The prenex form doesn't seem equivalent to the original sentence

Question

I was reading this stackexchange post and am confused about the following logical equivalence. I have tried to generalize what they've done as follows:

∀x[∀y[f(y)] → g(x)] ; for all x, if for all y, f(y) is true, then g(x) is true
∀x[~∀y[f(y)] V g(x)]
∀x[∃y~[f(y)] V g(x)]
∀x∃y[~[f(y)] V g(x)]
∀x∃y[f(y) → g(x)] ; for all x and at least 1 y, if f(y) is true, then g(x) is true

I've also written them out in plain English. Aren't their meanings significantly different? If you give them some intuitive meanings, they still remain different:

∀x[∀y[veggie(y) → dislikes(x,y)] → smart(x)] ; If all people dislike all vegetarians, then they are smart
∀x[~∀y[veggie(y) → dislikes(x,y)] V smart(x)]
∀x[∃y~[veggie(y) → dislikes(x,y)] V smart(x)]
∀x∃y[~[veggie(y) → dislikes(x,y)] V smart(x)]
∀x∃y[(veggie(y) → dislikes(x,y)) → smart(x)] ; If all people dislike at least one vegetarian, then they are smart

I'm confused as to why the scope of the quantifier affects the statement at all.

Answer 1

$∀x\Big(∀y f(y) → g(x)\Big)\tag{A1}$ for all x, if for all y, f(y) is true, then g(x) is true

$∀x∃y\Big(f(y) → g(x)\Big)\tag{A2}$ for all x and at least 1 y, if f(y) is true, then g(x) is true

Yes, the four sentences above are equivalent to one another. I've written a proof in the appendix of this answer.

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$∀x\Big(∀y \big(\operatorname{Veggie}(y) → \operatorname{Dislikes}(x,y)\big) → \operatorname{Smart}(x)\Big)\tag{B1}$ If all people dislike all vegetarians, then they are smart $\color\red{\textbf{✗}}$

Your translation means $$∀x∀y \Big(\operatorname{Veggie}(y) → \operatorname{Dislikes}(x,y)\Big) → ∀\color\red z\operatorname{Smart}(\color\red z),$$ whereas statement B1 means:
Everyone who dislikes every vegetarian is smart. (B1)

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The following is indeed logically equivalent to statement B1:

$∀x∃y\Big(\big(\operatorname{Veggie}(y) → \operatorname{Dislikes}(x,y)\big) → \operatorname{Smart}(x)\Big)\tag{B2}$ If all people dislike at least one vegetarian, then they are smart $\color\red{\textbf{✗}}$

You translation has two possible meanings: $$∀x∃y \Big(\operatorname{Veggie}(y) \color\red \land \operatorname{Dislikes}(x,y)\Big) → ∀\color\red z \operatorname{Smart}(\color\red z)\\ ∃y∀x \Big(\operatorname{Veggie}(y) \color\red \land \operatorname{Dislikes}(x,y)\Big) → ∀\color\red z \operatorname{Smart}(\color\red z).$$

On the other hand, statement B2 is best translated mechanistically and means:
For each person x, some person y exists such that the following holds: if x dislikes y if y is vegetarian, then x is smart. (B2)

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Reply to the OP's comment

Then what does does this mean? $∀x∀y\Big(\big(\operatorname{Veggie}(y)→\operatorname{Dislikes}(x,y)\big)→\operatorname{Smart}(x)\Big)\tag{C1}$

Statement C1 means (note that $x$ and $y$ can refer to the same person):
For every couple (x,y), if x dislikes y if y is vegetarian, then x is smart. (C1)

Statement C1 is a stronger assertion than statement B1 (Everyone who dislikes every vegetarian is smart / $∀x\Big(∀y \big(\operatorname{Veggie}(y) → \operatorname{Dislikes}(x,y)\big) → \operatorname{Smart}(x)\Big)$ ). For example, in the following scenario, B1 is true but C1 is false:

• the universe comprises Adam and Eve, each of whom likes themself, and between whom the smart carnivore likes the stupid vegetarian but not vice versa.

Statement C1 is in fact logically equivalent to:
For every couple (x,y), either x likes y and y is vegetarian or x is smart. (C2)

I'm struggling to intuitively grasp how B1 is logically equivalent to B2.

Okay, let's simplify the issue: you're asking how these two are equivalent sentences: $\Big(\forall y\,\big(Vy\to aDy\big)\Big) \to Sa\tag{D1}$ $\exists y\Big((Vy\to aDy)\to Sa\Big)\tag{D2}.$ Existentially quantified conditionals are unnatural to parse in natural language, so I agree that the sentence construction D2 is hard to intuitively grasp. If the contrapositive argument in the abovementioned appendix doesn't convince you that D1 ≡ D2, you could alternatively use the equivalence $(P\to R\:\equiv\: R \lor \lnot P)$ to convert each conditional to a disjunction.