Negation of existential quantifiers, functions in first order logic I know the following rule:
    $\lnot \forall a (P(a)$ means $\exists a \lnot P(a)$ and $\lnot \exists a (P(a)$ means $\forall a \lnot P(a)$
But what if the problem is something like: $\lnot \forall x \lnot \exists y (P(x) \land P(y))$? Will the negation cancel out? To something like: $ \exists x \forall y (P(x) \land P(y))$? 
And furthermore, what happens with functions $f$ then?
If there is given: $\lnot \forall a \lnot \exists b(P(a) \land P(f(b)) ) \implies Q(f(f(b))))$ and I want to find a Model which satisfies the formula, how can I?
 A: They don't cancel out in the way you write it, no. Don't forget, the rule in the form you cite is equivalent to:
$$
\neg\, \forall x \,\neg\, \varphi(x) \iff \exists x\, \varphi(x) \\
\,\neg \,\exists x \,\neg\, \varphi(x) \iff \forall x\, \varphi(x) \\
$$
Even more concisely for all four equivalences,
$$\begin{align}
\neg \exists &= \forall \neg \\
\neg \forall &= \exists \neg \\
\neg \forall \neg &= \exists \\
\neg \exists \neg &= \forall .
\end{align}$$
With your example, $\lnot \forall x\, \lnot \exists y\, (P(x) \land P(y))$ is equivalent to $\exists x\, \exists y \,(P(x) \land P(y))$.
If you have function symbols as well, the same quantifier rules still apply to variables, of course. Your second example sentence, plus a missing initial parenthesis, is $\lnot \forall a\, \lnot \exists b \left((P(a) \land P(f(b)) ) \to Q(f(f(b)))\right)$, which is equivalent to $\exists a\, \exists b\,((P(a) \land P(f(b)) ) \to Q(f(f(b))))$. 
The question about how you then find a model is a very different question.
