Bracketing and multiple negated quantifiers in predicate logic

Question

Do bracketing and placement affect quantifiers in predicate logic?

I.e., are the following two propositions equivalent (where x and y are variables and P and T predicates)

¬∃x (¬∃y Pxy → (∀z ¬(Pzx → Tz) ∨ Tx))

¬∃x ¬∃y ∀z (Pxy → ¬(Pzx → Tz) ∨ Tx))

Additionally, can the negation of first two quantifiers be dealt with at the same time? I.e.:

¬∃x (¬∃y Pxy → (∀z ¬(Pzx → Tz) ∨ Tx)) = 1

∃x (∃y Pxy → (∀z ¬(Pzx → Tz) ∨ Tx)) = 0

Edit: I should clarify, I mean in the specific case that a quantifier has scope over the entire preposition, can it be moved to the beginning/outside of brackets?

The second part of the question refers to the truth value of the proposition, where 1 is True and 0 is False.

Another way to think about it: does ¬∃x¬∃y = ¬∃y¬∃x ?

Answer 1

Regarding the first question, the answer is YES.

See Prenex Normal Form : we can move $∀z$ in front of "$Pxy →$" because $z$ is not free in it.

Regarding the second question, the answer is NO.

The negation of :

$¬∃x (¬∃y Pxy → (∀z ¬(Pzx → Tz) ∨ Tx))$

will be : $¬¬∃x (¬∃y Pxy → (∀z ¬(Pzx → Tz) ∨ Tx))$, i.e.

$∃x (¬∃y Pxy → (∀z ¬(Pzx → Tz) ∨ Tx))$.

Answer 2

Your first question is not one of bracketing, but of moving quantifiers to a different position in the formula.

The answer is in this case yes, you are allowed to move the quantifier from the succedent of the implication out of the implication to the front, under the restriction that $z$ does not occur free in the antecedent and would thereby be accidentally bound by the quantifier:

$(\phi \to \forall z \psi) \equiv \forall z (\phi \to \psi)$, if $z$ not free in $\phi$
$(\phi \to \exists z \psi) \equiv \exists z(\phi \to \psi)$, if $z$ not free in $\phi$

If you move the quantifier out of antecedent position, it changes to its dual ($\forall$ becomes $\exists$ and $\exists$ becomes $\forall$) - again, this is only allowed under the condition that you don't scope on previously free occurrences of $z$ in the succedent:

$(\forall z \phi \to \psi) \equiv \exists z (\phi \to \psi)$, if $z$ not free in $\psi$
$(\exists z \phi \to \psi) \equiv \forall z (\phi \to \psi)$, if $z$ not free in $\psi$

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Cf. your second question, no, you can not move around negations between quantifiers.
$\neg \exists x \neg \exists y \phi$ says "There exists no x such that there exists no y such that $\phi$", which is logically equivalent to $\forall x \exists y$, "For all x exist y such that $\phi$.
$\neg \neg \exists x \exists y \phi$ is logically equivalent to $\exists x \exists y \phi$ and says "There exist x and y such that $\phi$".

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The question you ask in your last paragraph is yet a different one, namely whether one can exchange quantifiers with each other.
The answer is: You can exchange quantifiers of the same type as long as you don't change the structure of negations, implications or other logical symbols, but you can't exchange quantifiers of different types:

$\exists x \exists y \equiv \exists y \exists x\\ \neg \exists x \neg \exists y \equiv \neg \exists y \neg \exists x\\ \forall x \forall y \equiv \forall y \forall x\\ \neg \forall x \neg \forall y \equiv \neg \forall y \neg \forall x$
(swapping quantifiers of the same type, keeping negations where they are: possible)
but
$\exists x \forall y \not \equiv \forall y \exists x\\ \forall x \exists y \not \equiv \exists y \forall x$
(swapping $\exists$ and $\forall$ quantifiers: not possible; same if there is negation involved).

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Concerning your truth values, the way you wrote it down is syntactically incorrect: Writing "$... \lor Tx)) = 0$" or "$... \lor Tx)) = 1$" would mean that "$=0$"\"$=1$" belongs to the formula, but it doesn't. What you want to say is "The formula ¬∃x (¬∃y Pxy → (∀z ¬(Pzx → Tz) ∨ Tx)) has the truth value 0", which you have to express as $⟦ ¬∃x (¬∃y Pxy → (∀z ¬(Pzx → Tz) ∨ Tx))⟧_v =0$" (for some assignment function $v$ - truth values are always defined relative to assignment functions).