Is the following schema a derived rule of our logic? Is the following schema a derived rule of our logic?
$$ A \rightarrow B \vdash A \rightarrow (\forall x)B, \text{ provided $x$ is not free in $A$ }$$
If yes, then give a proof. if no, show why by proving the invalid "strong generalization" using the above formula.
This question is in Ex 2 of Section 8.3 of Mathematical Logic by G.Tourlakis, page 208.
I don't understand what proving invalid "Strong generalization" has to do with proving invalidity of a rule.
The following is the solution for this question which I don't understand.
$$ (1)\ A \ \ \ \ \ \ \ \langle\text{hypothesis}\rangle $$
$$ (2)\ T \rightarrow A \ \ \ \ \ \ \ \ \langle(1) + \vdash A \equiv (T \rightarrow A) + Eqn\rangle$$
$$ (3)\ T \rightarrow  (\forall x)A \ \ \ \ \ \ \ \ \langle(2) + \text{above rule}\rangle$$
$$(4)\ (\forall x)A \ \ \ \ \ \ \ \ \langle(3) + \vdash X \equiv (T \rightarrow X)\rangle$$
 A: The statement of the Exercise gives us two hints :


If you do not think [that it is a derived rule of our logic] then give a definitive reason as to why — for example, using a concrete interpretation, or by proving the invalid "strong generalization" using [it] as a lemma.


We may try with the counter-example, i.e. finding an interpretation [see page 197] in the domain $\mathbb N$ of the natural numbers such that the above formula is false [see Example 8.1.7, page 200, and also page 143].
Let $A$ is $0 = 0$ ($x$ is not free in it), and let $B$ is $x = 0$.
We have that $A \rightarrow B$ is :

$0 = 0 \rightarrow (x = 0)$.

If we assign to $x$ the denotation $0$, we have that $A \rightarrow B$ is true, but of course $A \rightarrow (\forall x)B$ (which is : $0 = 0 \rightarrow (\forall x)(x = 0)$) is obviously false in $\mathbb N$.
Now we have to apply 2.6.1 Metatheorem. (Deduction Theorem) : 

if $\Gamma \cup \{ A \} \vdash B$, then also $\Gamma \vdash A \rightarrow B$ [page 81]

and 8.2.3 Metatheorem. (Soundness in First-Order Logic) : 

if $\vdash A$, then $\vDash A$.

If our formula would be a "sound" rule, we would have (by the Deduction Theorem, with $\Gamma = \emptyset$) that :

$\vdash (A \rightarrow B) \rightarrow (A \rightarrow (\forall x)B)$.

Thus, by Soundness :

$\vDash (A \rightarrow B) \rightarrow (A \rightarrow (\forall x)B)$.

But we have found a counter-example, i.e. an interpreation where the formula is false; thus, it cannot be universally valid (that is: true in all interpretations - see page 201).
