$\Pi^1\_{\infty}\text{-}\mathsf{CA}\_0$ proves existence of models of ATR$\_0$. But I think it does not imply ATR$\_0$, because Axiom Beta is a kind of replacement axiom. Is that right?



Does $\Pi^1_{\infty}$ comprehension imply ATR$_0$?
Yes, in fact $\Pi^1\_1$-CA0 suffices to prove ATR0. The simplest way to see this is that ATR0 is equivalent to $\Sigma^1\_1$-separation: if $\phi(n)$ and $\psi(n)$ are $\Sigma^1\_1$ formulas then
$$\forall n (\lnot \phi(n) \lor \lnot\psi(n)) \rightarrow \exists C \forall n ((\phi(n) \rightarrow n \in C) \land (\psi(n) \rightarrow n \notin C)).$$
Assuming $\Pi^1\_1$-CA0 one can simply take $C = \lbrace n : \lnot\psi(n)\rbrace$, for example, to satisfy the conclusion. Details can be found in Simpson's *Subsystems of Second-Order Arithmetic*.


