Zbierski "Models for Higher Order Arithmetics" (BULL. DE L'ACAD. POLONAISE DES SCIENCES Serie des sciences math., astr. et phys. - Vol. XIX, No. 7, 1971) defines ZF$\_n$ as ZFC with the power set axiom limited to $n$ successive power sets starting from the natural numbers $\mathbb{N}$. Note this includes the axiom of choice. He proves ZF$\_n$ is a conservative extension of $n+2$ order arithmetic with the axiom scheme of choice.. He notes Gandy had proved already ZF$\_2$ is not conservative over $4$th order arithmetic without the axiom scheme of choice.


But what is known about the case where the axiom of choice is also removed from ZF$\_2$? 


Sochor "Constructibility in higher order arithmetics" (Arch. Math. Logic (1993) 32:381-389) shows that $n$-th order arithmetic without choice can interpret $n$-th order arithmetic with choice.though obviously it is not a conservative extension. So the issue is not equiconsistency. It is conservativity.



Higher order arithmetic and fragments of ZFC
The usual way to prove something like this is to take a model of $(n+1)$-st order arithmetic $\mathcal{A}$ and interpret a model of set theory $\mathcal{V}$ in $\mathcal{A}$ such a the usual interpretation of $(n+1)$-st order arithmetic in $\mathcal{V}$ leads to a model isomorphic to $\mathcal{A}$. There are a few recipes to do this, using graph, trees and other convenient encodings of sets. See the discussion [here](http://ncatlab.org/nlab/show/pure+set), I will use wellfounded accessible pointed graphs (WAPGs), though Zbierski probably uses trees (I don't know since I haven't found the paper).


The model $\mathcal{A}$ of $(n+1)$-st order arithmetic has sorts $N = S\_0,S\_1,\ldots,S\_n$, where $N$ is the number sort and each $S\_{i+1}$ consists of subsets of the previous sort $S\_i$. The number sort $N$ has the usual arithmetic structure with full induction for the implied language. Each sort $S\_i$ ($i \geq 1$) has full comprehension within the implied language. I'll refrain from making choice assumptions to see how far things go. Each sort has a definable pairing function, so we can talk about tuples, relations and functions. In particular, we can talk about WAPGs coded in $S\_n$.


The model $\mathcal{V}$ is obtained by collecting all WAPGs in the top sort $S\_n$. Equality $G \equiv H$ between two WAPGs $G$ and $H$ is interpreted as the existence of a bisimulation between $G$ and $H$, and membership $G \in H$ is interpreted as the existence of a bisimulation between $G$ and an immediate subgraph $H/x$ of $H$. After quotienting by $\equiv$ if desired, the result is an interpretation of the language of set theory which is automatically extensional and wellfounded (in the internal sense). Furthermore, since WAPGs coded in $S\_n$ and bisimulations between them are definable in $(n+1)$-th order arithmetic, every formula $\phi$ in the language of set theory has a translation $\phi^V$ in the language of $(n+1)$-th order arithmetic such that $\mathcal{V} \vDash \phi \iff \mathcal{A} \vDash \phi^V$.


Furthermore, we can define canonical WAPGs for each sort $\omega,\mathcal{P}(\omega),\ldots,\mathcal{P}^{n-1}(\omega)$. Specifically, the WAPG for $\omega$ is the graph of the ordering relation on the sort $N$; the WAPG for $\mathcal{P}(\omega)$ is built on top of that one by adding a node for each $X \in \mathcal{S}\_1$ and linking it to its elements; etc. Using the fact that $\mathcal{A}$ has full comprehension and the translation, we see that each $\mathcal{P}^{i+1}(\omega)$ really is the powerset of the previous one as understood in $\mathcal{V}$. So the natural interpretation of the arithmetical sorts $\omega,\mathcal{P}(\omega),\ldots,\mathcal{P}^{n-1}(\omega)$ in $\mathcal{V}$ are canonically isomorphic to the original sorts $N,S\_1,\ldots,S\_{n-1}$ from $\mathcal{A}$. The top sort $\mathcal{P}^n(\omega)$ is not necessarily a set in $\mathcal{V}$ but it is a definable class. Again, full comprehension can be used to argue that $\mathcal{P}^n(\omega)$ is canonically isomorphic to the original top sort $S\_n$.


So far, we haven't needed any choice; I only used full comprehension in $\mathcal{A}$. It remains to check which axioms $\mathcal{V}$ satisfies. Many are straightforward to check. We already checked extensionality, foundation and infinity. Pairing and union correspond to simple combinatorial manipulations of WAPGs. Comprehension follows from comprehension in $\mathcal{A}$. Replacement/collection are problematic without further assumptions about $\mathcal{A}$. There is a good reason for that since we have just shown one half of:


**Theorem.** *The theory $\mathrm{Z}^-\_{n-1}$ (extensionality, foundation, pairing, union, comprehension and the existence of $\omega,\mathcal{P}(\omega),\ldots,\mathcal{P}^{n-1}(\omega)$) is a conservative extension of $(n+1)$-th order arithmetic (without choice).*


The missing half is the straightforward proof that $\mathrm{Z}^-\_{n-1}$ is an extension of $(n+1)$-th order arithmetic. In other words, that the interpretation $N = \omega, S\_1 = \mathcal{P}(\omega), \ldots, S\_n = \mathcal{P}^n(\omega)$ always yields a model of $(n+1)$-th order arithmetic.




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Here is a reason why some choice principles might be necessary. For simplicity, I will assume we are working in $2$-nd order arithmetic. I will show that if collection holds in $\mathcal{V}$ then countable choice holds in $\mathcal{A}$. Say $\phi(n,x)$ is a formula in $2$-nd order arithmetic that relates each natural number $n$ with a set $x$ of natural numbers. We can translate $\phi(n,x)$ to a synonymous formula $\psi(n,x)$ in $\mathcal{V}$ such that $(\forall n \in \omega)(\exists x \subseteq \omega)\psi(n,x)$. Every set in $\mathcal{V}$ is countable, by construction, so if there is a set $b \in \mathcal{V}$ such that $(\forall n \in \omega)(\exists x \in b)\psi(n,x)$, then back in $\mathcal{A}$ there is a countable sequence $\langle x\_n \rangle$ coded in $S\_1$ such that $\phi(n,x\_n)$ holds for every $n$.


In a similar fashion, for any $n$, if $S\_{n-1}$ is wellorderable in a model $\mathcal{A}$ of $(n+1)$-st order arithmetic, then collection in $\mathcal{V}$ gives choice for definable $S\_{n-1}$-indexed families in $\mathcal{A}$.


