Do ${\forall}x{\forall}y{\exists}z$, ${\forall}x({\forall}y{\exists}z)$ and $({\forall}x{\forall}y){\exists}z$ all mean the same thing?
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1Yes, but there is a difference between $\forall x (\forall y A\wedge B)$ and $(\forall x \forall y) (A\wedge B)$ but that's probably clear. – Peter Franek Mar 23 '16 at 18:28
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Why would it be clear? – user285146 Mar 23 '16 at 18:34
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Consider these formulas concerning real numbers: (1) $\forall x ((\forall y ,,x>y) \vee (y=0))$, and (2) $(\forall x \forall y) ((x>y) \vee (y=0))$. The second one has no free variables and is false: the first has a free variable $y$ and is true/false depending on $y$. – Peter Franek Mar 23 '16 at 18:41
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Yes, order matters A LOT. Try to read some book of logic, order is essential in quantifiers. – Masacroso Mar 23 '16 at 18:43
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@Masacroso It's confusing but he doesn't really ask about order, but about parenthesis. The way he wrote it, it looks like being the same thing. – Peter Franek Mar 23 '16 at 18:44
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Yes, but the second two are confusing. The analogy with an operation like $+$ is misleading. You always just take the quantifiers in the order they are given. The parentheses make people wonder what is going on. – almagest Mar 23 '16 at 19:25