positive integer $y$ such that $1-y+z=0$ for every positive integer $z$. Since $1-y+z=0$ if and only if $z=y-1$, this is clearly false: no matter what $y$ I pick, you can choose for $z$ any positive integer except $y-1$ and ensure that $x^2-y+z$ is not $0$.

The modified statement

$$\forall x\forall z\exists y\left(x^2-y+z=0\right)\;,$$

on the other hand, says something very different: it says that no matter what positive integers you choose for $x$ and $z$, I can find a positive integer $y$ such that $x^2-y+z=0$. And of course I can: I just pick $y=x^2+z$.

Since one statement is true and the other false, they certainly cannot be equivalent.

Brian M. Scott
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  • reason I get so confused is choosing values for each variable and in which order. If the for loop starts at $x=1$ and choose $y=1$ as my choice, then the for loop for $z$ would start at $z=1$? If so then $(1-1+z)$ which is $z=0$ but zero in no a positive integer...right? – Dimitri Sep 14 '13 at 03:10
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    @Dimitri: These expressions really don’t have anything to do with FOR loops, and you shouldn’t try to think of them in those terms. Take a look at this question and my answer to it, and see whether it helps a bit; the logical expressions involved are a little simpler. – Brian M. Scott Sep 14 '13 at 03:16
  • I looked at the question and the statement is clearer to see when there are two variables; for some reason it makes sense. So, given the fact I chose $x=1$, which would give $1-y+z=0$. So far so good. Now, I re-read the expression with my new expression $1-y+z=0$ like you said: there exists a $y$ such that $1-y+z=0$? At this point, person $y$ would choose his number of the game, right? (I do appologize for any repetitiveness) – Dimitri Sep 14 '13 at 03:29
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    @Dimitri: No matter how many quantifiers there are, think of it as a game between two players, one called $\forall$ and the other, $\exists$. You’re $\exists$, and your opponent is $\forall$. You play in the order in which the quantifiers appear. If it’s $\forall\exists\forall$, he plays first, then you play, then he plays. If it’s $\forall\forall\exists$, he makes two plays, and then you get to play. His goal is to make the statement false; yours is to make it true. If you can always do so, no matter how he makes his choices, you win, and the statement is true. If not, you lose, and ... – Brian M. Scott Sep 14 '13 at 03:43
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    ... it’s false. With $$\forall x\exists y\forall z\left(x^2-y+z=0\right);,$$ for instance, you cannot always make $x^2-y+z=0$ true. In fact in this case your opponent can always make it false: he picks any positive integer for $x$, and no matter what positive integer you pick for $y$, he can choose $z$ to be some positive integer other than $y-x^2$, thereby ensuring that $x^2-y+z$ isn’t $0$. – Brian M. Scott Sep 14 '13 at 03:45
  • I see now. I like the idea of the game. Thank you very much for the explanation and over-clarification. I'll try to do some more using the game concept. – Dimitri Sep 14 '13 at 03:48
  • @Dimitri: You’re welcome. Don’t feel too bad if it takes a little while to get the hang of this: nested quantifiers are a stumbling block for quite a few students. – Brian M. Scott Sep 14 '13 at 03:50
  • Yea, they're a bit tricky when more variables are added and I think most students in my class are a little confused as well... – Dimitri Sep 14 '13 at 03:55
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The original sentence says that for any $x$, there is a $y$, such that whatever $z$ we pick, we have $x^2-y+z=0$.

So the $y$ has to work for all $z$. But that's impossible. If it works for $z=1000$, it fails for all other $z$. The sentence is (very) false.

We cannot interchange the quantifiers $\forall$ and $\exists$ without altering the meaning of the sentence.

André Nicolas
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  • So, in another words, whenever an Existential quantifier exists, we can always pick an arbitrary number for it and loop the rest? – Dimitri Sep 14 '13 at 03:04
  • The number is not necessarily arbitrary we can take account of the value of $x$ when choosing it, because the formula has the $\forall x$ before the $\exists y$. But if we have universal quantifiers like the $\forall z$ to the right of $\exists y$, we cannot change our minds about the value of $y$: it has to work for all $z$. Note that the unmodified sentence is (very) false in the integers, while your modification is true. – André Nicolas Sep 14 '13 at 03:08
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Consider $$\forall y. \exists x. x\text{ is the mother of }y$$ means that for every person $y$, there is some person $x$ who is $y$'s mother, which is true; every person has a mother. But $$\exists x.\forall y. x\text{ is the mother of }y$$ says that there is some person $x$, so that for every person $y$, $x$ is $y$'s mother, which is false; there is no person who is everyone's mother.

The order is important.

MJD
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