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In the following 2 statements -

1. Anyone who eats any pumpkin is a nutrition fanatic. 
&forall x ((&exist y (PUMPKIN(y) &and EAT(x,y))) &rarr FANATIC(x))

2. Anyone who buys any pumpkin either carves it or eats it. 
&forall x &forall y (PUMPKIN(y) &and BUY(x,y) &rarr CARVE(x,y) &or EAT(x,y))

In the first case any pumpkin is given by &exist and

in the second case any pumpkin is given by &forall

Why the second case is also not given by &exist?

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I think your answer is that because the second case is written in prenex normal form while the first one isn't. You can read about prenex normal form in logic more for example here and here.

If I want to explain your cases briefly we have followings in the formal predicate calculus (from page 84 of Logics for mathematicians, Hamilton)

Let $\mathcal{A}$ and $\mathcal{B}$ be two statements and $x_i$ be the statements variables

First, if $x_i$ does not occur free in $\mathcal{B}$, then we have

$$\vdash ((\forall x_i) (\mathcal{A} \rightarrow \mathcal{B})\iff ((\exists x_i)\mathcal{A} \rightarrow \mathcal{B})) \quad*\\ \vdash ((\exists x_i) (\mathcal{A} \rightarrow \mathcal{B})\iff ((\forall x_i)\mathcal{A} \rightarrow \mathcal{B})) $$

second, if $x_i$ doesn't occur free in $\mathcal{A}$ we have

$$\vdash ((\forall x_i) (\mathcal{A} \rightarrow \mathcal{B})\iff (\mathcal{A} \rightarrow (\forall x_i)\mathcal{B}))\\ \vdash ((\exists x_i) (\mathcal{A} \rightarrow \mathcal{B})\iff (\mathcal{A} \rightarrow (\exists x_i)\mathcal{B}))$$

Let $x_i$ be the $y$ you've denoted as a pumpkin. Also, let $\mathcal{A}$ be $\mathbf{pumpkin}(y)\, \wedge \mathbf{buy}(x,y)$ and $\mathcal{B}$ be $\mathbf{carve(x,y)} \wedge \mathbf{eat}(x,y)$. We have following statement by ignoreing the part anyone who

$$(\exists y) \mathcal{A} \rightarrow \mathcal{B} ,$$

that we can rewrite it by above proposition $*$ as follows (Note $y$ is not free in $\mathcal{B}$)

$$(\forall y) (\mathcal{A} \rightarrow \mathcal{B}) $$

Therefore two following statements are equivalent in logic by considering the first part we have ignored, that is

$$(\forall x)(\forall y) (\mathcal{A} \rightarrow \mathcal{B}) \equiv (\forall x) ((\exists y) \mathcal{A} \rightarrow \mathcal{B}) .$$

Note that you are not allowed to write your first statement in prenex normal form because $y$ is free in $\mathcal{B}=\mathbf{fanatic}(x)$.

As a practice, you can prove the four above propositions.

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  • $\begingroup$ According to the other answer given, the second statement(exists version) contradicts "Anyone who buys any pumpkin either carves it or eats it.". Also when i converted it to clause to prove a resolution-refutation problem, both the statements gave correct answer, so they should be equivalent. So the second statement is correct or not? $\endgroup$ – trim24 Dec 3 '17 at 12:30
  • $\begingroup$ I didn't understand what you asked in your comment, but about the other answer I can draw your attention to the parenthesis of statements. $\forall x \exists y (A\rightarrow B)$ has different meaning from $\forall x (\exists y A \rightarrow B)$. qivian is considering the $\forall x \exists y (A\rightarrow B)$ statement while I considered $\forall x (\exists y A \rightarrow B)$. $\endgroup$ – Doralisa Dec 3 '17 at 12:38
  • $\begingroup$ @trim24 I think your second statement is correct. As I have explained your second statement is the prenex normal form of $\forall x (\exists y A\rightarrow B)$. $\endgroup$ – Doralisa Dec 3 '17 at 12:42
  • $\begingroup$ sorry for being unclear. I was confused with the parenthesis, but its clear now. $\endgroup$ – trim24 Dec 3 '17 at 14:59
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So I am assuming your question is: Why is the formula to "Anyone who buys any pumpkin either carves it or eats it."

$\forall x \forall y (\text{pumpkin}(y) \wedge \text{buy}(x, y) \implies \text{carve}(x, y) \vee \text{eat}(x, y)) $

instead of

$\forall x \exists y (\text{pumpkin}(y) \wedge \text{buy}(x, y) \implies \text{carve}(x, y) \vee \text{eat}(x, y)) $

The problem with your version is that there might be someone who buys a pumpkin and decides not to eat or carve it and this would contradict the statement that every bought pumpkin (by a person $x$) will be either carved or eaten.

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