I have this formula:
$\forall x,y(R(x,y) \wedge \exists z R(x,z))\qquad(1)$
I remember that it is correct to push down the universal quantifier in this way:
$\forall x(\forall y R(x,y) \wedge \exists z R(x,z))\qquad(2)$
I also think the first conjunct of $(1)$ implies the second one, then $(1)$ can also be semplified in just
$\forall x,y(R(x,y))\qquad(3)$
My problem is that I don't think an analogous implication would hold for $(2)$, then I'm not able to reach $(3)$ passing from $(2)$ (while, if the three formulas are equivalent, it should be possible).
Where am I wrong? I think the error is in the implication leading to $(3)$, but I don't see exactly why.
The Wiki article prenex normal form for some more examples of rules for shifting quantifiers around $\land$ and $\lor$ and whether they work when the domain is empty.
$\forall x \mathop. \varphi(x) \land \psi$ is equivalent to $(\forall x \mathop. \varphi(x)) \land \psi$ where $\psi$ does not contain a free occurrence of $x$ holds if we constrain the domain of discourse to be non-empty, but fails otherwise. (For example, consider what happens when $\varphi = \top$ and $\psi = \bot$.)
However, $\forall x y \mathop. R(x, y)$ does imply $\forall x \exists y \mathop. R(x, y)$, always. It holds even if the domain of discourse is empty because it is headed by a universal quantifier.
Additionally, $\forall x \mathop. \varphi(x) \land \psi(x)$ is equivalent to $(\forall x \mathop. \varphi(x)) \land (\forall x \mathop. \psi(x))$ whether empty models are allowed or not.