In Enderton's "A Mathematical Introduction to Logic" page 112 it gives as an axiom of First Order Logic :
$$ \forall x \; (\alpha (x) \implies \beta (x)) \implies ( \forall x \alpha (x) \implies \forall x \beta (x) ) \tag{1}$$
This is in addition to :
$$ \forall x \; \alpha(x) \implies \alpha^x _ t \tag{2}$$
with t substitutable for x in $\alpha$ and
$$ \alpha \implies \forall x \; \alpha(x) \tag{3}$$
where x is not free in $\alpha$.
My question is : Why is Axiom (1) above needed given axioms (2) and (3) ?
I ask this because it looks like Axiom (2) can be used to show, for any integer n :
$$ \forall x \; (\alpha (x) \implies \beta (x)) \implies ( \forall x \; \alpha (x) \implies \beta (t_1) \land \beta (t_2) ... \land \beta (t_n) ) \tag{1a}$$
however deductions and expressions are finite in first order logic, so $\beta (t_1) \land \beta (t_2) ... \land \beta (t_n) $ can not be replaced by $\forall x \beta (x)$ using only Axiom (2). Axiom (3) can't be used either. So Enderton's Axiom (1) above is used to "plug this infinity gap" ?
