Mitchell, Steel. FSIT. Lemma 2.8: Is $k$-solidity actually needed?

Question

Consider the following result (which is Lemma 2.8 in Mitchell and Steel's paper on Fine Structure and Iteration Trees):

Lemma 2.8 Let $\pi \colon \mathcal{H} \to \mathcal{M}$ be generalized $r \Sigma_{k}$ elementary, where $\mathcal M$ is a ppm (not of type III) and $1 \leq k < \omega$. Suppose that $\rho_{k}^{\mathcal{M}} \subseteq \mathcal{H}$ and $\pi \restriction \rho_{k}^{\mathcal{M}} = \operatorname{id}$. Suppose also that $\pi(r)$ is the $k$th standard parameter of $(\mathcal{M}, \pi(q))$ and that $\pi(r)$ is $k$-solid and $k$-universal over $(\mathcal{M}, \pi(q))$. Then

1. $\rho_{k}^{\mathcal{H}} = \rho_{k}^{\mathcal{M}}$,
2. $r$ is the $k$th standard parameter of $(\mathcal{H},q)$ and
3. $r$ is $k$-universal over $(\mathcal{H}, q)$.

The proofs of items $1.$ and $2.$ are included in this paper and in both cases it seems that the $k$-solidity of $\pi(r)$ over $(\mathcal{M}, \pi(q))$ isn't actually needed. Hence I decided to prove item $3.$ to see how $k$-solidity comes into play. However, if my argument is correct, it doesn't rely on $k$-solidity either.

Here is my proof of item $3.$:

Proof (of $3.$). Let $A \in \mathcal{H}$ be such that $A \subseteq \rho_{k}^{\mathcal{H}}$. Then $\pi(A) \cap \rho_{k}^{\mathcal{M}} \in \mathcal M$. By the $k$-universality of $\pi(r)$ over $(\mathcal{M}, \pi(q))$ there is hence some generalized Skolem term $\tau \in S_{\kappa}$ and some $\vec{\alpha} \in ^{< \omega} \rho_{k}^{\mathcal{M}}$ s.t. $$ \pi(A) \cap \rho_{k}^{\mathcal{M}} = \tau^{\mathcal{M}}[\vec{\alpha}, \pi(r), \pi(q)] \cap \rho_{k}^{\mathcal{M}}. $$ Let $B := \tau^{\mathcal{H}}[\vec{\alpha},r,q] \cap \rho_{k}^{\mathcal{H}}$. Combining the fact that $\pi$ is generalized $r \Sigma_{k}$-elementary, $\rho_{k}^{\mathcal{H}}= \rho_{k}^{\mathcal{M}} \subseteq \mathcal{H}$ and $\pi \restriction \rho_{k}^{\mathcal M} = \operatorname{id}$ we have that \begin{align*} \mathcal{H} \models A \cap \rho_{k}^{\mathcal{H}} = B &\iff \mathcal{H} \models A \cap \rho_{k}^{\mathcal{H}} = \tau^{\mathcal H}[\vec{\alpha}, r,q] \cap \rho_{k}^{\mathcal{H}} \\ &\iff \mathcal{M} \models \pi(A) \cap \rho_{k}^{\mathcal{M}} = \tau^{\mathcal{M}}[\vec{\alpha},\pi(r),\pi(q)] \cap \rho_{k}^{\mathcal{M}}. \end{align*} Since the last line is true, it follows that $A \cap \rho_{k}^{\mathcal{H}} =\tau^{\mathcal{H}}[\vec{\alpha},r,q] \cap \rho_{k}^{\mathcal{H}}$. Thus $r$ is indeed $k$-universal over $(\mathcal{H}, q)$. Q.E.D.

Question: Did I miss something and this result actually relies on the $k$-solidity of $\pi(r)$ over $(\mathcal{M},\pi(q))$ or can this assumption be dropped?

If $k$-solidity is needed, I'd like to understand where exactly in the proof it is used and ideally I'd like to see an example in which Lemma 2.8 fails without $k$-solidity.

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PS: I am aware that this question isn't exactly 'ongoing research' and I strongly considered posting it over at MSE. However, since the group of people able to answer this question is more likely to be encountered here and since a somewhat similar question has been asked and well-received here, I decided to go with mathoverflow.

Answer 1

In my question I already verified that (assuming 1. and 2.) item 3. is provable without assuming that $\pi(r)$ is $k$-solid over $(\mathcal{M},q)$. To see that we can actually drop $k$-solidity in this lemma, it hence suffices to see that 1. and 2. also don't require that $k$-solidity of $\pi(r)$.

If $\rho_{k}^{\mathcal{M}} = \operatorname{Ord}^{\mathcal{M}}$, then $\pi = \operatorname{id}$ and the lemma trivially holds. Thus assume that $\rho_{k}^{\mathcal{M}} < \operatorname{Ord}^{\mathcal{M}}$.

1. Let $\alpha \leq \rho_{k}^{\mathcal M}$. Since $\pi \restriction \rho_{\kappa}^{\mathcal{M}} = \operatorname{id}$ and since $\pi$ is generalized $r \Sigma_{k}$-elementary, we have - up to a slight abuse of notation - \begin{align*} \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{ s \}) &= \{(\phi, \vec{a}, s) \mid \vec{a} \in ^{< \omega}{\alpha} \wedge \mathcal{H} \models \phi[\vec{a}, s] \} \\ &= \{(\phi, \vec{a}, s) \mid \vec{a} \in ^{< \omega}{\alpha} \wedge \mathcal{M} \models \phi[\vec{a}, \pi(s)] \} \end{align*} In particular, for any $\phi \in r \Sigma_{k}$ and any $\vec{a} \in ^{< \omega}{\alpha}$, we have $$ (\phi, \vec{a}, s) \in \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{s \}) \iff (\phi, \vec{a}, \pi(s)) \in \operatorname{Th}_{k}^{\mathcal{M}}(\alpha \cup \{ \pi(s) \}). $$ By enlarging $\alpha$, if necessary, we may assume that $\alpha$ is primitive recursively closed and hence uniformly code \begin{align*} \{(\phi, \vec{a}) \mid (\phi, \vec{a}, s) \in \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{ s \}) \} \\ = \{(\phi, \vec{a}) \mid (\phi, \vec{a}, \pi(s)) \in \operatorname{Th}_{k}^{\mathcal{M}}(\alpha \cup \{ \pi(s) \}) \} \end{align*} as a subset $A \subseteq \alpha$. Since $\alpha < \rho_{k}^{\mathcal{M}}$, we have $\operatorname{Th}_{k}^{\mathcal{M}}(\alpha \cup \{ \pi(s) \}) \in \mathcal{M}$ and hence $A \in \mathcal{M}$. By the strong acceptability of $\mathcal{M}$ - observing that $\rho_{k}^{\mathcal{M}}$ is an $\mathcal{M}$-cardinal - this yields $$ A \in \mathcal{J}_{\rho_{k}^{\mathcal{M}}}^{\mathcal M} = \left( H_{\rho_{k}^{\mathcal{M}}} \right)^{\mathcal{M}} \overset{\pi \restriction \rho_{k}^{\mathcal{M}} = \operatorname{id}}{=} \left( H_{\rho_{k}^{\mathcal{M}}}\right)^{\mathcal{H}} \subseteq \mathcal{H}. $$ Therefore $$ \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{ s \}) = \{ (\phi, \vec{a}, s) \mid \langle \phi, \vec{a} \rangle \in A \} \in \mathcal{H} $$ and $\rho_{k}^{\mathcal{M}} \leq \rho_{k}^{\mathcal{H}}$.

   On the other hand, suppose that $\rho_{k}^{\mathcal{M}} < \rho_{k}^{\mathcal{H}}$. Then $\operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{M}} \cup \{(r,q)\}) \in \mathcal{H}$. Let $A \subseteq \rho_{k}^{\mathcal{M}}$, $A \in \mathcal{H}$ be a uniform code for $$ \{ (\phi, \vec{a}) \mid (\phi, \vec{a}, (r,q)) \in \operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{M}} \cup \{(r,q)\})\}. $$ Then $A = \pi(A) \cap \rho_{k}^{\mathcal{M}} \in \mathcal{M}$ witnesses (as above) that $\operatorname{Th}_{k}^{\mathcal{M}}(\rho_{k}^{\mathcal{M}} \cup \{ \pi(r), \pi(q)\}) \in \mathcal{M}$. This contradicts the fact that $\pi(r)$ is the $k$th standard parameter of $(\mathcal{M},\pi(q))$!

2. The proof above also shows that $\operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{H}} \cup \{ (r,q)\}) \not \in \mathcal{H}$. Hence it suffices to show that for all $s <_{\operatorname{lex}} r$ $$ \operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{H}} \cup \{ (s,q)\}) \in \mathcal{H}. $$ So, fix $s <_{\operatorname{lex}} r$. Then $\pi(s)<_{\operatorname{lex}} \pi(r)$ and hence $$ \operatorname{Th}_{k}^{\mathcal{M}}(\rho_{k}^{\mathcal{H}} \cup \{ (\pi(s),\pi(q))\}) \in \mathcal{M}. $$ Let $A \subseteq \rho_{k}^{\mathcal{M}}$, $A \in \mathcal{M}$ be the code of this fact as above. Since $\pi(r)$ is $k$-universal over $(M, \pi(q))$ there is some $\tau \in \operatorname{Sk}_{k}$ and $\vec{a} \in ^{< \omega}{\rho_{k}^{\mathcal{M}}}$ such that $A = \tau^{\mathcal{M}}[\vec{a}, \pi(r), \pi(q)] \cap \rho_{k}^{\mathcal{M}}$. Let $$ B = \tau^{\mathcal{H}}[\vec{a}, \pi(r), \pi(q)] \cap \rho_{k}^{\mathcal{M}}. $$ Since $\pi$ is generalized $r \Sigma_{k}$-elementary we have, for all $\xi < \rho_{k}^{\mathcal{M}} = \rho_{k}^{\mathcal{H}}$ $$ \mathcal{M} \models \xi \in \tau^{\mathcal{M}}[\vec{a},\pi(r),\pi(q)] \iff \mathcal{H} \models \xi \in \tau^{\mathcal{H}}[\vec{a},r,q]. $$ Thus $B = A \in \mathcal{H}$ witnesses that $\operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{H}} \cup \{(s,q)\}) \in \mathcal{H}$ and hence the $<_{\operatorname{lex}}$-minimality of $r$.