Initializing non-deterministic variables in QBF
It looks to me like there is some confusion here about what QBF is and when it is needed.  You don't need QBF for this problem; SAT is fine.

Here's how to solve this using a SAT solver. Notice that the condition counter &gt; 0 is true at the end if and only if there was at least one loop iteration where f output 1.  In other words, the assertion will be true if and only if

$$\exists \mathtt{loop\_aux\_0},\dots,\mathtt{loop\_aux\_n} . \mathtt{f}(\mathtt{loop\_aux\_0},\dots,\mathtt{loop\_aux\_n})=1.$$

That can be expressed as a SAT instance.  There are no nested/alternating quantifiers; the formula has the form $\exists x. \varphi(x)$, so it can be solved by SAT.  Simply express the condition f(loop_aux_0, ..., loop_aux_n)=1 in CNF form, and then use a SAT solver to check whether that condition can ever be true (for any value of the loop variables).



QBF is useful when you have nested quantifiers, e.g., $\exists x . \forall y . \exists z . \varphi(x,y,z)$.  The result will be either true or false.

It's also worth knowing about #SAT.  #SAT is useful when you want to count the number of satisfying assignments to some formula $\varphi(x)$.  In other words, instead of checking whether $\exists x . \varphi(x)$ is true, suppose we want to know the number of values $x$ such that $\varphi(x)$ is true; then this is an instance of #SAT.

QBF and #SAT are not the same thing.  They're different problems.  And QBF isn't some kind of magic wand that lets you compress any SAT instance by some enormous factor; QBF is for a specific type of logical formula.