Does $\forall x \exists y (P(x) \land Q(y))$ imply $\forall x P(x)$?

Question

I am not sure if the $x$ in $\forall x \exists y (P(x) \land Q(y))$ carries some kind of "context" from the $\exists y$ that would make $\forall x P(x)$ not neccesarily true.

Answer 1

Let $c$ be arbitrary. Then by the hypothesis we get

$$\exists y[P(c)\land Q(y)] $$ Hence we get $$P(c)\land Q(d)$$ for some $d$. Hence we get $P(c)$. Since $c$ is arbitrary, thus we get $\forall xP(x)$

Answer 2

Yes, that follows. First, here are some relevant equivalence principles:

Prenex Laws

Where $\varphi$ is any formula and where $x$ is not a free variable in $\psi$:

$ \forall x \ \varphi \land \psi \Leftrightarrow \forall x (\varphi \land \psi)$

$ \psi \land \forall x \ \varphi \Leftrightarrow \forall x (\psi \land \varphi)$

$ \exists x \ \varphi \land \psi \Leftrightarrow \exists x (\varphi \land \psi)$

$ \psi \land \exists x \ \varphi \Leftrightarrow \exists x (\psi \land \varphi)$

$ \forall x \ \varphi \lor \psi \Leftrightarrow \forall x (\varphi \lor \psi)$

$ \psi \lor \forall x \ \varphi \Leftrightarrow \forall x (\psi \lor \varphi)$

$ \exists x \ \varphi \lor \psi \Leftrightarrow \exists x (\varphi \lor \psi)$

$ \psi \lor \exists x \ \varphi \Leftrightarrow \exists x (\psi \lor \varphi)$

So with those (well, the first and third):

$$\forall x \exists y (P(x) \land Q(y))$$

$$\Leftrightarrow$$

$$\forall x (P(x) \land \exists y \ Q(y))$$

$$\Leftrightarrow$$

$$\forall x \ P(x) \land \exists y \ Q(y)$$

And clearly, this last statement implies $\forall x \ P(x)$