Question on the non-existence of a satisfaction formula in $\mathbb{L}$

Question

I know this may sound trivial, but I'm having trouble figuring out what the problem is. Let $\mathbb{L}$ be the class of constructible sets. We know that $(\mathbb{L}_{\omega + \omega}, \in)$ is a countable structure and that $\omega \in \mathbb{L}_{\omega + \omega}$. So we can build complex things involving $\omega$ and therefore code a lot of things inside $(\mathbb{L}_{\omega + \omega}, \in)$. Since every element of $\mathbb{L}_{\omega + \omega}$ can be obtained from some $\phi$ and some finite number of parameters, why can't we describe the whole construction of $\mathbb{L}_{\omega + \omega}$ inside $\mathbb{L}_{\omega + \omega}$ (at a finite level $\mathbb{L}_{\omega+n}$) resulting in a definition of a formula $\mathrm{Sat}$ that defines satisfiability in $\mathbb{L}_{\omega + \omega}$? Parameter $u$ from higher levels can be reduced to level $\mathbb{L}_{\omega}$, since $u = \{ x \in \mathbb{L}_{\omega+n} : \phi(x,\bar{v})\}$, where $\mathbb{L}_{\omega + n}$ and every $\bar{u}$ can be substituted by their defining formulas and finite number of parameters from lower levels (finitely many levels down). $\mathbb{L}_{\omega}$ can be defined using such formula also, so everything in $\mathbb{L}_{\omega + \omega}$ can be defined using parameters from $\mathbb{L}_{\omega}$. The same question holds for any structure though.

Answer 1

You can't even construct $M$ in $L_{\omega+\omega}$ elementary equivalent to $L_{\omega+\omega}$ because of Tarski's undefinability theorem:

We have that the set of hereditaly finite sets, $\mathsf {HF}$, is a subset of $L_{\omega+\omega}$. Let $T$ be the set of all sentences that $L_{\omega+\omega}$ satisfies. Now suppose there is some $M\in L_{\omega+\omega}$ such that $(M,\in)$ and $(L_{\omega+\omega},\in)$ are elementary equivalent, then as $\mathsf{HF}\subseteq L_{\omega+\omega}$ we can define in $L_{\omega+\omega}$ whether for any formula $\varphi(\bar x)$ and any $\bar a\in M$, $M\models\varphi(\bar a)$, thus as $M\in L_{\omega+n}$ for some $n<\omega$, we have that $T\in L_{\omega+n+1}$, contradicting Tarski's undefinability theorem.