Reference Requestion: what's the proof of elementary equivalence of hierarchies raised over sets of quine atoms using Ehrenfeucht-Fraisse games? In a prior posting Noah Schweber had presented this claim:


$(*)\quad$ Suppose $A,B$ are infinite sets of Quine atoms. Then for each ordinal $\alpha$ we have $$L_{\alpha,A}\equiv L_{\alpha,B}.$$ $(*)$ can be proved via Ehrenfeucht-Fraisse games.


Where $L_{\alpha,A}$ means $L$ hierarchy on set $A$, built as:
$L_{0,A}=A$,
$L_{\lambda,A}=\bigcup_{\alpha<\lambda} L_{\alpha,A}$ for $\lambda$ limit, and
$L_{\alpha+1,A}=L_{\alpha,A} \cup\mathcal{P}_{def}(L_{\alpha,A})$.
The argument posed makes no mention of difference in cardinality between $A$ and $B$, so it's applicable when $A$ is countable and $B$ is not.

My question: what is the proof of that result? Or if it's long then can one point me to a reference of that result? preferably one that is reachable online openly.

 A: In retrospect, EF-games are overkill here; we can just use Lowenheim-Skolem + a bit of constructibility theory.
First, note that any bijection $b:A_0\rightarrow A_1$ extends to a (unique) isomorphism $\hat{b}: L_{A_0,\alpha}\cong L_{A_1,\alpha}$.
Next, let $B$ be an uncountable "starting set" and $\alpha$ any countable ordinal. By downward Lowenheim-Skolem, there is a countable elementary $\mathcal{X}\subseteq L_{B,\alpha}$ with $\alpha\subseteq \mathcal{X}$. Let $A=B\cap\mathcal{X}$; clearly $A$ is countable, and by the usual absoluteness-of-$L$-hierarchy argument we get $\mathcal{X}=L_{A,\alpha}$.
(Specifically, we argue by induction on $\beta<\alpha$ that the thing $\mathcal{X}$ thinks is $L_{A,\beta}$ actually is $L_{A,\beta}$. Since $\mathcal{X}$ thinks every set is in some $L_{A,\beta}$, this gives the desired result. Note that we're crucially using here the fact that $\mathcal{X}\cap Ord$ is downwards-closed.)
We can put these together as follows. Let $A,B$ be arbitrary infinite sets of Quine atoms and $\alpha$ a countable ordinal. By the previous paragraph, we get countably infinite $A_0\subseteq A,B_0\subseteq B$ such that $$L_{A_0,\alpha}\preccurlyeq L_{A,\alpha}\quad\mbox{and}\quad L_{B_0,\alpha}\preccurlyeq L_{B,\alpha}.$$ But since there is a bijection between $A_0$ and $B_0$ we get an isomorphism between the corresponding $L$-structures, and putting this all together we have $$L_{A,\alpha}\equiv L_{A_0,\alpha}\cong L_{B_0,\alpha}\equiv L_{B,\alpha}.$$

