Example
Let $\Gamma$ the set of first-order Peano axioms: no variables free.
1) $\Gamma \vdash \exists x (x = 0)$ --- easily provable
2) $\Gamma, x=0 \vdash x=0$ --- obvious
3) $\Gamma \vdash x=0$ --- from 1) and 2) by $\exists$-elim : wrong !
4) $\Gamma \vdash \forall x (x=0)$ --- from 3) by $\forall$-intro,
1) $\Gamma, x=0 \vdash x = 0$
2) $\Gamma, x=0 \vdash \forall x (x=0)$ --- by $\forall$-intro: wrong !
3) $\Gamma \vdash x=0 \to \forall x (x=0)$ --- from 2) by $\to$-intro
4) $\Gamma \vdash \forall x [x=0 \to \forall x (x=0)]$ --- from 3) by $\forall$-intro
5) $\Gamma \vdash 0=0 \to \forall x (x=0)$ --- from 4) by $\forall$-elim.
The ground for the restriction on $\forall$-intro are linked to the "generalization principle":
what holds for any, holds for all.
Thus, in order to formalize this principle with a rule of inference, we read it as:
if something holds for an "arbitrary object", then it holds for all objects.
We have to capture the informal concept of “arbitrary object” by way of a syntactic criterion.
Consider now a variable $x$ in the context of a derivation: we shall call $x$ arbitrary if nothing has been assumed concerning $x$. In other words, $x$ is arbitrary at its particular occurrence in a derivation if the part of the derivation above it contains no hypotheses containing $x$ free.
The premises $\forall x (P(x) \land Q(x))$ and $\exists x \, \lnot P(x)$ are contradictory, so according to the principle of explosion (which corresponds to the rule $\bot_e$, a.k.a. ex falso quodlibet in natural deduction) from them you can derive everything, in particular $\exists x \, \lnot Q(x)$, as showed by the following derivation.
$$
\dfrac{\exists x \, \lnot P(x) \qquad \dfrac{\dfrac{[\lnot P(x)]^* \qquad \dfrac{\dfrac{\forall x (P(x) \land Q(x))}{P(x) \land Q(x)}\forall_e}{P(x)}\land_e}{\bot}\lnot_e}{\exists x \, \lnot Q(x)}\bot_e}{\exists x \, \lnot Q(x)}\exists_e^*
$$
Note that there is no need for the rules RAA and $\lnot_i$.
Best Answer
A priori, you are right, there might be a problem in the rules $\land_E$ and $\to_E$, because the . But in fact, the problem is easily solved because there is another nice property for natural deduction:
This lemma can be easily proved by induction on the derivation of $\Gamma \vdash \varphi$ (if you prefer, you can prove it simultaneously to the proof of van Dalen's Theorem 2.8.3.(i)). Note that you are in language where the only connectives are $\land, \to, ⊥$ and $\forall$ (p. 91).
Thanks to the lemma above, you do not have any problem in the proof of Theorem 2.8.3.(i) with the cases $\land_E$ and $\to_E$. For instance, for $\land_{E_i}$ (with $i \in \{1,2\}$), you have that \begin{align} \dfrac{\quad\Gamma\\\quad \ \vdots\\\varphi_1 \land \varphi_2}{\varphi_i}\land_{E_i} \end{align} According to your hypothesis, $x$ does not occur in $\Gamma$ or $\varphi_i$, but what about $\varphi_j$ with $j\neq i$? By the lemma above, if $x$ occurred free in $\varphi_j$ then $x$ would occur free in $\Gamma$, which contradicts the hypothesis. So, $x \notin FV(\varphi_j)$. Moreover, if $x$ occurred bound in $\varphi_j$ then you could rename bound variables in $\varphi_j$ so that $x$ does not occur bound in $\varphi_j$. Therefore, $x$ do not occur in $\Gamma$ or $\varphi_1$ or $\varphi_2$. Hence, you can apply the inductive hypothesis to the derivation of $\Gamma \vdash \varphi \land \varphi_2$. Then, you can easily conclude by yourself.