First off, it's although the general satisfaction relation (for both classes and sets) is not immediately clear formalizable in ZFC, the satisfaction relation for set models is formalizable. That is, there is an $\mathcal{L}$-formula $\mathrm{Sat}$ such that $\mathrm{Sat}(M,E,\ulcorner \varphi \urcorner,x_1,\dots ,x_n) \leftrightarrow (M,E)\vDash \varphi (x_1, \dots , x_n)$. Just as you could unpack the collection definition of definable subsets what it means for a real function to be continuous at a point and express it in ZFC, you can unpack the definition of what it means for a set exists, because unless being definable is itself model to satisfy a first-order definable propertyformula. That said, we can't apply this would typically be quite tedious to write down, and so in particular you probably wouldn't want to prove the $L_{\alpha+1}$ exist by applying Separation to the power set of $L_{\alpha}$, although you could.As you'll see in Jech
Rather, there's another the typical approach is to defining the find an alternative yet provably equivalent way of characterizing $L_{\alpha +1}$, \mathrm{Def}(L_{\alpha}) = L_{\alpha+1}$in terms of something other than definable subsets, and then prove that is the thing satisfying this other characterization exists. The approach you'll see in terms Jech uses the notion of closures under those 10 Godel operations, cf. This formulation (in terms of closure Corollary 13.8 and the preceding lemmas and theorems. Kunen also presents an alternative approach which also has to do with closing off under Godel certain operations) is formalizable, and provably equivalent I feel his approach is better motivated, so I recommend checking out Chapter VI, Section 1 in Kunen's "Set Theory: An Introduction to Independence Proofs." Let me add parenthetically that yes, you can deal with the original formulation$\exists a_1, \dots a_n \in L_{\alpha}$so that you do get a formula of ZFC. One way is to say$\exists f \exists n \exists r (definable subsets)f$is a function,$n = \mathrm{dom}(f)$,$n$is natural,$r = \mathrm{ran}(f)$,$r \subset L_{\alpha})$. 1 As you point out, it's not immediately clear that the collection of definable subsets of a set exists, because unless being definable is itself a first-order definable property, we can't apply Separation. As you'll see in Jech, there's another approach to defining the$L_{\alpha +1}\$, and that is in terms of closures under those 10 Godel operations. This formulation (in terms of closure under Godel operations) is formalizable, and provably equivalent to the original formulation (definable subsets).