Iterated projecta and their respective iterated Skolem functions So I have been trying to study some fine structure from M. Zeman and R. Schindler's article in the handbook, and I have run into a few questions regarding the iterated Skolem functions for iterated projecta.
My first question might seem a little bit nitpicky, but it actually plays a considerable role in my misunderstandings. So here it goes: $\bf{(I)}$ on page $633$, above lemma $5.4$ the functions $h^{n+1}_M$ are defined. And the domain of each $h^{n+1}_M$ is $$\omega^{<\omega} \times |M^{n+1,p}|^{<\omega},$$ and we also have that $\mbox{On}\cap M^{n+1,p} = \rho_{n+1}(M)$. On the other hand, in lemma $5.4$, it is claimed that $$M = h^{n+1}_M"(\rho_{n+1}(M)\cup \{p\}),$$ for some $p\in R^{n+1}_M$. But then $p$ is a finite seq. of finite seq.s of ordinals above $\rho_{n+1}(M)$ and so is not in the domain of $h^{n+1}_M$!
Am I right? Or am I missing something? As a note I can say that I do understand the recursive phenomenon described a few lines above this definition, but there, we can't feed $p$ directly to each $h$ function and we break it up to each individual $p(i)$ and use it at the $i$-th step, but I don't see how we can do it formally with the given definition.
My next $2$ questions involve lemma $5.5$:
$\bf{(II)}$ The statement of lemma $5.5$ is odd. Meaning that I think if we take $M = (J_\alpha^B, D)$, an acceptable $J$-structure, then this lemma seems to imply that if $R^n_M \neq \emptyset$, then $\rho_0(M) = \rho_1(M) = \dots = \rho_n(M) = \alpha$. Meaning that $M$ is actually very rigid up to $n$, which seems in some sense very counter-intuitive, because then the only possibility for $p \in R^n_M$ is for it to be $p = \emptyset$, which seems odd. To see the above fact we show it for $\rho_1(M)$ and for $m\le n$, the result follows by iterating the proof for case $m =1$ by induction. Now if $\rho_1(M) < \alpha$, let $T \in {\bf \Sigma}_1^M\cap P(\rho_1(M))$ such that $T \not\in M$. But since $T \in {\bf \Sigma}_1^M \subset {\bf \Sigma}_\omega^{M^{1, p|1}}$, by $5.5$, and since $P(J_{\rho_1(M)})\cap \mbox{rud}_{B, A^{1,p|1}_M}(J_{\rho_1(M)}\cup\{J_{\rho_1(M)}\}) = P(J_{\rho_1(M)}) \cap {\bf \Sigma}_\omega^{M^{1, p|1}}$, then $T\in M$ by the fact that $\rho_1 < \alpha$ and so $\mbox{rud}_{B, A^{1,p|1}_M}(J_{\rho_1(M)}\cup\{J_{\rho_1(M)}\}) \subset M$. Which is a contradiction by the choice of $T$. So $\rho_1(M) = \alpha$.
Is my above reasoning correct? If not, I would appreciate it if someone would kindly point out my mistake.
$\bf{(III)}$ I also have trouble with the proof of lemma $5.5$. The main one is the last equivalence in the proof. My main concern is that when using $h^n_M$, I don't know how we handle the $p\in R^n_M$ as a parameter to generate the members we want, and also the abuse of notation of writing $h^n_M$ with only one input rather than the tuple it needs, makes the proof hard for me to understand. Can someone kindly elaborate more on the details of this proof?
 A: It seems to me that the definition of $h^{n+1}_M$ is slightly inaccurate in this article. I will write $h^{n+1, p}_M$ to make the dependence on the parameter $p$ clear. $p$ should be fed into the iterated $\Sigma_1$-Skolem functions at the apropriate places: Instead of
$$h^{n+1}_M(\langle\vec i, i_0,\dots i_k\rangle, \langle \vec x_{i_0},\dots, \vec x_{i_k}\rangle) $$
$$=h^n_M(\vec i, \langle h_{M^{n, p\upharpoonright n}}(i_0, \vec x_{i_0}),\dots, h_{M^{n, p\upharpoonright n}}(i_k, \vec x_{i_k})\rangle)$$
it should be
$$h^{n+1, p}_M(\langle\vec i, i_0,\dots i_k\rangle, \langle \vec x_{i_0},\dots, \vec x_{i_k}\rangle) $$
$$=h^{n, p\upharpoonright n}_M(\vec i, \langle h_{M^{n, p\upharpoonright n}}(i_0, \vec x_{i_0}^\frown p(n)),\dots, h_{M^{n, p\upharpoonright n}}(i_k, \vec x_{i_k}^\frown p(n))\rangle)$$
(with $h^{0, \emptyset}_M=h_M$). Both authors published books in which the function $h^{n, p}_M$ is, modulo some technicalities, defined in this way (Ralf Schindler's  "Set Theory" p.252 and Martin Zeman's "Inner Modles and Large Cardinals" p.29, note that the function is called $\tilde h^{n}_M$ in the latter).
One can also think of $p$ as an additional argument of the function $h^n_M$, i.e. $h^n_M(\vec i, \vec x, p)=h^{n, p}_M(\vec i, \vec x)$. This is in fact a feature of $\tilde h^n_M$ in Zeman's book. This also explains
$$M=h^{n+1}_M"(\rho_{n+1}(M)\cup\{p\})$$
by slight abuse of notation.
Regarding $\mathbf{(II)}$: It is definately possible that $R^n_M\neq\emptyset$ and $\rho_n(M)< \alpha$. For example, all levels of the $J$-hierachy are acceptable and sound so that $R^n_{J_\alpha}=P^n_{J_\alpha}\neq\emptyset$ for all $n$ and there are plenty of $\alpha$ for which $\rho_1(J_\alpha)<\alpha$.
The problem in your argument is that
$$\mathrm{rud}_{B, A^{1, p\upharpoonright 1}_M}(J^B_{\rho_1(M)}\cup\{J^B_{\rho_1(M)}\})\subseteq M$$
cannot be justified. This is true if the additional $A^{1, p\upharpoonright 1}_M$ is dropped from the subscript of $\mathrm{rud}$, but more cannot be said. Note that $A^{1, p\upharpoonright 1}_M\notin M$ but is an element of the rudimentary closure on the left hand side!
On $\mathbf{(III)}$, I believe the fog should settle when every instance of $h^n_M(z')$ in the proof of Lemma 5.5 is replaced by $h^{n, p}_M(z')$ with the definition from above. That this works is a consequence of $M=h^{n, p}_M"M^{n, p}$.
