I’ve been studying universal generalization recently and, according to the textbooks, $\forall x Q(x)$ can be derived from $Q(a)$, if the variable $a$ is arbitrary. A variable is arbitrary, when it does not appear in any of the undischarged assumptions throughout the derivation.
Example $1$: Prove that $\forall x Q(x)$ derives from $\forall x [P(x) \rightarrow Q(x)]$ and $\forall x P(x)$.
$$\begin{array}{lll} 1. & \forall x [P(x) \implies Q(x)] & \\ 2. & \forall x P(x) & \\ \hline 3. & P(a) \implies Q(a) & (1.; \text{U.I.}) \\ 4. & P(a) & (2.; \text{U.I.}) \\ 5. & Q(a) & (3.,4.;\text{M.P.})\\ 6. & \forall xQ(x) & (5.; \text{U.G.}) \end{array}$$
We could generalize $Q(a)$ in step $6$, because the variable $a$ was not in the premises of our proof (steps $1$ and $2$).
However, consider the following example:
Example $2$: Prove that $\forall x Q(x)$ derives from $\forall x [P(a) \rightarrow Q(x)]$ and $P(a)$.
$$\begin{array}{lll} 1. & \forall x[P(a) \implies Q(x)] & \\ 2. & P(a) & \\ \hline 3. & P(a) \implies Q(a) & (1.; \text{U.I.}) \\ 4. & Q(a) & (2.,3.; \text{M.P.}) \\ 5. & \forall xQ(x) & (4.; \text{U.G.}) \\ & & \text{MISTAKE: $a$ appears in $P(a)$} \end{array}$$
In this case, according to the definition of arbitrariness presented above, we are actually not able to universally generalize $Q(a)$ in step $5$, since the variable $a$ does appear in one of the premises (step $2$). Nevertheless, $\forall x Q(x)$ does derive from $\forall x [P(a) \rightarrow Q(x)]$ and $P(a)$, so universal generalization should be possible here. Where is my reasoning flawed?
Prove that $(\forall x)Q(x)$ derives from $(\forall x)[P(a) \implies Q(x)]$ and $P(a)$.
$$\begin{array}{lll} 1. & \forall x[P(a) \implies Q(x)] & \\ 2. & P(a) & \\ \hline 3. & P(a) \implies Q(x_0) & (1.; \text{U.I.}) \\ 4. & Q(x_0) & (2.,3.; \text{M.P.}) \\ 5. & \forall xQ(x) & (4.; \text{U.G.}) \end{array}$$
Note that, when we use the U.I. rule, we are considering that $P(a) \implies Q(x_0)$, for any chosen $x_0$. So, in the end, you have $Q(x_0)$ for all $x_0$. Then, we only have to use the U.G. rule. $\square$