Pulling universal quantifier out of parenthesis makes nonequivalent statement?

Question

Assuming I have the statement ∀x(∀y¬Q(x,y)∨P(x)), can I pull the universal quantifier ∀y out of the parenthesis? Meaning, is this statement equivalent to ∀x∀y(¬Q(x,y)∨P(x)) ?

An approach I tried so far:

1. ∀x((∃y Q(x,y) ) => P(x)). (original eq.)
2. ∀x((∀y¬Q(x,y))∨P(x)) (De Morgan's application)
3. ∀x∀y(¬Q(x,y)∨P(x)). (Working off the assumption that taking out the ∀y is a valid operation).
4. ∀x∀y (Q(x,y) => P(x)) (Going backwards from the ¬P v Q definition of implication)

Statement 4 does not seem to be equivalent to statement 1, which suggests that pulling out the universal quantifier is not acceptable. I would greatly appreciate any confirmation of whether this is the case, and if so, what governs when quantifiers can be brought to the outside of the parenthesis.

Answer 1

The original expression: $\forall x~((\exists y~Q(x,y))\to P(x))$ says "For any $x$ it holds that if some $y$ satisfies $Q(x,y)$, then $P(x)$ is satisfied."

Now either consequent is true for all $x$ or, whenever it is false, the antecedent is also false (ie for that $x$ no $y$ can satisfy $Q(x,y)$). Thus the expression equates to: $\forall x~(\neg P(x)\to\forall y~\neg Q(x,y))$

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The final expression: $\forall x~\forall y~(Q(x,y)\to P(x))$ says: "For any $x$ and $y$, it holds that if $Q(x,y)$ then $P(x)$."

Now either consequent is true for all $x$ or, whenever it is false, the antecedent is also false; moreover false for all $y$ when $P(x)$ is false for some $x$. Thus the expression equates to: $\forall x~\forall y~(\neg P(x)\to \neg Q(x,y))$

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Therefore the original and final expressions are equivalent.

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Answer 2

They're equivalent.

Here is a proof:

This tree was generated here.

Answer 3

For universal quantifier. In general, if $x$ apear in both $A$ and $B$ we have $$\exists xA(x)\to \forall xB(x)\Rightarrow\forall x(A(x)\to B(x))\tag{1}$$ $$\forall x(A(x)\to B(x))\not\Rightarrow \exists xA(x)\to \forall xB(x)\tag{2}$$ However, if $x$ not apear in $B$ we have $$\forall x(A(x)\to B)\Leftrightarrow\exists xA(x)\to \forall x B\tag{3}$$ The statement in question is similar to $(3)$, which is also valid. $$∀x∀y(Q(x,y)→P(x))\Leftrightarrow∀x(∃y Q(x,y)→P(x))\tag{4}$$ And we can formulate a direct proof for $(4)$ by nature deduction $$\def\fitch#1#2{\hspace{2ex}\begin{array}{|l}#1\\\hline#2\end{array}} \fitch{\forall x\forall y(Q(x,y)\to P(x))} {\fitch{\boxed{a}} {\forall y(Q(a,y)\to P(a))\\ \fitch{\exists y~Q(a,y)} {\fitch{\boxed{b}~Q(a,b)} {Q(a,b)\to P(a)\\ P(a)}\\ P(a)}\\ \exists y~Q(a,y)\to P(a)}\\ \forall x~(\exists y~Q(x,y)\to P(x))}\\ $$ Hence $\forall x\forall y(Q(x,y)\to P(x))\Rightarrow\forall x~(\exists y~Q(x,y)\to P(x))$. For the another direction we have $$ \fitch{\forall x(\exists y~Q(x,y)\to P(x))} {\fitch{\boxed{a}} {\exists y~Q(a,y)\to P(a)\\ \fitch{\boxed{b}~Q(a,b)} {\exists y~Q(a,y)\\ P(a)}\\ \forall y~(Q(a,y)\to P(a))}\\ \forall x\forall y~(Q(x,y)\to P(x))}$$ Therefore $\forall x~(\exists y~Q(x,y)\to P(x))\Rightarrow \forall x\forall y(Q(x,y)\to P(x))$. This proves $(4)$.