Consider first-order logic with some fixed, relational vocabulary $\tau$. A sentence is a formula in this logic with no free variables.


A sentence is in [prenex normal form](http://en.wikipedia.org/wiki/Prenex_normal_form), if all quantifiers are moved to the front. For example $\exists x\exists y(P(x)\to P(y))$ is in prenex normal form whereas $\exists x( P(x)\lor \exists y(P(x)\to P(y)))$ is not.


Let quantifier rank of a sentence be the maximum number of nested quantifications in it.


Now, we know that quantifier rank is a measure of complexity of a first-order formula in the sense that there are only a finite number of sentences of a fixed quantifier rank up to logical equivalence (lets consider only finite models). 


If we have a sentence of the form $\varphi \equiv \exists x(\exists y \alpha(x,y)\lor\exists y \beta(x,y))$, where $\alpha$ and $\beta$ are formulas with two free variables $x$ and $y$, it is straightforward to transform it to a prenex form by simply renaming the occurences of $y$s and shifting quantifiers in front, i.e. $\exists x\exists y\_1\exists y\_2(\alpha(x,y\_1)\lor \beta(x,y\_2))$. But now the originating formula had quantifier rank of $2$ when the prenex form formula has quantifier rank $3$.


How would one transform a formula to prenex normal form in a way that would keep the quantifier rank untouched? 



Prenex normal form vs. quantifier rank
If you begin with $\psi \equiv (\exists x)R(x) \land (\exists x)(\lnot R(x)) \land (\forall y)S(y)$, the usual prenex form will have depth 3, while the original formula has depth 1. So you are asking how to find an equivalent formula to $\psi$ that only has one quantifier. 


Let's assume our language only has two unary relation symbols $R$ and $S$. Then the formula $\psi$ above cannot be equivalent to any one quantifier formula $\phi$. First assume $\phi$ is existential, and true in some model. Then we can make a new model $M$ by adding one more element for which $S$ does not hold. Then $\phi$ is still true in $M$ (it will be preserved under extensions) but $\psi$ is not true in $M$ because $(\forall y)S(y)$ is not true. 


Suppose $\phi$ is universal. Take any model with more than 2 elements in which $\psi$ holds, and call one of the elements $c$. Take the submodel $N$ containing just $c$. Then $\phi$ is still true in $N$, because it is preserved under taking substructures, but $\psi$ is not true because $\psi$ requires at least two elements in the domain. 


