The definability of the extender sequence $\mathbb{E}$ from $\mathbb{E}{\upharpoonright}\aleph_1$ in $L[\mathbb{E}]$

TL;DR

This paper demonstrates that in certain models of set theory, the entire extender sequence can be defined from its restriction to the first uncountable ordinal, leading to insights about the structure of mice and models where V=HOD.

Contribution

It proves the definability of the extender sequence from initial segments in models with short extenders and establishes conditions under which the universe is equal to HOD.

Findings

The extender sequence is definable from its restriction to leph_1 in models with short extenders.

Models with a Woodin limit of Woodin cardinals satisfy V=HOD.

Various local and generic extension versions of the definability result are established.

Abstract

Let $M$ be a short extender mouse. We prove that if $E\in M$ and $M$ satisfies "$E$ is a countably complete short extender whose support is a cardinal $\theta$ and $\mathcal{H}_\theta\subseteq\mathrm{Ult}(V,E)$", then $E$ is in the extender sequence $\mathbb{E}^M$ of $M$. We also prove other related facts, and use them to establish that if $\kappa$ is an uncountable cardinal of $M$ and $\kappa^{+M}$ exists in $M$ then $(\mathcal{H}_{\kappa^+})^M$ satisfies the Axiom of Global Choice. We prove that if $M$ satisfies the Power Set Axiom then $\mathbb{E}^M$ is definable over the universe of $M$ from the parameter $X=\mathbb{E}^M\upharpoonright\aleph_1^M$, and $M$ satisfies "every set is $\mathrm{OD}_{\{X\}}$". We also prove various local versions of this fact in which $M$ has a largest cardinal, and a version for generic extensions of $M$. As a consequence, for example, the minimal proper class mouse with a Woodin limit of Woodin cardinals models "$V=\mathrm{HOD}$". This adapts to many other similar examples. We also describe a simplified approach to Mitchell-Steel fine structure, which does away with the parameters $u_n$.

Equations

$${\mathbb{E}}^M\text{ is }{\Delta }_2^{\lfloor M\rfloor }(\{\text{mathbbm}m\})\text{-definable}.$$

$$\mathrm{Ult}(M,E)|{\nu }_E=M|{\nu }_E.$$

$$w_{k+1}^H(q,\alpha ){=}_{\mathrm{def}}{\mathrm{Th}}_{k+1}^H(\alpha \cup \{q,{\vec{p}}_k^H\}).$$

$$w_{k+1}^H(q){=}_{\mathrm{def}}\{w_{k+1}^H(q\upharpoonright i,q_i){\}}_{i<\mathrm{lh}(q)}$$

$$w_{k+1}^H{=}_{\mathrm{def}}{w}_{k+1}^H(p_{k+1}^H).$$

$$\bar{H}={\mathrm{cHull}}_{k+1}^H(\theta \cup \{{\vec{p}}_k^H,q\}),$$

$$\gamma \notin {\mathrm{Hull}}_{k+1}^H(\gamma \cup \{{\vec{p}}_k^H,x,r\backslash \{\gamma \}\}).$$

$$q=(q_0,q_1,\dots ,q_{m-1}),$$

$$u_i,r\in {\mathrm{Hull}}_{k+1}^H({\gamma }_i\cup \{{\vec{p}}_k^H,q\upharpoonright i\}),$$

$$u_i,r\in H_i{=}_{\mathrm{def}}{\mathrm{Hull}}_{k+1}^H((\eta +1)\cup \{{\vec{p}}_k^H,q\upharpoonright i\})$$

$$\eta \notin {\mathrm{Hull}}_{k+1}^H(\eta \cup \{{\vec{p}}_k^H,q\upharpoonright i\}).$$

$$H_i⊊H_i^{\prime }={\mathrm{Hull}}_{k+1}^H({\gamma }_i\cup \{{\vec{p}}_k^H,q\upharpoonright (i+1)\}),$$

$$\rho >{\rho }_{k+1}^W=\gamma =\mathrm{lgcd}(W|\rho )$$

$$W={\mathrm{Ult}}_k(J,F).$$

$$\eta \notin {\mathrm{Hull}}_{k+1}^W((\gamma +1)\cup \{{\vec{p}}_k^W,p_{k+1}^W\backslash \rho \}),$$

$$H_{\gamma }={\mathrm{Hull}}_{k+1}^H(\gamma \cup \{{\vec{p}}_{k+1}^H\})$$

$$W_{\gamma }={\mathrm{Hull}}_{k+1}^{W_{\gamma }}(\kappa \cup \{\eta ,x\})$$

$$\mathrm{lh}(E_{\alpha }^{\mathcal{T}})\le \mathrm{lh}(E_{\beta }^{\mathcal{T}})\text{ for all }\alpha +1<\beta +1<\mathrm{lh}(\mathcal{T}),$$

$$\nu (E_{\alpha }^{\mathcal{T}})\le \nu (E_{\beta }^{\mathcal{T}})\text{ for all }\alpha +1<\beta +1<\mathrm{lh}(\mathcal{T}).$$

$${\psi }_{\pi }:\mathrm{Ult}(M|({\mu }^{+}{)}^M,F^M)\to \mathrm{Ult}(N|({\kappa }^{+}{)}^N,F^N)$$

$${\rho }_{m+1}^M=\omega <\kappa =\mathrm{cr}(E)<\tau ={\rho }_m^M={\rho }_{n+1}^M.$$

$$U={\mathrm{Ult}}_m(M,E)\text{ is wellfounded and }(m,{\omega }_1+1)\text{-iterable},$$

$$\bar{M}={\mathrm{cHull}}_m^M(\{t\})$$

$$\bar{U}={\mathrm{Ult}}_{m-1}(\bar{M},\bar{E}).$$

$$\mathfrak{P}=((\bar{M},m-1,\bar{\theta }),(\bar{U},m-1),\bar{\theta }).$$

$$M^{\prime }={\mathrm{cHull}}_m^M(\gamma \cup \{t^{\prime }\})◃M,$$

$${\pi }^{\prime }:\bar{M}\to M^{\prime }$$

$${\pi }^{\prime }\upharpoonright \bar{\theta }=\pi \upharpoonright \bar{\theta }=\psi \upharpoonright \bar{\theta }.$$

$${\pi }_{\alpha }:M_{\alpha }^{\mathcal{T}}\to M_{\alpha }^{\mathcal{U}}$$

$$\mathrm{cr}(i_{\beta \alpha }^{\mathcal{T}})<i_{0\beta }^{\mathcal{T}}(\bar{\gamma })\text{ for all }\beta \in (-1,\alpha {)}_{\mathcal{T}}$$

Full text

We prove that if $E\in M$ and $M\models$“$E$ is a countably complete short extender whose support is a cardinal $\theta$ and $\mathcal{H}_{\theta}\subseteq\mathrm{Ult}(V,E)$”, then $E$ is in the extender sequence $\mathbb{E}^{M}$ of $M$. We also prove other related facts, and use them to establish that if $\kappa$ is an uncountable cardinal of $M$ and $(\kappa^{+})^{M}$ exists in $M$ then $(\mathcal{H}_{\kappa^{+}})^{M}$ satisfies the Axiom of Global Choice.

We then prove that if $M$ satisfies the Power Set Axiom then $\mathbb{E}^{M}$ is definable over the universe of $M$ from the parameter $X=\mathbb{E}^{M}\!\upharpoonright\!\aleph_{1}^{M}$, and therefore $M$ satisfies “Every set is $\mathrm{OD}_{X}$”. We also prove various local versions of this fact in which $M$ has a largest cardinal, and a version for generic extensions of $M$.

As a consequence, for example, the minimal proper class mouse with a Woodin limit of Woodin cardinals models “$V=\mathrm{HOD}$”. This adapts to many other similar examples.

We also describe a simplified approach to Mitchell-Steel fine structure, which does away with the parameters $u_{n}$.

1 Introduction

Let $M$ be a mouse. Write $\mathbb{E}^{M}$ for the extender sequence of $M$, not including the active extender $F^{M}$ of $M$. Write $\left\lfloor M\right\rfloor$ for the universe of $M$. Write $\mathbbm{m}^{M}=M|\aleph_{1}^{M}$. Write $\mathsf{PS}$ for the Power Set Axiom.

We consider here the following questions:

1. –

Given $E\in M$ such that $M\models$“$E$ is an extender”, is $E\in\mathbb{E}^{M}$? 2. –

(Steel) Suppose $M\models\mathsf{ZFC}$. Does $M\models$“There is $X\subseteq\aleph_{1}$ such that $V=\mathrm{HOD}_{X}$”? 3. –

Is $\mathbb{E}^{M}$ definable over $\left\lfloor M\right\rfloor$, possibly from some (small) parameter?

The main theorem of the paper is the following, which answers Steel’s question above positively, in fact with $X=\mathbbm{m}^{M}$.

Theorem 1.0**.**

Let $M$ be a $(0,\omega_{1}+1)$-iterable premouse satisfying $\mathsf{PS}$ and $\mathbbm{m}=\mathbbm{m}^{M}$. Then

[TABLE]

Therefore if $\left\lfloor M\right\rfloor\models\mathsf{ZFC}$ then $\left\lfloor M\right\rfloor\models$“$V=\mathrm{HOD}_{\mathbbm{m}}$” and $M\models\mathsf{ZFC}$.111Note that by writing “$M\models\mathsf{ZFC}$” we mean the structure $(\left\lfloor M\right\rfloor,\in,\mathbb{E}^{M})$, and so the assumption that $\left\lfloor M\right\rfloor\models\mathsf{ZFC}$ does not trivially imply $M\models\mathsf{ZFC}$.

The first proof we give of this theorem, in §3, will actually yield a more general and local version, in which the mouse can have a largest cardinal, but in which case we must allow somewhat higher complexity in the definition of $\mathbb{E}^{M}$ from the parameter $\mathbbm{m}^{M}$. We will also give a variant proof in §4, which uses the same main idea, but is a little simpler, and is in some ways more general, but in other ways less. Related results have been known for some time. Kunen proved that $L[U]$ satisfies “$U$ is the unique normal measure”, and therefore satisfies “$V=\mathrm{HOD}$”. Recall that $M_{n}$ is the minimal proper class mouse with $n$ Woodin cardinals. Steel proved [4] that for $n\leq\omega$, $\mathbb{E}^{M_{n}}$ is definable over $\left\lfloor M_{n}\right\rfloor$ without parameters.222He in fact showed that $M_{n}$ is its own “core model” (this must be defined appropriately). The author proved similar results for larger, sufficiently self-iterable mice in [7] and [9]. The proofs of these earlier results depended on the mice in question being sufficiently self-iterable. But non-meek mice typically fail to have such self-iterability. To prove Theorem 1, we use a different approach, which avoids any use of self-iterability, and is more focused on condensation properties. We easily get the following corollary:

Corollary**.**

Let $M$ be a $(0,\omega_{1}+1)$-iterable premouse. Suppose $M$ satisfies $\mathsf{PS}+$“Every countable segment of me is $(\omega,\omega_{1}+1)$-iterable”. Then $\mathbb{E}^{M}$ is (lightface) $\Delta_{2}^{M}$-definable and $\left\lfloor M\right\rfloor\models$“Every set is $\mathrm{OD}$”.333Here and elsewhere, for premice modelling $\mathsf{PS}$, we say that $x\in\mathrm{OD}^{M}$ iff there is $\alpha<\mathrm{OR}^{M}$ such that $\{x\}$ is definable from ordinal parameters over $(\mathcal{H}_{\alpha})^{M}$.

A natural variant of Steel’s question mentioned above is whether the same holds but with $X\in\mathbb{R}^{M}$. We do not know the answer here in general, but in a separate paper [5], we will extend the results and methods here to answer the question affirmatively for tame mice. We will also establish some structural facts regarding $\mathrm{HOD}^{\left\lfloor M\right\rfloor}$ for arbitrary (short extender) mice $M$ modelling $\mathsf{PS}$.444These facts, however, leave the full analysis of $\mathrm{HOD}^{\left\lfloor M\right\rfloor}$ very much open, in particular in the case of $L[x]$ for a cone of reals $x$.

To prove the more local version of Theorem 1 above we will use certain extender maximality properties of $\mathbb{E}^{M}$, Theorems 1 and 1 below, which are refinements of results from [7]. Their proofs are very similar.

Theorem 1.0** (Steel, Schlutzenberg).**

Let $M$ be an $(0,\omega_{1}+1)$-iterable premouse. Let $E\in M$ be such that $M\models$“$E$ is a short, total, countably complete extender”, $\nu_{E}$ is a cardinal of $M$ and $\mathcal{H}_{\nu_{E}}^{M}\subseteq\mathrm{Ult}(M,E)$. Then (the trivial completion of) $E$ is in $\mathbb{E}^{M}$.

Remark 1.0**.**

Steel first proved that $E\in\mathbb{E}^{M}$ under the assumptions of 1 together with the added assumptions that $M\models\mathsf{PS}+$“$\nu_{E}$ is regular” and $E$ coheres $\mathbb{E}^{M}$ below $\nu_{E}$; that is,

[TABLE]

The author then generalized Steel’s proof to obtain 1.

Note that if $M\models$“$E$ is a normal measure”, then $\nu_{E}=(\mathrm{cr}(E)^{+})^{M}$, so the requirement that $\mathcal{H}_{\nu_{E}}^{M}\subseteq\mathrm{Ult}(M,E)$ holds automatically, and therefore $E\in\mathbb{E}^{M}$ (given $M$ is iterable).

Theorem 1.0**.**

Let $M$ be a $(0,\omega_{1}+1)$-iterable premouse. Let $E,R\in M$ and $\tau\in\mathrm{OR}^{M}$ be such that $\tau$ is a cardinal of $M$, $R$ is a premouse and $\rho_{\omega}^{R}=\tau$ and $M\models$“$E$ is a short extender, $\mathcal{H}_{\tau}\subseteq\mathrm{Ult}(M,E)$ and $R\triangleleft\mathrm{Ult}(M,E)$”. Then $R\triangleleft M$.

Slightly less general versions of 1 and 1 were obtained by the author in 2006. Prior to this, Woodin had conjectured that if $M$ is a mouse, $\kappa$ is uncountable in $M$ and $(\kappa^{+})^{M}$ exists, then $L(\mathcal{P}(\kappa)^{M})\models\mathsf{AC}$. Woodin’s conjecture follows immediately from the following corollary to the preceding theorems. Steel noticed that corollary follows from 1 combined with an argument of Woodin’s.555 The corollary appeared first in [7]. It can now be deduced trivially from 1. However, we will give its original proof (from [7]), as this constitutes a significant part of the proof of 1, and so it serves as a useful warm-up.

Corollary 1.0**.**

Let $M$ be a $(0,\omega_{1}+1)$-iterable premouse and $\kappa\in\mathrm{OR}^{M}$ be such that $M\models$“$\kappa$ is uncountable” and $(\kappa^{+})<\mathrm{OR}^{M}$. Then $M|(\kappa^{+})^{M}$ is definable from parameters over $\mathcal{H}_{(\kappa^{+})^{M}}^{M}$.

The extender maximality theorems proven here are refinements of results obtained by the author in [7]. The inductive condensation stack argument in §3 was obtained in 2015, and presented by the author in the Oberseminar für Mengenlehre at the Institut für Mathematische Logik und Grundlagenforschung, Universität Münster, in Spring 2016, and also at the 1st Irvine conference on descriptive inner model theory and hod mice, in July 2016. The direct condensation stack argument in §4 was obtained in 2019.

In §5 we also describe a simplification to Mitchell-Steel fine structure, making do without the parameters $u_{n}$. This simplification was observed by the author in 2012/13, while visiting John Steel at UC Berkeley. One could just use the standard fine structure, but it simplifies certain definitions, and we will officially make use of it here and in the future.

1.1 Conventions and Notation

Most non-standard conventions, in particular in connection with premice, extenders, fine structure, iteration trees and phalanxes, are as in [9, §1.1]. However, there are two main differences. Firstly, in this paper we use the term premouse slightly differently: as in [9] we use Mitchell-Steel indexing; however, we allow extenders of superstrong type in the extender sequence $\mathbb{E}_{+}^{M}$ of a premouse $M$. There are some small changes that this introduces, as explained in [8, Remark 2.44***]. Secondly, we adopt a simplified version of Mitchell-Steel fine structure, explained in §5, which avoids the parameters $u_{n}$. By 5, this change actually has no impact on the fine structural notions such as $p_{k}$, $k$-soundness, etc. Because of the change, we use the notation $\mathrm{Hull}_{k+1}^{M}(X)$ and $\mathrm{cHull}_{k+1}^{M}(X)$ as in 5.1, not [9].

For a structure $M$, $\left\lfloor M\right\rfloor$ denotes the universe of $M$.

Let $N$ be a premouse and $\mathbb{E}=\mathbb{E}^{N}$.

We write $\left\lfloor N\right\rfloor$ for the universe of $N$, and $\mathbbm{e}^{N}=\mathbb{E}\!\upharpoonright\!\omega_{1}^{N}$ and $\mathbbm{m}^{N}=N|\omega_{1}^{N}$.

If $N$ is passive and $\alpha=\mathrm{OR}^{N}$, then $\mathcal{J}^{\mathbb{E}}_{\alpha}$ denotes $N$, and when working inside $N$, $\mathcal{J}^{\mathbb{E}}$ also denotes $N$. And $\mathcal{J}(P)$ denotes the rudimentary closure of $P\cup\{P\}$.

Let $n<\omega$. We say that $N$ satisfies $(n+1)$-condensation iff $N$ is $n$-sound and whenever $H$ is $(n+1)$-sound and $\pi:H\to N$ is $n$-lifting (see [6, Definition 2.1]) and $\rho_{n+1}^{H}\leq\mathrm{cr}(\pi)$, then either $H\trianglelefteq N$ or, letting $\rho=\rho_{n+1}^{H}$, then $N|\rho$ is active with extender $E$ and $H\triangleleft\mathrm{Ult}(N|\rho,E)$. (See [6, Theorem 4.2***].) We say $N$ satisfies $\omega$-condensation iff it satisfies $(n+1)$-condensation for all $n<\omega$.

We say that $N$ is an $\omega$-premouse iff $N$ is $\omega$-sound and $\rho_{\omega}^{N}=\omega$; in this case we let $\deg(N)$ denote the least $n$ such that $\rho_{n+1}^{N}=\omega$. An $\omega$-mouse is an $(\omega,\omega_{1}+1)$-iterable $\omega$-premouse. If $N$ is an $\omega$-mouse, we write $\Sigma_{N}$ for the unique $(\omega,\omega_{1}+1)$-strategy for $N$.

For $\alpha<\mathrm{OR}^{N}$, recall that $\alpha$ is a cutpoint of $N$ iff for all $E\in\mathbb{E}_{+}^{N}$, if $\mathrm{cr}(E)<\alpha$ then $\mathrm{lh}(E)\leq\alpha$.

For an extender $E$, $t_{E}$ and $\tau_{E}$ denote the Dodd parameter and Dodd projectum of $E$ respectively, if they are defined.

2 Extender maximality

In this section we prove Theorems 1 and 1. The proofs are refinements of less general results proved in [7]. Toward these proofs, we begin with a lemma which helps us to find sound hulls of premice; the proof is basically as in [9, Lemma 3.1***], but here we use the fact that condensation follows from normal iterability in order to reduce our assumptions.

Definition 2.1**.**

Let $k<\omega$, let $H$ be $k$-sound, $q\in[\rho_{0}^{H}]^{<\omega}$ and $\alpha\in\mathrm{OR}^{H}$, with $\alpha\leq\min(q)$ if $q\neq\emptyset$. The $(k+1)$-solidity witness for $(H,q,\alpha)$, (or just for $(q,\alpha)$), is

[TABLE]

Letting $q=\{q_{0},\ldots,q_{\mathrm{lh}(q)-1}\}$ with $q_{i}>q_{i+1}$, the (set of all) $(k+1)$-solidity witnesses for $(H,q)$ (or just for $q$) is

[TABLE]

where $q\!\upharpoonright\!i=\{q_{0},\ldots,q_{i-1}\}$. The (set of all) $(k+1)$-solidity witnesses for $H$ is

[TABLE]

Note that in the preceding definition, we are not assuming that the solidity witnesses in consideration are in $H$.

Definition 2.2**.**

Let $k<\omega$, let $H$ be $(k+1)$-sound, $q\in\mathfrak{C}_{0}(H)$, $\theta<\rho_{0}^{H}$,

[TABLE]

$\pi:{\bar{H}}\to H$ be the uncollapse and $\pi({\bar{q}})=q$. We say that $(\theta,q)$ is $(k+1)$-self-solid (for $H$) iff ${\bar{H}}$ is $k+1$-sound and $\rho_{k+1}^{\bar{H}}=\theta$ and $p_{k+1}^{\bar{H}}=\bar{q}$.

Let $x\in\mathfrak{C}_{0}(H)$ and $r\in[\rho_{0}^{H}]^{<\omega}$. We say that $r$ is an $\mathrm{r}\Sigma_{k+1}^{H}(\{x\})$-generator iff for every $\gamma\in r$, we have

[TABLE]

Lemma 2.2**.**

Let $k<\omega$ and let $H$ be $(k+1)$-sound and $(k,\omega_{1}+1)$-iterable. Let $r\in H$ and $\theta\leq\rho_{k+1}^{H}$ be a cardinal of $H$. Then there is $q\in H$ such that:

1. –

$(\theta,q)$* is $(k+1)$-self-solid for $H$,* 2. –

$p_{k+1}^{H}=q\backslash\min(p_{k+1}^{H})$, 3. –

$r\in\mathrm{Hull}_{k+1}^{H}(\theta\cup\{\vec{p}_{k}^{H},q\})$, and 4. –

${\bar{H}}=\mathrm{cHull}_{k+1}^{H}(\theta\cup\{\vec{p}_{k}^{H},q\})\trianglelefteq H$.

Proof.

We may assume $H$ is countable and $\theta<\rho_{k+1}^{H}$. We will define $m<\omega$ and

[TABLE]

with $q_{i}>q_{i+1}$ for $i+1<m$. Let $p=p_{k+1}^{H}$. We start with $q\!\upharpoonright\!\mathrm{lh}(p)=p$. We define $q_{i}$ for $i\geq\mathrm{lh}(p)$ by induction on $i$, with $q_{i}<\rho_{k+1}^{H}$. We simultaneously define an $H$-cardinal $\gamma_{i}$, with $\gamma_{\mathrm{lh}(p)}=\rho_{k+1}^{H}$ and $\theta\leq\gamma_{i}\leq q_{i-1}$ for $i>\mathrm{lh}(p)$, and

[TABLE]

where $u_{i}$ is the set of $(k+1)$-solidity witnesses for $(H,q\!\upharpoonright\!i)$. Now let $i\geq\mathrm{lh}(p)$, and let $q\!\upharpoonright\!i$, $\gamma_{i}$ be given. If $\gamma_{i}=\theta$ then we set $m=i$, so $q=q\!\upharpoonright\!i$ and we are done. So suppose $\gamma_{i}>\theta$. Let $\eta<\gamma_{i}$ be least such that $\eta>\theta$ and $\eta$ is not a cardinal of $H$ and

[TABLE]

and

[TABLE]

Let $q_{i}=\min(\mathrm{OR}\backslash H_{i})$ and let $\gamma_{i}=\operatorname{card}^{H}(q_{i})=\operatorname{card}^{H}(\eta)$.

Clearly

[TABLE]

so it suffices to see that $u_{i+1}\in H_{i}^{\prime}$. Note that the transitive collapse $W_{i}$ of $H_{i}$ is (equivalent to) the $(k+1)$-solidity witness for $(q\!\upharpoonright\!i,q_{i})$, so it suffices to see that $W_{i}\in H_{i}^{\prime}$. For this it suffices to see that $W_{i}\triangleleft H$, since then $W_{i}$ is the least segment $W$ of $H$ such that $\mathrm{OR}^{W}\geq q_{i}$ and $\rho_{\omega}^{W}=\gamma_{i}=\mathrm{lgcd}(H|q_{i})$.

Let $\rho=q_{i}$ and $\gamma=\gamma_{i}$ and $W=W_{i}$. Let $\pi:W\to H$ be the uncollapse. Then $\pi(p_{k+1}^{W}\backslash\rho)=q\!\upharpoonright\!i$ and $W$ is $\rho$-sound and $\mathrm{cr}(\pi)=\rho$ and

[TABLE]

(we have $\rho_{k+1}^{W}\geq\gamma$ because $W\in H$). So by condensation as stated in [6, Theorem 4.2***], either (a) $W\triangleleft H$ or (b) letting $J\triangleleft H$ be least such that $q_{i}\leq\mathrm{OR}^{J}$ and $\rho_{\omega}^{J}=\gamma$, then $\rho_{k+1}^{J}=\gamma<\rho_{k}^{J}$ and there is a type 1 extender $F$ over $J$ with $\mathrm{cr}(F)=\gamma$ and

[TABLE]

But since $\eta>\gamma$ and because of line (1), we have

[TABLE]

and therefore (b) is false. So $W\triangleleft H$, as required.

Since $\gamma_{i+1}<\gamma_{i}$, the construction terminates successfully.

Finally, the fact that ${\bar{H}}\trianglelefteq H$ (where ${\bar{H}}$ is defined in the statement of the theorem) follows from condensation. ∎

Related calcuations also give the following:

Lemma 2.2**.**

Let $k<\omega$ and let $H$ be $(k+1)$-sound and $(k,\omega_{1}+1)$-iterable. Suppose $\rho=\rho_{k+1}^{H}=(\kappa^{+})^{H}>\omega$ and $\kappa$ is an $H$-cardinal. For $\gamma<\rho$ let

[TABLE]

and $W_{\gamma}$ be the transitive collapse of $H_{\gamma}$. Then:

1. (i)

For all sufficiently large $\gamma\in(\kappa,\rho)$, either:

1. –

$W_{\gamma}\triangleleft H$, or 2. –

$(\kappa^{+})^{W_{\gamma}}=\rho_{\omega}^{W_{\gamma}}=\gamma$, $H|\gamma$ is active with $E$666Note then that $\mathrm{cr}(E)<\kappa$ and $E$ is $H$-total. and $W_{\gamma}\triangleleft\mathrm{Ult}(H,E)$. 2. (ii)

For cofinally many $\gamma<\rho$, we have $W_{\gamma}\triangleleft H$ and $\rho_{k+1}^{W_{\gamma}}=\kappa$.

Proof.

For $\gamma<\rho$, say that $\gamma$ is a generator iff $\gamma\notin H_{\gamma}$. We say that a generator is a limit generator iff it is a limit of generators, and is otherwise a successor generator. Note that the set of generators above $\kappa$ is club in $\rho$. Let $\gamma>\kappa$ be a generator. Then note that $W_{\gamma}\in H$ and $\gamma=(\kappa^{+})^{W_{\gamma}}$ and hence, either $\rho_{k+1}^{W_{\gamma}}=\kappa$ or $\rho_{k+1}^{W_{\gamma}}=\gamma$; moreover, if $\gamma$ is a successor generator then $\rho_{k+1}^{W_{\gamma}}=\kappa$, since then

[TABLE]

for some $\eta<\gamma$ and some $x$.

Now let $\eta_{0}$ be the least generator $\gamma>\kappa$ such that $w_{k+1}^{H}\in H_{\gamma}$. We claim that the conclusion of (i) holds for all generators $\gamma>\eta_{0}$. We proceed by induction on $\gamma$.

First suppose that $\gamma$ is a limit generator. Then by induction, for eventually all successor generators $\gamma^{\prime}<\gamma$, we have $W_{\gamma^{\prime}}\triangleleft H$ and $\gamma^{\prime}=(\kappa^{+})^{W_{\gamma^{\prime}}}$ and $W_{\gamma^{\prime}}$ projects to $\kappa$. It follows that $W_{\gamma^{\prime}}\in H_{\gamma}$, so $W_{\gamma^{\prime}}\in W_{\gamma}$, which implies that $\rho_{k+1}^{W_{\gamma}}=\gamma$, and therefore $W_{\gamma}$ is $(k+1)$-sound. So the conclusion for $W_{\gamma}$ follows from $(k+1)$-condensation.

Now suppose that $\gamma$ is a successor generator. Then there is a largest generator $\eta<\gamma$, and we have $\kappa<\eta_{0}\leq\eta<\gamma$, and $W_{\gamma}$ projects to $\kappa$. So using condensation as stated in [6, Theorem 4.2***] as in the proof of 2, we get $W_{\gamma}\triangleleft H$.

Part (ii) now easily follows; in fact its conclusion holds for every sufficiently large successor generator. ∎

Remark 2.2**.**

Let $M$ be an $m$-sound premouse. Recall that a (putative) iteration tree on $M$ is $m$-maximal given that (i) $\mathcal{T}$ satisfies the monotone length condition

[TABLE]

and for each $\alpha+1<\mathrm{lh}(\mathcal{T})$, (ii) $\gamma=\mathrm{pred}^{\mathcal{T}}(\alpha+1)$ is least such that $\mathrm{cr}(E^{\mathcal{T}}_{\alpha})<\nu(E^{\mathcal{T}}_{\alpha})$, (iii) $M^{*\mathcal{T}}_{\alpha+1}\trianglelefteq M^{\mathcal{T}}_{\gamma}$ is as large as possible, and (iv) $k=\deg^{\mathcal{T}}(\alpha+1)$ is as large as possible (with $k\leq\deg^{\mathcal{T}}(\gamma)$ if $M^{*\mathcal{T}}_{\alpha+1}=M^{\mathcal{T}}_{\gamma}$) subject to (iii).

Definition 2.3**.**

Let $M$ be an $m$-sound premouse. An essentially $m$-maximal tree on $M$ satisfies the requirements of $m$-maximality, except that we drop the monotone length condition, replacing it with montone $\nu$ condition, that is, that

[TABLE]

Remark 2.3**.**

It is easy to see that, for example, $(m,\omega_{1}+1)$-iterability is equivalent to essential-$(m,\omega_{1}+1)$-iterability.

Definition 2.4**.**

Let $\pi:\mathfrak{C}_{0}(M)\to\mathfrak{C}_{0}(N)$ be $\Sigma_{0}$-elementary between premice $M,N$ of the same type.

If $M,N$ are passive then $\psi_{\pi}$ denotes $\pi$. If $M,N$ are active, $\mu=\mathrm{cr}(F^{M})$ and $\kappa=\mathrm{cr}(F^{N})$, then

[TABLE]

denotes the embedding induced by the Shift Lemma from $\pi$. So in both cases, $\pi\subseteq\psi_{\pi}$ and $\psi_{\pi}$ is fully elementary.

Now we say that $\pi$ is:

1. –

$\nu$-low iff $M,N$ are type 3 and $\psi_{\pi}(\nu^{M})<\nu^{N}$, 2. –

$\nu$-preserving iff, if $M,N$ are type 3 then $\psi_{\pi}(\nu^{M})=\nu^{N}$, and 3. –

$\nu$-high iff $M,N$ are type 3 and $\psi_{\pi}(\nu^{M})>\nu^{N}$.

Remark 2.4**.**

Suppose $\pi,M,N$ are as above and $M,N$ are type 3. It is easy to see that if $\pi$ is $\mathrm{r}\Sigma_{2}$-elementary then $\pi$ is $\nu$-preserving, and if $\pi$ is $\mathrm{r}\Sigma_{1}$-elementary then $\pi$ is non-$\nu$-low. Moreover, one can show that if $\pi=\pi_{0}$ is a $\nu$-preserving neat $k$-embedding, then the copying construction with $\pi$ preserves tree order, and for each $\alpha$, $\pi_{\alpha}$ is a $\nu$-preserving neat $\deg^{\mathcal{T}}(\alpha)$-embedding. (Here if $M^{\mathcal{T}}_{\alpha}$ is type 3 and $\rho_{0}(M^{\mathcal{T}}_{\alpha})<\mathrm{lh}(E^{\mathcal{T}}_{\alpha})<\mathrm{OR}(M^{\mathcal{T}}_{\alpha})$ then we copy $E^{\mathcal{T}}_{\alpha}$ to $E^{\mathcal{U}}_{\alpha}=\psi_{\pi_{\alpha}}(E^{\mathcal{T}}_{\alpha})$.)

We will deduce Theorems 1 and 1 from the following:

Theorem 2.4**.**

Let $N$ be a $(0,\omega_{1}+1)$-iterable pm, $F\in N$ and $\mu,\sigma\in\mathrm{OR}^{N}$, $\dot{t}\in V_{\omega}$ and $W$ be such that:

1. –

$\sigma$* is an $N$-cardinal,* 2. –

$F$* is a short $N$-extender with support $\sigma\cup\dot{t}$, weakly amenable to $N$, coded as a subset of $N|\sigma$, such that $N\models$“$F$ is countably complete”,* 3. –

$W=\mathrm{Ult}(N,F)$, $\mu=\mathrm{cr}(F)<\sigma$ and $\mathcal{H}_{\sigma}^{N}\subseteq W$.

Then (i) $W|(\sigma^{+})^{W}=N||(\sigma^{+})^{W}$ and if $\dot{t}=\emptyset$ then (ii) $F\in\mathbb{E}_{+}^{N}$.

Proof.

We may assume that $N=\mathcal{J}(R)$ where $F$ is definable from parameters over $R$ and $\rho_{\omega}^{R}=\sigma$. Say that $F$ is $\mathrm{r}\Sigma_{n+1}^{R}(\{r\})$. We may also assume inductively that all segments of $R$ satisfy the theorem.

Let $n\ll m<\omega$ and $M=\mathrm{cHull}_{m+1}^{R}(\{s\})$ where $(\omega,s)$ is $(m+1)$-self-solid for $R$ and $r\in\mathrm{rg}(\pi_{MR})$ where $\pi_{MR}:M\to R$ is the uncollapse.

Let $\pi_{MR}(E)=F$. So $E$ is an $M$-extender over generators $\tau\cup\dot{t}$, where either $\tau<\rho_{0}^{M}$ and $\pi_{MR}(\tau)=\sigma$, or $\tau=\rho_{0}^{M}$ and $\rho_{0}^{R}=\sigma$. And $E$ is $\utilde{\mathrm{r}\Sigma}_{n+1}^{M}$-definable, $M$ is $(m+1)$-sound, $n+10<m$ and

[TABLE]

Other relevant properties of $(R,F)$ also reflect to $(M,E)$. Moreover,

[TABLE]

by the countable completeness of $F$ in $N$ and the $(\omega,\omega_{1}+1)$-iterability of $R$.

Now $\tau<\mathrm{OR}^{M}$. For suppose $\tau=\mathrm{OR}^{M}$. Since $\rho_{n+10}^{M}=\tau$, therefore $M$ is passive. If $\tau=(\kappa^{+})^{M}$, i.e. $\kappa$ is the largest cardinal of $M$, then we have $U|(\kappa^{+})^{M}=M|(\kappa^{+})^{M}=M$ (by condensation for $M$), but then $E\in U$, which is impossible. So $\tau>(\kappa^{+})^{M}$. Then $E\!\upharpoonright\!\eta\in M$ for all $\eta<\tau$ (since $\rho_{n+10}^{M}=\tau$), so by induction (with conclusion (i) of the theorem), $U|\tau=M|\tau=M$, so again, $E\in U$, a contradiction.

Let $t$ be $(m,\omega)$-self-solid for $M$, and such that letting

[TABLE]

and $\pi:\bar{M}\to M$ be the uncollapse, then $r,s\in\mathrm{rg}(\pi_{MR}\circ\pi)$. Let $\pi(\bar{t})=t$, etc, and $\bar{E}=\pi^{-1}(E)$, etc. So $\bar{E}$ is defined over $\bar{M}$ from $\bar{t}$ just as $E$ is over $M$ from $t$, and the relevant properties of $(M,E,U)$ reflect to $(\bar{M},\bar{E},\bar{U})$, where

[TABLE]

Let $\theta$ be the largest $M$-cardinal $\leq\tau$ such that $M|\tau=U|\tau$.777We will show that $\theta=\tau$. Let $\pi(\bar{\theta})=\theta$ if $\theta<\rho_{0}^{M}$, and otherwise $\bar{\theta}=\rho_{0}^{\bar{M}}$. So $\bar{\theta}$ has the same defining property with respect to $\bar{M},\bar{U}$. Define the phalanx (see [9, §1.1] for the notation)

[TABLE]

Claim 1.

$\mathfrak{P}$* is $(\omega_{1}+1)$-iterable.*

Proof.

We will lift trees on $\mathfrak{P}$ to essentially $m$-maximal trees on $U$, which by 2 and line (2) suffices. Let $\psi:\bar{U}\to U$ be the Shift Lemma map. Let $\theta^{\prime}=\sup\pi``\bar{\theta}$.

Case 1.

$\theta^{\prime}<\theta$.

Let $\gamma=\operatorname{card}^{M}(\theta^{\prime})$, so $\gamma<\theta$. Let $t^{\prime}$ be such that $(\gamma,t^{\prime})$ is $m$-self-solid for $M$, with

[TABLE]

and $\widetilde{\pi}:M^{\prime}\to M$ be the uncollapse, such that $t\in\mathrm{rg}(\widetilde{\pi})$. Let

[TABLE]

be $\pi^{\prime}=\widetilde{\pi}^{-1}\circ\pi$. So

[TABLE]

Note that $\mathrm{OR}^{M^{\prime}}<\theta$, so $M^{\prime}\triangleleft U$.

We can use $(\pi^{\prime},\psi)$ to lift trees $\mathcal{T}$ on $\mathfrak{P}$ to essentially $m$-maximal trees $\mathcal{U}$ on $U$. In case $\theta$ is a limit cardinal of $M$ then everything here is routine (and we actually get $m$-maximal trees on $U$). So assume that $\theta=(\gamma^{+})^{M}$. Most of the details of the copying process are routine, but we explain enough that we can point out how the wrinkles are dealt with. Let $\pi(\bar{\gamma})=\gamma$. For $\alpha<\mathrm{lh}(\mathcal{T})$ with $\alpha>0$, we say that $\mathrm{root}^{\mathcal{T}}(\alpha)=0$ if $M^{\mathcal{T}}_{\alpha}$ is above $\bar{U}$, and $\mathrm{root}^{\mathcal{T}}(\alpha)=-1$ if above $\bar{M}$. Let $\alpha<\mathrm{lh}(\mathcal{T})$. If $\mathrm{root}^{\mathcal{T}}(\alpha)=0$ then the copy map

[TABLE]

is produced routinely. Suppose $\mathrm{root}^{\mathcal{T}}(\alpha)=-1$. If $(-1,\alpha]_{\mathcal{T}}$ does not drop in model and

[TABLE]

(note we might have $\bar{\gamma}=\mathrm{cr}(i^{\mathcal{T}}_{-1,\alpha})$), then $[0,\alpha]_{\mathcal{U}}$ does not drop in model or degree and

[TABLE]

and $\pi_{\alpha}$ is produced in the obvious manner via the Shift Lemma. Otherwise, $(0,\alpha]_{\mathcal{U}}$ drops in model, and

[TABLE]

again produced in the obvious manner. We copy extenders using these maps. There is a wrinkle when $\mathrm{pred}^{\mathcal{T}}(\alpha+1)=-1$ and $\mathrm{cr}(E^{\mathcal{T}}_{\alpha})=\bar{\gamma}$, so consider this case. We have then $\mathrm{cr}(E^{\mathcal{U}}_{\alpha})=\gamma$. Because

[TABLE]

and

[TABLE]

we get $M^{*\mathcal{T}}_{\alpha+1}=M$, and $M^{*\mathcal{U}}_{\alpha+1}=U$ (not $M^{\prime}$), and $M^{\prime}\triangleleft U|\theta$. Now if $E^{\mathcal{U}}_{\alpha}$ is not of superstrong type then

[TABLE]

and

[TABLE]

and things are standard. However, if $E^{\mathcal{U}}_{\alpha}$ is of superstrong type, then

[TABLE]

so when we lift $E^{\mathcal{T}}_{\alpha+1}$, we get $\mathrm{lh}(E^{\mathcal{U}}_{\alpha+1})<\mathrm{lh}(E^{\mathcal{U}}_{\alpha})$. However,

[TABLE]

Now we claim that $E^{\mathcal{T}}_{\alpha}$ is also superstrong, and therefore $\nu(E^{\mathcal{T}}_{\alpha})=\lambda(E^{\mathcal{T}}_{\alpha})$ and $\nu(E^{\mathcal{U}}_{\alpha})=\lambda(E^{\mathcal{U}}_{\alpha})$, and then it follows that

[TABLE]

as required for the monotone $\nu$-condition.

So suppose $E^{\mathcal{T}}_{\alpha}$ is not superstrong. So $\nu(E^{\mathcal{T}}_{\alpha})<\lambda(E^{\mathcal{T}}_{\alpha})$, so

[TABLE]

which implies that $E^{\mathcal{T}}_{\alpha}=F(M^{\mathcal{T}}_{\alpha})$ and $\pi_{\alpha}$ is $\nu$-low. In particular, $\pi_{\alpha}$ is not $\mathrm{r}\Sigma_{1}$-elementary, so is not a near [math]-embedding. Let $j=\mathrm{root}^{\mathcal{T}}(\alpha)\in\{-1,0\}$. By the proof that the copying construction propagates near embeddings (see [3]), $[j,\alpha]_{\mathcal{T}}$ does not drop in model, and so $\bar{M},\bar{U}$ are active. But because $\bar{U}=\mathrm{Ult}_{m-1}(\bar{M},\bar{E})$ and

[TABLE]

we have $\bar{\gamma}\neq\mathrm{cr}(F^{\bar{U}})$, and then similarly, as $\bar{\theta}\leq\lambda(E^{\mathcal{T}}_{0})$, it easily follows that $j=-1$. But then $\mathrm{cr}(i^{\mathcal{T}}_{j\alpha})\leq\bar{\gamma}$ and $\bar{\theta}\leq\lambda(E^{\mathcal{T}}_{0})$, so $\bar{\gamma}\neq\mathrm{cr}(F(M^{\mathcal{T}}_{\alpha}))$, contradiction.

So $\nu(E^{\mathcal{T}}_{\alpha})\leq\nu(E^{\mathcal{T}}_{\alpha+1})$, as desired. This is the only situation in which the monotone length condition can fail. We leave the remaining details of the lifting process to the reader.

Case 2.

$\pi``\bar{\theta}$* is unbounded in $\theta$.*

In this case we do not see how to produce a single map lifting $\bar{M}$, and instead produce a sequence of maps. Note that $\theta$ is a limit cardinal of $M$ (by the case hypothesis we have an $\utilde{\mathrm{r}\Sigma}_{m}^{M}$-singularization of $\theta$, and if $\theta=(\gamma^{+})^{M}$ this routinely implies that $\rho_{m}^{M}<\theta$, a contradiction), and so $\bar{\theta}$ is a limit cardinal of $\bar{M}$. For each $\bar{M}$-cardinal $\gamma<\bar{\theta}$, let $(M^{\prime}_{\gamma},\sigma_{\gamma})$ be such that $M^{\prime}_{\gamma}\triangleleft M|\theta$ and

[TABLE]

is a near $(m-1)$-embedding and

[TABLE]

and

[TABLE]

we get such pairs by taking appropriate hulls much as in the previous case.

Now for each $\gamma$ we have $M^{\prime}_{\gamma}\triangleleft U$. So we can use $(\left<\sigma_{\gamma}\right>_{\gamma<\bar{\theta}},\psi)$ to lift trees on $\mathfrak{P}$ to $m$-maximal trees on $U$. This is much as in the previous case, but this time when $\mathrm{cr}(E^{\mathcal{T}}_{\alpha})=\gamma<\bar{\theta}$, then we define $Q_{\alpha+1}=i^{\mathcal{U}}_{0,\alpha+1}(M^{\prime}_{\gamma})$ and define $\pi_{\alpha+1}$ via the Shift Lemma from $\sigma_{\gamma}$ and $\pi_{\alpha}$. We get the monotone length condition here, because

[TABLE]

The details are left to the reader.∎

Using the claim, we can now complete the proof. We get a successful comparison $(\mathcal{T},\mathcal{U})$ of $(\bar{M},\mathfrak{P})$. Note that all extenders used in the comparison have length $>\bar{\theta}$. Standard fine structural arguments show that $b^{\mathcal{U}}$ is above $\bar{U}$ and both $b^{\mathcal{T}},b^{\mathcal{U}}$ do not drop in model,

[TABLE]

and $\deg^{\mathcal{T}}(\infty)=m-1=\deg^{\mathcal{U}}(\infty)$. So $\bar{\theta}\leq\mathrm{cr}(i^{\mathcal{U}})$, so

[TABLE]

and since $\mathrm{lh}(E^{\mathcal{T}}_{0})>\bar{\theta}$, therefore

[TABLE]

But if $\bar{\theta}<\tau$ then because $\mathcal{H}_{\tau}^{\bar{M}}\subseteq\bar{U}$, we get

[TABLE]

which contradicts the choice of $\bar{\theta}$. So $\bar{\theta}=\tau$, which with line (3) gives the statement of conclusion (i) of the theorem but with $\bar{M}$ instead of $N$. However, this statement is preserved by $\pi,\pi_{MR}$, so part (i) for $N$ follows.

Assuming also that $\dot{t}=\emptyset$, so $\bar{E}$ is generated by $\tau$, then standard arguments show that $\bar{E}$ is just the $(\mu,\tau)$-extender derived from $i^{\mathcal{T}}$, and therefore that in fact $\bar{E}\in\mathbb{E}^{\bar{M}}$. But this reflects back to $N$, giving part (ii). ∎

Proof of Theorems 1 and 1.

Theorem 2 directly implies 1. For 1 note that we may replace the given extender with a sub-extender with generators of the form $\tau\cup\dot{t}$, and then appeal to 2. ∎

From 1 we immediately get:

Corollary 2.4**.**

Let $N$ be a $(0,\omega_{1}+1)$-iterable premouse and $\mu,\delta,\kappa\in N$. Then:

1. –

If $N\models$“$\mu$* is a normal measure” then $\mu\in\mathbb{E}^{N}$.* 2. –

If $N\models$“$\delta$* is Woodin” then $N\models$“$\delta$ is Woodin via extenders in $\mathbb{E}^{N}$”.* 3. –

If $N\models\mathsf{PS}+$“$\kappa$* is strong” then $N\models$“$\kappa$ is strong via extenders in $\mathbb{E}^{N}$”.*

We next prove a finer variant of Theorem 2. However, we do not actually need the variant in later sections of the paper.

Definition 2.5**.**

Let $M$ be an active premouse, $F=F^{M}$ and $\kappa=\mathrm{cr}(F)$. We say $F$ is of superstrong type iff $i^{M}_{F}(\kappa)$ is the largest cardinal of $M$. We say a premouse $N$ is below superstrong iff no $E\in\mathbb{E}_{+}^{N}$ is of superstrong type.

Recall the Dodd projectum and parameter $\tau_{E}$ and $t_{E}$ of a short extender $E$; see [2] or [9, §2] for background. The most important fact we use in this section regarding this notion is the following:

Fact 2.5** (Steel).**

Let $M$ be a $1$-sound, $(0,\omega_{1}+1)$-iterable premouse which is below superstrong. Then every $E\in\mathbb{E}_{+}^{M}$ is Dodd-sound.

Because of the “below superstrong” restriction above, Theorem 2 below is similarly restricted. Note that 2 is for Mitchell-Steel indexing. An analogous theorem has been proven by Zeman for mice with Jensen indexing, without the superstrong restriction (see [11]). Moreover, we believe that Steel’s proof for Mitchell-Steel indexing generalizes so as to allow superstrongs, but this has not been published.888For the generalization of the other standard fine structural facts, such as the solidity of the standard parameter, the proof “below superstrong” adapts to the superstrong case with very little modification. However, for the proof of Dodd-soundness, the proof requires significant extra work. So we believe that Theorem 2 actually holds without the superstrong restriction. Note that in 2, we allow $E$ itself to ostensibly be of “superstrong type”, but then it follows that $t=\emptyset$ and $E\in\mathbb{E}_{+}^{M}$, so in fact, $E$ is not of such type.

Definition 2.6**.**

Let $M$ be a premouse and $E$ a short extender, weakly amenable to $M$. Let $U=\mathrm{Ult}_{0}(M,E)$ (we don’t assume $U$ is wellfounded). Let $\tau=\tau_{E}$ and $t=t_{E}$ (the Dodd projectum and parameter of $E$). We say that $E\!\upharpoonright\!(\tau\cup t)$ is amenably $\utilde{\mathrm{r}\Sigma}_{m+1}^{M}$ iff $\tau<\rho_{0}^{M}$ and $(\tau^{+})^{U}$ is wellfounded and $U|(\tau^{+})^{U}\subseteq M$, and the standard coding of $E$ as an amenable subset of $U|(\tau^{+})^{U}$ is $\utilde{\mathrm{r}\Sigma}_{m+1}^{M}$. Here the coding consists of tuples $(\xi,\alpha_{\xi},E_{\xi})$ where $\xi<(\kappa^{+})^{M}$ and $E_{\xi}$ is the natural coding of the extender fragment

[TABLE]

as a subset of $M|\tau$999That is, represent $t$ with a finite set of integers, and $\alpha_{\xi}$ is the least $\alpha$ such that $E_{\xi}\in U|(\alpha_{\xi}+\omega)$. (By the usual proof (see [1, §2]), $E_{\xi}\in U$ and the $\alpha_{\xi}$’s are cofinal in $(\tau^{+})^{U}$.)

Theorem 2.6**.**

Let $m<\omega$ and let $M$ be an $(m+1)$-sound, $(m,\omega_{1}+1)$-iterable premouse which is below superstrong. Let $E$ be a short $M^{\mathrm{sq}}$-extender weakly amenable to $M$ with $\kappa=\mathrm{cr}(E)<\rho_{m}^{M}$ (we actually assume more below). Let $U=\mathrm{Ult}_{m}(M,E)$. Let $\tau=\tau_{E}$ and $t=t_{E}$. Suppose that:

1. –

$E\!\upharpoonright\!(\tau\cup t)$* is amenably $\utilde{\mathrm{r}\Sigma}_{m+1}^{M}$ (hence, $(\tau^{+})^{U}$ is wellfounded).* 2. –

$\tau\leq\rho_{m+1}^{M}$* and $\tau$ is an $M$-cardinal,* 3. –

$\mathcal{H}_{\tau}^{M}\subseteq U$.

Then (i) $U|(\tau^{+})^{U}=M||(\tau^{+})^{U}$ and (ii) if $E$ is Dodd-solid101010That is, $E\!\upharpoonright\!(\alpha\cup(t\backslash(\alpha+1)))\in U$ for each $\alpha\in t$. then $E\in\mathbb{E}_{+}^{M}$.

Proof.

Let $j:M\to U$ be the ultrapower map.

If $U||\tau=M||\tau$ then let $\theta=\tau$, and otherwise let $\lambda$ be least such that $U|\lambda\neq M|\lambda$ and let $\theta=\operatorname{card}^{M}(\lambda)$. So $\theta$ is an $M$-cardinal and $\theta\leq\tau$. Note that if $E$ is Dodd-solid then $E$ is Dodd-sound. (For suppose $(\kappa^{+})^{M}<\tau$. As $E\!\upharpoonright\!(\tau\cup t)$ is amenably $\utilde{\mathrm{r}\Sigma}_{m+1}^{M}$ and $\tau\leq\rho_{m+1}^{M}$, then $E\!\upharpoonright\!(\alpha\cup t)\in M$ for each $\alpha<\tau$. But $\mathcal{H}_{\tau}^{M}\subseteq U$, so $E\!\upharpoonright\!(\alpha\cup t)\in U$.) Let $e\in M^{<\omega}$ be such that:

$\theta,\tau\in e$. 2. 2.

If $\mathfrak{C}_{0}(M)$ has largest cardinal $\Omega$ then $\Omega\in e$. 3. 3.

The amenable coding of $E\!\upharpoonright\!(\tau\cup t)$ (described in 2.6) is $\mathrm{r}\Sigma_{m+1}^{M}(\{e\})$. 4. 4.

If $\theta<\tau$ then $\lambda\in e$ where $\lambda$ is least such that $U|\lambda\neq M|\lambda$. 5. 5.

If $(\tau^{+})^{U}<(\tau^{+})^{M}$ then $(\tau^{+})^{U}\in e$. 6. 6.

If $(\tau^{+})^{U}=(\tau^{+})^{M}$ but $U|(\tau^{+})^{U}\neq M||(\tau^{+})^{M}$ then $\lambda\in e$ where $\lambda$ is least such that $U|\lambda\neq M|\lambda$. 7. 7.

If $E$ is Dodd-solid then there are $a,f\in e$ such that $a\in[\tau]^{<\omega}$ and $[a\cup t,f]^{M,m}_{E}$ is the (finite) set of Dodd-solidity witnesses (for $t$). 8. 8. 111111This condition is only relevant at the very end of the proof, and its motivation will only become clear there; the reader can ignore it until that point.

If $E$ is Dodd-solid (hence Dodd-sound) and $\theta=\tau$ and

[TABLE]

and $U|\lambda=M||\lambda$ but $M|\lambda$ is active with an extender $F$ such that $\kappa<\mathrm{cr}(F)$, then there are $a,f\in e$ with $a\in[\tau]^{<\omega}$ and such that

[TABLE]

Let ${\bar{M}}=\mathrm{cHull}_{m+1}^{M}(\{\vec{p}_{m}^{M},q\})$ and $\pi:{\bar{M}}\to M$ be the uncollapse, where $q$ is such that $(\omega,q)$ is $(m+1)$-self-solid for $M$ and $e\in\mathrm{rg}(\pi)$ ($q$ exists by 2).

Let $\pi({\bar{q}})=q$, $\pi({\bar{\theta}})=\theta$, etc. So ${\bar{M}}$ is $(m+1)$-sound with $\rho_{m+1}^{\bar{M}}=\omega$ and ${\bar{q}}=p_{m+1}^{{\bar{M}}}$. Let ${\bar{E}}\!\upharpoonright\!{\bar{\tau}}\cup{\bar{t}}$ be defined over ${\bar{M}}$ from ${\bar{e}}$ as $E\!\upharpoonright\!\tau\cup t$ is defined over $M$ from $e$. Then the usual proof that $\Sigma_{1}$-substructures of premice are premice121212As $\Sigma_{1}$ includes a constant symbol for the largest initial segment of the active extender. and some similar considerations show that most of the facts reflect to ${\bar{M}},{\bar{E}}$, etc, and in particular:

1. 1’.

${\bar{E}}\!\upharpoonright\!{\bar{\tau}}\cup{\bar{t}}$ is a weakly amenable extender over ${\bar{M}}$ with ${\bar{\kappa}}=\mathrm{cr}({\bar{E}})<\rho_{m}^{\bar{M}}$. Let

[TABLE] 2. 2’.

${\bar{E}}$ is generated by ${\bar{\tau}}\cup{\bar{t}}$. 3. 3’.

If $M$ has largest cardinal $\Omega$ then ${\bar{M}}$ has largest cardinal $\bar{\Omega}$. 4. 4’.

${\bar{\theta}},{\bar{\tau}}$ are ${\bar{M}}$-cardinals, $\mathcal{H}^{\bar{M}}_{{\bar{\tau}}}\subseteq{\bar{U}}$ and ${\bar{M}}|{\bar{\theta}}={\bar{U}}|{\bar{\theta}}$. 5. 5’.

If $\theta<\tau$ then ${\bar{M}}|({\bar{\theta}}^{+})^{{\bar{M}}}\neq{\bar{U}}|({\bar{\theta}}^{+})^{{\bar{U}}}$, and $\bar{\lambda}$ is least such that ${\bar{M}}|\bar{\lambda}\neq{\bar{U}}|\bar{\lambda}$. 6. 6’.

If $\theta=\tau$ and $(\tau^{+})^{U}<(\tau^{+})^{M}$ then $({\bar{\tau}}^{+})^{\bar{U}}<({\bar{\tau}}^{+})^{\bar{M}}$ and $\pi(({\bar{\tau}}^{+})^{\bar{U}})=(\tau^{+})^{U}$. 7. 7’.

If $\theta=\tau$ and $(\tau^{+})^{U}=(\tau^{+})^{M}$ then $({\bar{\tau}}^{+})^{\bar{U}}=({\bar{\tau}}^{+})^{\bar{M}}$. 8. 8’.

If $\theta=\tau$ and $U|(\tau^{+})^{U}=M||(\tau^{+})^{U}$ then ${\bar{U}}|({\bar{\tau}}^{+})^{\bar{U}}={\bar{M}}||({\bar{\tau}}^{+})^{\bar{U}}$. 9. 9’.

If $\theta=\tau$ and $U|(\tau^{+})^{U}\neq M||(\tau^{+})^{U}$ then ${\bar{\tau}}<\bar{\lambda}<({\bar{\tau}}^{+})^{\bar{U}}$ and $\bar{\lambda}$ is least such that ${\bar{U}}|\bar{\lambda}\neq{\bar{M}}|\bar{\lambda}$. 10. 10’.

If $E$ is Dodd-solid then ${\bar{E}}$ is Dodd-solid with respect to ${\bar{t}}$. That is, for each $\alpha\in{\bar{t}}$, we have

[TABLE] 11. 11’.

If $E$ is Dodd-solid (hence Dodd-sound) and $U,M,\lambda,F$ are as in condition 8, then $\bar{\lambda}=({\bar{\tau}}^{+})^{\bar{U}}$ and ${\bar{M}}|\bar{\lambda}$ is active with ${\bar{F}}$ and the Dodd-soundness witness

[TABLE]

(We do not yet know that ${\bar{U}}$ is wellfounded. And because $\rho_{m+1}^{\bar{M}}=\omega$, it does not yet seem clear that if $E$ is Dodd-sound then ${\bar{E}}$ is Dodd-sound; however, we will eventually see that this is true.) Let ${\bar{j}}:{\bar{M}}\to{\bar{U}}$ the ultrapower map. Let $\psi:{\bar{U}}\to U$ be the Shift Lemma map. Define the phalanx $\mathfrak{P}=(({\bar{M}},m,{\bar{\theta}}),({\bar{U}},m),{\bar{\theta}})$.

Claim 1.

${\bar{U}}$* is wellfounded and $\mathfrak{P}$ is $(m,\omega_{1}+1)$-iterable.*

Proof.

The argument is mostly similar to that in the proof of 2. We will lift $m$-maximal trees $\mathcal{T}$ on $\mathfrak{P}$ to essentially $m$-maximal trees on $M$. For this we will find embeddings from ${\bar{M}}$ and ${\bar{U}}$ into segments of $M$ with appropriate agreement. As before, in one case we only see how to find an infinite sequence of embeddings from $\bar{M}$ into various segments of $M$, and use of all these together as base copy maps. We will initially find such a system of maps inside $U$, and then deduce that there is also such a system in $M$ via the elementarity of $j$. We first make some general observations that will lead finding the system of embeddings in $U$.

Let $R\triangleleft M$. Note that $M$ satisfies condensation with respect to premice embedded into $R$; in particular, $M\models$“For every $s<\omega$ and every premouse $S\in R$ such that $S$ is $(s+1)$-sound and $\pi:S\to R$ is $s$-lifting and and $\mathrm{cr}(\pi)\geq\rho_{s+1}^{S}$, either (i) $S\triangleleft R$ or (ii) $\alpha=_{\mathrm{def}}\mathrm{cr}(\pi)=\rho_{s+1}^{S}$ and $R|\alpha$ is active and $S\triangleleft\mathrm{Ult}(R|\alpha,F^{R|\alpha})$”. Therefore $U$ satisfies the same statement regarding its proper segments.

Let $M_{\kappa}=\mathrm{cHull}_{m+1}^{M}(\kappa\cup\{q\})$. Let $\sigma_{\kappa}:{\bar{M}}\to M_{\kappa}$ and $\pi_{\kappa}:M_{\kappa}\to M$ be the natural maps and $\pi_{\kappa}(q_{\kappa})=q$. Note that $M_{\kappa}$ is sound and $M_{\kappa}\in M$. So $\rho_{m+1}^{M_{\kappa}}=\kappa\leq\mathrm{cr}(\pi_{\kappa})$. By condensation, $M_{\kappa}\triangleleft M$.

Now $\tau$ is a $U$-cardinal with $\kappa<\tau\leq j(\kappa)$. Working in $U$, let

[TABLE]

with $r\in\mathrm{OR}[j(M_{\kappa})]^{<\omega}$ chosen such that $U\models$“$(\tau,r)$ is $(m+1)$-self-solid for $j(M_{\kappa})$ and letting $\varrho_{\tau}:U^{\prime}\to j(M_{\kappa})$ be the uncollapse, then $j(q_{\kappa}),t\in\mathrm{rg}(\varrho_{\tau})$”. Such an $r$ exists by the elementarity of $j$ and by 2. (Note that $U\models$“$j(M_{\kappa})$ is wellfounded”; the transitive collapse $U^{\prime}$ is computed inside $U$, where it is well-defined.) Note that if $\tau=j(\kappa)$ then $t=\emptyset$ and $U^{\prime}=j(M_{\kappa})$ and $r=j(q_{\kappa})$. And if $\tau<j(\kappa)$ then $\rho_{m+1}^{U^{\prime}}=\tau$, so $U^{\prime}\triangleleft j(M_{\kappa})$ by condensation in $U$. In fact, $U^{\prime}\triangleleft U|(\tau^{+})^{U}$, and we assumed that $U|(\tau^{+})^{U}$ is wellfounded, so $U^{\prime}$ is wellfounded.

Let $\psi:{\bar{U}}\to j(M_{\kappa})$ be the Shift Lemma map. So $\psi$ is $\mathrm{r}\Sigma_{0}$-elementary. Now $\mathrm{rg}(\psi)\subseteq\mathrm{rg}(\varrho_{\tau})$, because if131313Here for a premouse $R$, $f^{R}_{\tau,r}$ is the partial function $f:\mathfrak{C}_{0}(R)^{2}\to\mathfrak{C}_{0}(R)$ given by $f(a^{\prime},t^{\prime})=\tau^{R}(r,a^{\prime},t^{\prime})$.

[TABLE]

where $\tau$ is an $\mathrm{r}\Sigma_{m}$ term and $q\in\mathfrak{C}_{0}({\bar{M}})$ and $a\in\bar{\tau}^{<\omega}$, then

[TABLE]

so

[TABLE]

so $\psi(x)\in\mathrm{rg}(\varrho_{\tau})$. So define $\psi^{\prime}:{\bar{U}}\to U^{\prime}$ by $\psi^{\prime}=\varrho^{-1}_{\tau}\circ\psi$. Then $\psi^{\prime}$ is $m$-lifting, because if $\varphi$ is $\mathrm{r}\Sigma_{m+1}$ and ${\bar{U}}\models\varphi(x)$ then easily

[TABLE]

so $U\models\text{``}U^{\prime}\models\varphi(\psi^{\prime}(x))\text{''}$, so $U^{\prime}\models\varphi(\psi^{\prime}(x))$. Also $\psi^{\prime}\!\upharpoonright\!{\bar{\tau}}=\pi\!\upharpoonright\!{\bar{\tau}}$. Also $\psi^{\prime}$ is c-preserving; if $m=0$ and $M$ has largest cardinal $\Omega$, this follows easily from commutativity and the fact that we put $\Omega\in\mathrm{rg}(\pi)$, and if $m=0$ and $M$ has no largest cardinal then it is because then for any $M$-cardinal $\lambda$, we have $M|\lambda\preccurlyeq_{1}M$ by condensation, and hence, $\kappa<\max(q)$ (as $\kappa\in\mathrm{rg}(\pi)$), and so ${\bar{M}},M_{\kappa}$ have largest cardinals $\bar{\Omega},\Omega_{\kappa}$ respectively, with $\pi(\bar{\Omega})=\pi_{\kappa}(\Omega_{\kappa})=\operatorname{card}^{M}(\max(q))$.

For $\eta<\theta$, let

[TABLE]

and $\pi_{\eta}:M_{\eta}\to M$ be the uncollapse and $\sigma_{\eta}:{\bar{M}}\to M_{\eta}$ the natural map, so $\pi_{\eta}\circ\sigma_{\eta}=\pi$. Since $\eta<\theta\leq\tau\leq\rho_{m+1}^{M}$, we have $M_{\eta}\in M$. Note that if $\eta$ is an $M$-cardinal then $M_{\eta}$ is $(m+1)$-sound with $\eta=\rho_{m+1}^{M_{\eta}}$ and $p_{m+1}^{M_{\eta}}=\sigma_{\eta}({\bar{q}})\backslash\eta$, so $M_{\eta}\triangleleft M|\theta$.

Now as before, we consider two cases.

Case 3.

$\pi``{\bar{\theta}}$* is bounded in $\theta$.*

Let $\eta=\sup\pi``{\bar{\theta}}$. We have $M_{\eta}$, etc, as above. Note that either:

1. –

$\eta$* is a limit cardinal of $M$ (hence the comments above apply), or* 2. –

$M||\eta$* has largest cardinal $\xi$ where $\xi$ is an $M$-cardinal and $\xi\in\mathrm{rg}(\pi)$, and*

[TABLE]

because $\mathrm{rg}(\pi)=\mathrm{Hull}_{m+1}^{M}(\{q\})$ is cofinal in $\eta$; therefore, $\rho_{m+1}^{M_{\eta}}=\xi$ and $p_{m+1}^{M_{\eta}}=\sigma_{\eta}({\bar{q}})\backslash\xi$.

It follows that $M_{\eta}$ is sound, and of course $\mathrm{cr}(\pi_{\eta})\geq\eta$. Since $\eta<\theta\leq\rho_{m+1}^{M}$, condensation (as stated in [6]) gives $M_{\eta}\triangleleft M|\theta$. Note that $\sigma_{\eta}\!\upharpoonright\!{\bar{\theta}}=\pi\!\upharpoonright\!{\bar{\theta}}$ and $\sigma_{\eta}\in M|\theta$. Since $M|\theta=U|\theta$, therefore $M_{\eta}\triangleleft U|\theta$ and $\sigma_{\eta}\in U|\theta$. Note that $M_{\eta}\triangleleft\mathfrak{C}_{0}(U^{\prime})$ as $\eta<\tau$.

Now $\sigma_{\eta}\!\upharpoonright\!{\bar{\theta}}=\pi\!\upharpoonright\!{\bar{\theta}}=\psi^{\prime}\!\upharpoonright\!{\bar{\theta}}$ and $\sigma_{\eta},{\bar{U}},U^{\prime}\in U$, with ${\bar{U}}\in\mathrm{HC}^{U}$, and moreover, $U|((\mathrm{OR}^{U^{\prime}})^{+})^{U}$ is wellfounded. So by absoluteness, in $U$ there is some c-preserving $m$-lifting embedding $\widetilde{\psi}:{\bar{U}}\to U^{\prime}$ with $\widetilde{\psi}\!\upharpoonright\!{\bar{\theta}}=\sigma_{\eta}\!\upharpoonright\!{\bar{\theta}}$.

So $U\models\varphi^{+}({\bar{M}},{\bar{U}})$, where $\varphi^{+}({\bar{M}},{\bar{U}})$ asserts “There are proper segments $M^{*}$ and $U^{*}$ of me, with $M^{*}\triangleleft\mathfrak{C}_{0}(U^{*})$, and there are c-preserving $m$-lifting embeddings

[TABLE]

with $\pi^{*}\!\upharpoonright\!{\bar{\theta}}=\psi^{*}\!\upharpoonright\!{\bar{\theta}}$”.

So by elementarity, $M\models\varphi^{+}({\bar{M}},{\bar{U}})$. Let $M^{*},U^{*},\pi^{*},\psi^{*}$ witness this in $M$. These embeddings are enough to copy $m$-maximal trees on $\mathfrak{P}$ to essentially $m$-maximal trees on $M$. The point of the requirement that $M^{*}\triangleleft U^{*}$ is as follows. Suppose that ${\bar{\theta}}=({\bar{\kappa}}^{+})^{\bar{M}}$. Then when iterating $\mathfrak{P}$, extenders $G$ with $\mathrm{cr}(G)={\bar{\kappa}}$ apply to ${\bar{M}}$. Let $\kappa^{*}=\pi^{*}({\bar{\kappa}})$ and $G^{*}$ be the lift of $G$. Then $G^{*}$ will measure all subsets of $\kappa^{*}$ in $U^{*}$. Because $M^{*}\triangleleft\mathfrak{C}_{0}(U^{*})$, we can define a copy map

[TABLE]

as usual; this suffices. Although the lifted tree can fail the monotone length condition, it will be essentially $m$-maximal; this works much as in 2.

Case 4.

$\pi``{\bar{\theta}}$* is unbounded in $\theta$.*

Then $\theta$ is a limit cardinal of $M$, because $\theta$ is an $M$-cardinal ${\leq\rho_{m+1}^{M}}$ and there is an $\utilde{\mathrm{r}\Sigma}_{m+1}^{M}$-definable cofinal partial map $\omega\to\sup\pi``{\bar{\theta}}$. For each $M$-cardinal $\mu<\theta$ we have $M_{\mu},\sigma_{\mu}\in M|\theta=U|\theta$. We have $M_{\mu},\sigma_{\mu},U^{\prime}\in U|(\tau^{+})^{U}$.

Let $C$ be the set of ${\bar{M}}$-cardinals ${<{\bar{\theta}}}$. Working in $U$, let $T$ be the tree searching for $\widetilde{\psi},\widetilde{U}$ and a sequence $\left<\widetilde{M}_{{\bar{\mu}}},\widetilde{\sigma}_{{\bar{\mu}}}\right>_{{\bar{\mu}}\in C}$ such that:

1. –

$\widetilde{U}\triangleleft U|(j(\kappa)^{+})^{U}$** 2. –

$\widetilde{\psi}:{\bar{U}}\to\widetilde{U}$* is c-preserving $m$-lifting,* 3. –

$\widetilde{U}|\psi({\bar{\mu}})\trianglelefteq\widetilde{M}_{\bar{\mu}}\triangleleft\widetilde{U}$. 4. –

$\widetilde{\sigma}_{\bar{\mu}}:{\bar{M}}\to\widetilde{M}_{\bar{\mu}}$* is c-preserving $m$-lifting,* 5. –

$\widetilde{\sigma}_{\bar{\mu}}\!\upharpoonright\!({\bar{\mu}}+1)\subseteq\widetilde{\psi}$.

Now $U\models$“$T$ is illfounded”; in fact $U\models$“$T\cap\mathcal{J}(U^{\prime})$ is illfounded”, because $\psi^{\prime},U^{\prime},\left<M_{\mu},\sigma_{\mu}\right>_{\mu}$ exist, and $U|(\tau^{+})^{U}$ is wellfounded and models $\mathsf{ZFC}^{-}$.

Now $T=j(T^{M})$ for some $T^{M}\in M$, so $M\models$“$T^{M}$ is illfounded”. But then letting $\widetilde{U},\widetilde{\psi},\left<\widetilde{M}_{\bar{\mu}},\widetilde{\sigma}_{\bar{\mu}}\right>_{{\bar{\mu}}\in C}$ witness this, these objects allow us to lift $m$-maximal trees on $\mathfrak{P}$ to $m$-maximal trees on $M$ (here when we use an extender $G$ with $\mathrm{cr}(G)={\bar{\gamma}}<{\bar{\theta}}$, we apply it to ${\bar{M}}$, and our next lifting map is of the form

[TABLE]

where ${\bar{\mu}}=({\bar{\gamma}}^{+})^{\bar{M}}$ and where $i$ is the upper ultrapower map, and $\varphi$ is defined as usual using $\widetilde{\sigma}_{{\bar{\mu}}}$.).

This completes both cases, and hence, the proof that $\mathfrak{P}$ is iterable. ∎

We have ${\bar{M}}|{\bar{\theta}}={\bar{U}}|{\bar{\theta}}$. So comparison of $(\mathfrak{P},M)$ uses only extenders indexed above ${\bar{\theta}}$. So by the claim, there is a successful such comparison $(\mathcal{U},\mathcal{T})$.

Claim 2.

We have:

$M^{\mathcal{U}}_{\infty}=M^{\mathcal{T}}_{\infty}$, $b^{\mathcal{U}},b^{\mathcal{T}}$ do not drop in model or degree, $b^{\mathcal{U}}$ is above ${\bar{U}}$ and $i^{\mathcal{U}}\circ{\bar{j}}=i^{\mathcal{T}}$. 2. 2.

${\bar{\theta}}={\bar{\tau}}$, so $\theta=\tau$. 3. 3.

${\bar{U}}|({\bar{\tau}}^{+})^{\bar{U}}={\bar{M}}||({\bar{\tau}}^{+})^{\bar{U}}$, so $U|(\tau^{+})^{U}=M||(\tau^{+})^{M}$. 4. 4.

If $E$ is Dodd-solid then ${\bar{E}}\in\mathbb{E}_{+}^{{\bar{M}}}$, so $E\in\mathbb{E}_{+}^{M}$.

Proof.

Because ${\bar{M}}$ is $(m+1)$-sound and $\rho_{m+1}^{\bar{M}}=\omega$, standard arguments give part 1.

Part 2: Suppose that ${\bar{\theta}}<{\bar{\tau}}$. Then since $\mathcal{H}_{\bar{\tau}}^{\bar{M}}\subseteq{\bar{U}}$, we have $({\bar{\theta}}^{+})^{\bar{U}}=({\bar{\theta}}^{+})^{\bar{M}}$. But then since $b^{\mathcal{U}}$ is above ${\bar{U}}$ and does not drop,

[TABLE]

contradicting the choice of $\theta$ (and hence ${\bar{\theta}}$).

Part 3: Much as in part 2, but now with ${\bar{\tau}}={\bar{\theta}}$, so $\mathrm{cr}(i^{\mathcal{U}})\geq{\bar{\tau}}$. The conclusion that $U|(\tau^{+})^{U}=M||(\tau^{+})^{U}$ follows from the reflection between ${\bar{M}}$ and $M$ discussed earlier.

Part 4: If ${\bar{E}}\in\mathbb{E}^{\bar{M}}$, note that ${\bar{E}}\in\mathbb{E}(\mathfrak{C}_{0}({\bar{M}}))$, since ${\bar{\tau}}<\rho_{0}^{\bar{M}}$; it easily follows then that $E=\pi({\bar{E}})$, just by the elementarity of $\pi$. Similarly if ${\bar{E}}=F^{\bar{M}}$ then $E=F^{M}$ by elementarity. So we just need to see that ${\bar{E}}\in\mathbb{E}_{+}^{\bar{M}}$, assuming that $E$ is Dodd-solid.

If $t=\emptyset$ then this follows from the ISC as in the proof of the ISC for pseudo-mice. Suppose instead that $E$ is Dodd-solid (hence Dodd-sound) and $t\neq\emptyset$. So as discussed earlier, ${\bar{E}}$ is Dodd-solid with respect to ${\bar{t}}$. Since ${\bar{M}}$ is $1$-sound and iterable, by 2 and as in [9, §2], we can analyse the Dodd-structure of the extenders used in $\mathcal{T}$, decomposing them into Dodd-sound extenders. As there, there is exactly one extender $G=E^{\mathcal{T}}_{\alpha}$ used along $b^{\mathcal{T}}$, $G$ has a largest generator $\gamma$, and $\gamma=i^{\mathcal{U}}(\max({\bar{t}}))$, and there is a unique $\beta\leq_{\mathcal{T}}\alpha$ such that the Dodd-core $D$ of $G$ is in $\mathbb{E}_{+}(M^{\mathcal{T}}_{\beta})$, $\tau_{D}\leq{\bar{\tau}}$, and that if $\beta<_{\mathcal{T}}\alpha$, then letting $\varepsilon+1=\mathrm{succ}^{\mathcal{T}}(\beta,\alpha)$, then $M^{*\mathcal{T}}_{\varepsilon+1}=M^{\mathcal{T}}_{\beta}|\mathrm{lh}(D)$, and letting $k=i^{*\mathcal{T}}_{\varepsilon+1,\alpha}$, then $\mathrm{cr}(k)\geq\tau_{D}$ and

[TABLE]

and

[TABLE]

Note that $\rho_{1}(M^{\mathcal{T}}_{\beta}|\mathrm{lh}(D))\leq\tau_{D}\leq{\bar{\tau}}$.

Suppose $D\neq F^{M^{\mathcal{T}}_{\beta}}$. Then $\beta=0$, as otherwise ${\bar{\tau}}<\mathrm{lh}(E^{\mathcal{T}}_{0})\leq\rho_{1}(M^{\mathcal{T}}_{\beta}|\mathrm{lh}(D))$, contradiction. So $D\in\mathbb{E}^{{\bar{M}}}$. Since ${\bar{\tau}}$ is an ${\bar{M}}$-cardinal, therefore $\tau_{D}={\bar{\tau}}$, so

[TABLE]

so $E=D$, as desired.

Now suppose instead that $D=F^{M^{\mathcal{T}}_{\beta}}$. Then again $\beta=0$, since otherwise ${\bar{\tau}}\leq\lambda(E^{\mathcal{T}}_{0})<\tau_{D}$, contradiction. So $D=F^{\bar{M}}$. We claim that $\alpha=0$, so $G=D$ is Dodd-sound, and it follows then (as in [9]) that $\mathcal{U}$ is trivial and we are done. So suppose $0<_{\mathcal{T}}\alpha$; so $(0,\alpha]_{\mathcal{T}}$ does not drop in model. Let $F^{*}$ be the first extender used along $(0,\alpha]_{\mathcal{T}}$. So ${\bar{\tau}}\leq\nu_{F^{*}}$. Note that

[TABLE]

where $G^{\prime}$ is the active extender of $\mathrm{Ult}_{0}({\bar{M}},F^{*}\!\upharpoonright\!{\bar{\tau}})$. Therefore ${\bar{U}}$ is the iterate of ${\bar{M}}$ given by the tree $\mathcal{T}^{\prime}$ which uses exactly two extenders, $E^{\mathcal{T}^{\prime}}_{0}=F^{*}\!\upharpoonright\!{\bar{\tau}}$ and $E^{\mathcal{T}^{\prime}}_{1}=G^{\prime}$. It follows that $\mathcal{T}=\mathcal{T}^{\prime}$, $\mathcal{U}$ is trivial, $E^{\mathcal{T}}_{0}=F^{*}$, $\nu_{E^{\mathcal{T}}_{0}}={\bar{\tau}}$,

[TABLE]

and $E^{\mathcal{T}}_{1}=G=G^{\prime}$. So $E^{\mathcal{T}}_{0}\neq F^{\bar{M}}$ (as ${\bar{\kappa}}=\mathrm{cr}(E^{\mathcal{T}}_{1})=\mathrm{cr}(D)$ and $D=F^{\bar{M}}$), so

[TABLE]

${\bar{M}}|\bar{\lambda}$ is active with $E^{\mathcal{T}}_{0}$, and ${\bar{\kappa}}<\mathrm{cr}(E^{\mathcal{T}}_{0})$. It follows that $E^{\mathcal{T}}_{0}={\bar{F}}$ from property 11’ above. But then by that property,

[TABLE]

Also ${\bar{t}}=k(t_{D}\backslash\mathrm{cr}(E^{\mathcal{T}}_{0}))$ and

[TABLE]

But then $D\in{\bar{M}}$, contradiction. ∎

This completes the proof of the theorem. ∎

3 Inductive condensation stack: $\mathbb{E}$ from $\mathbb{E}\!\upharpoonright\!\omega_{1}$

In this section we prove Theorem 1. We first give the proofs of some older results, as their methods are then used in the proof of 1. The first is an observation due to Jensen.

Fact 3.0** (Jensen).**

Let $N$ be a premouse of height $\kappa>\omega$, where $\kappa$ is regular. Let $P$ be a sound premouse such that $N\trianglelefteq P$, $\rho_{\omega}^{P}=\kappa$, and $\omega$-condensation holds for $P$. Let $Q$ be likewise. Then $P\trianglelefteq Q$ or $Q\trianglelefteq P$.

Proof.

Suppose not. Taking a hull of $V$, it is easy to find $\bar{P},\bar{Q}$ such that $\bar{P}\ntrianglelefteq\bar{Q}\ntrianglelefteq\bar{P}$ and fully elementary maps $\pi:\bar{P}\to P$ and $\sigma:\bar{Q}\to Q$ and $\bar{\kappa}$ such that

[TABLE]

and $\pi(\bar{\kappa})=\kappa=\sigma(\bar{\kappa})$. So by condensation, either

1. (i)

$\bar{P}\trianglelefteq M|\kappa$ and $\bar{Q}\trianglelefteq M|\kappa$, or 2. (ii)

$M|\bar{\kappa}$ is active and $\bar{P}\trianglelefteq U$ and $\bar{Q}\trianglelefteq U$ where $U=\mathrm{Ult}(M|\bar{\kappa},F^{M|\bar{\kappa}})$.

In either case, it follows that either $\bar{P}\trianglelefteq\bar{Q}$ or $\bar{Q}\trianglelefteq\bar{P}$, a contradiction. ∎

A slight adaptation gives:

Fact 3.0**.**

Let $M$ be a $(0,\omega_{1}+1)$-iterable premouse with no largest proper segment. Let $\kappa>\omega$ be a regular cardinal of $M$. Let $P\in M$ be a sound premouse such that $M|\kappa\trianglelefteq P$, $\rho_{\omega}^{P}=\kappa$, and $\omega$-condensation holds for $P$. Then $P\triangleleft M$.

Proof.

Use the proof above with $Q\trianglelefteq M$ such that $P\in Q$ and $\rho_{\omega}^{Q}=\kappa$. ∎

A slight refinement of this argument gives:

Fact 3.0**.**

Let $M$ be a $(0,\omega_{1}+1)$-iterable premouse. Let $\kappa>\omega$ be a regular cardinal of $M$. Let $P\in M$ be a $(n+1)$-sound premouse such that $M|\kappa\trianglelefteq P$, $\rho_{n+1}^{P}=\kappa$, and $(n+1)$-condensation holds for $P$. Then $P\triangleleft M$.

The second ingredient is an argument of Woodin’s, which is used in the proof of Corollary 1 below. Steel noticed that 1 follows from Theorem 1 combined with Woodin’s argument.

Proof of Corollary 1.

We have that $M$ is $(0,\omega_{1}+1)$-iterable, $\kappa$ is uncountable in $M$ and $(\kappa^{+})^{M}<\mathrm{OR}^{M}$. We want to see that $M|(\kappa^{+})^{M}$ is definable from parameters over $\mathcal{H}=(\mathcal{H}_{\kappa^{+}})^{M}$. There are two cases.

Case 1.

$M$* has no cutpoint in $[\kappa,(\kappa^{+})^{M})$.*

Then there are unboundedly many $\gamma<(\kappa^{+})^{M}$ indexing an $M$-total extender. So by 1, given a premouse $P\in\mathcal{H}$ such that $M|\kappa\trianglelefteq P$ and $\rho_{\omega}^{P}=\kappa$, we have

[TABLE]

iff there is $E\in\mathcal{H}$ such that $P\triangleleft\mathrm{Ult}(M|\kappa,E)$ and $\mathcal{H}\models$“$E$ is a countably complete short extender”. So $M|(\kappa^{+})^{M}$ is definable over $\mathcal{H}$ from the parameter $M|\kappa$, which suffices.

Case 2.

Otherwise ($M$ has a cutpoint $\gamma_{0}\in[\kappa,(\kappa^{+})^{M})$).

The proof in this case is due to Woodin, and was found earlier. Let $X$ be the set of all $H\in\mathrm{HC}^{M}$ such that there is $P\triangleleft M|(\kappa^{+})^{M}$ and $\pi\in M$ such that $\pi:H\to P$ is elementary. Since $(\kappa^{+})^{M}<\mathrm{OR}^{M}$, we have $X\in M$ and $X$ is essentially a subset of $\omega_{1}^{M}$ in $M$. So $X\in\mathcal{H}$. Let $P\in M$ be a sound premouse such that $M|\gamma_{0}\trianglelefteq P$, $\gamma_{0}$ is a cutpoint of $P$ and $\rho_{\omega}^{P}\leq\gamma_{0}$. Then we claim that (i) $P\triangleleft M$ iff (ii)

[TABLE]

it follows that $M|(\kappa^{+})^{M}$ is definable over $\mathcal{H}$ from the parameter $(X,M|\gamma_{0})$, which suffices. Now (i) implies (ii) by definition. So suppose (ii) holds. Let $P\in Q\triangleleft M$, with $\rho_{\omega}^{Q}\leq\gamma_{0}$. Working in $M$, let $Y\preccurlyeq Q$ be countable, with $P\in Y$. The transitive collapses ${\bar{P}}$ of $Y\cap P$ and ${\bar{Q}}$ of $Y$ are in $X$, so can be compared in $V$. But ${\bar{P}}|{\bar{\gamma}}_{0}={\bar{Q}}|{\bar{\gamma}}_{0}$ where ${\bar{\gamma}}_{0}$ is a cutpoint of both ${\bar{P}},{\bar{Q}}$, and ${\bar{P}},{\bar{Q}}$ are sound and project $\leq{\bar{\gamma}}_{0}$. So standard calculations give that ${\bar{P}}\trianglelefteq{\bar{Q}}$, so $P\trianglelefteq Q$.∎

Woodin’s argument above makes use of the parameter $X$. We can actually replace this parameter with $\mathbbm{m}^{M}$:

Lemma 3.0**.**

Let $N$ be an $(0,\omega_{1}+1)$-iterable premouse with no largest proper segment. Let $M\triangleleft N$ and $H\in\mathrm{HC}^{N}$ and $\pi:H\to M$ be elementary with $\pi\in N$. Then there is ${\bar{M}}\triangleleft N|\omega_{1}^{N}$ and an elementary $\bar{\pi}:H\to{\bar{M}}$ with $\bar{\pi}\in N$.

Proof.

Let $M\triangleleft P\triangleleft N$ be such that $\pi\in P$. Let $q\in(\mathrm{OR}^{P})^{<\omega}$ be such that $(\omega,q)$ is $1$-self-solid for $P$ and such that

[TABLE]

Let

[TABLE]

Then by 2, ${\bar{P}}\triangleleft N|\omega_{1}^{N}$. Let $\sigma:{\bar{P}}\to P$ be the uncollapse. Then $\sigma(H)=H$. Let $\sigma(\bar{\pi})=\pi$ and $\sigma({\bar{M}})=M$. Then ${\bar{M}}\triangleleft{\bar{P}}$ and $\bar{\pi}:H\to{\bar{M}}$ elementarily, so we are done. ∎

Similarly:

Lemma 3.0**.**

Let $N$ be a $(0,\omega_{1}+1)$-iterable premouse. Let $M\triangleleft N$ and $H\in\mathrm{HC}^{N}$ and $m<\omega$ and $\pi:H\to M$ be an $m$-lifting ((weak, near) $m$-embedding respectively)* with $\pi\in N$. Then there is ${\bar{M}}\triangleleft N|\omega_{1}^{N}$ and an $m$-lifting ((weak, near) $m$-embedding respectively*)* $\bar{\pi}:H\to{\bar{M}}$ with $\bar{\pi}\in N$.*

Proof.

Consider the case that $N=\mathcal{J}(M)$ and $\pi:H\to M$. Then there is $k<\omega$ and $x\in M$ such that $\pi$ is $\mathrm{r}\Sigma_{k}^{M}(\{x\})$. Argue as in the proof of 3, but at degree $n$ instead of $1$, with $n>k+m+5$. ∎

Woodin’s argument above is abstracted into the following definition:

Definition 3.1**.**

Let $M$ be a $(0,\omega_{1}+1)$-iterable premouse satisfying “$\omega_{1}$ exists”, with no largest proper segment. Then $\mathrm{css}^{M}$ (countable substructures) denotes the set of all $H\in\mathrm{HC}^{M}$ such that for some $P\triangleleft M$, there is $\pi\in M$ such that $\pi:H\to P$ is elementary. (So by 3, $\mathrm{css}^{M}$ is definable over $\mathbbm{m}^{M}$, uniformly in $M$.) Let $P,Q\in M$ be sound premice. Working in $M$, say that $P$ is $\mathbbm{m}^{M}$-verified iff every countable elementary substructure of $P$ is in $\mathrm{css}^{M}$, and say that $Q$ is an $(\mathbbm{m}^{M},P)$-lower part premouse iff $P\trianglelefteq Q$, $P$ is a cutpoint of $Q$, $Q$ projects to $P$ and $Q$ is $\mathbbm{m}^{M}$-verified. Working over $M$, the stack of all $(\mathbbm{m}^{M},P)$-lower part premice is denoted $\mathrm{Lp}^{M}_{\mathbbm{m}}(P)$.

Note that $\mathrm{Lp}^{M}_{\mathbbm{m}}(P)$ is definable over $\left\lfloor M\right\rfloor$ from $\mathbbm{m}^{M},P$; the fact that it forms a stack follows from the proof of 1.

In order to prove Theorem 1, it easily suffices to prove that if $M$ is passive, $(0,\omega_{1}+1)$-iterable and satisfies $\mathsf{ZFC}^{-}+$“$\omega_{1}$ exists”, then $\mathbb{E}^{M}$ is definable over $\left\lfloor M\right\rfloor$ from $\mathbb{E}^{M}\!\upharpoonright\!\omega_{1}^{M}$, uniformly in $M$. We will in fact prove a stronger fact, Theorem 3 below, making do with less than $\mathsf{ZFC}^{-}$. We may assume that $M$ has a largest cardinal $\theta$. The proof breaks into different cases, depending on the nature of $M$ above $\theta$. Clearly the cases are not mutually exclusive (Case (c)1 is in fact subsumed by Case (c)4); the cases describe situations in which certain methods of proof work.

Definition 3.2**.**

Let $M$ be a premouse. Let $\kappa<\theta$ be cardinals of $M$. We say that $\kappa$ is $\mathcal{H}_{\theta}$-strong in $M$ iff there is $E\in M$ such that $M\models$“$E$ is a countably complete short extender” and $\mathrm{cr}(E)=\kappa$ and $\mathcal{H}^{M}_{\theta}\subseteq\mathrm{Ult}(M,E)$.

Definition 3.3**.**

A premouse $M$ is eventually trivial iff $M=\mathcal{J}_{\alpha}(R)$ for some $R\triangleleft M$ and $\alpha>0$.

Remark 3.3**.**

In the theorem statement below, in each case we specify definability classes $\Gamma,\Lambda$. The case specification is $\Gamma^{\left\lfloor M\right\rfloor}(\{M|\theta\})$, meaning that there is a $\Gamma$ formula $\varphi$ such that for any $(0,\omega_{1}+1)$-iterable premouse $M$ satisfying “$\omega_{1}$ exists and $\theta$ is the largest cardinal”, the case hypothesis holds of $M$ iff $\left\lfloor M\right\rfloor\models\varphi(M|\theta)$. In the given case, the definition of $\mathbb{E}^{M}$ is $\Lambda^{\left\lfloor M\right\rfloor}(\{M|\theta\})$. (The fact that the case specification is definable is obviously used in defining $M|\theta$ from $\mathbbm{m}^{M}$ over $\left\lfloor M\right\rfloor$.)

Definition 3.4**.**

Let $M$ be a passive premouse with a largest cardinal $\theta\geq\omega_{1}^{M}$. We say that $M$ is tractable iff either (i) $\theta$ is regular in $M$, or (ii) $\theta$ is a cutpoint of $M$, or (iii) $M$ has no cutpoint in $[\theta,\mathrm{OR}^{M})$, or (iv) $\operatorname{cof}^{M}(\theta)>\omega$, or (v) $M\models$“$\theta$ is not a limit of cardinals which are $\mathcal{H}_{\theta}$-strong”, or (vi) $\operatorname{cof}^{\utilde{\Sigma}_{2}^{\left\lfloor M\right\rfloor}}(\mathrm{OR}^{M})>\omega$, or (vii) [$\operatorname{cof}^{\utilde{\Sigma}_{1}^{\left\lfloor M\right\rfloor}}(\mathrm{OR}^{M})>\omega$ and $M$ is eventually trivial].

Theorem 3.4**.**

Let $M$ be a passive $(0,\omega_{1}+1)$-iterable premouse satisfying “$\omega_{1}$ exists”. Then:

1. (a)

If $M$ is tractable then $\mathbb{E}^{M}$ is $\Sigma_{4}^{\left\lfloor M\right\rfloor}(\{\mathbbm{m}^{M}\})$, uniformly in such $M$. 2. (b)

If $\left\lfloor M\right\rfloor\models\mathsf{PS}$ then $\mathbb{E}^{M}$ is $\Sigma_{2}^{\left\lfloor M\right\rfloor}(\{\mathbbm{m}^{M}\})$, uniformly in such $M$. 3. (c)

In fact, suppose that $M$ has largest cardinal $\theta$ and either:

$\theta$* is regular in $M$; and let $(\Gamma,\Lambda)=(\Pi_{1},\Sigma_{1})$, or* 2. 2.

$\theta$* is a cutpoint of $M$; let $(\Gamma,\Lambda)=(\Pi_{2},\Sigma_{2})$, or* 3. 3.

$M$* has no cutpoint in $[\theta,\mathrm{OR}^{M})$; let $(\Gamma,\Lambda)=(\Pi_{3},\Sigma_{2})$, or* 4. 4.

$\operatorname{cof}^{M}(\theta)>\omega$; let $(\Gamma,\Lambda)=(\Pi_{1},\Sigma_{1})$, or 5. 5.

$M\models$“$\theta$* is not a limit of $\mathcal{H}_{\theta}$-strong cardinals”; let $(\Gamma,\Lambda)=(\Sigma_{3},\Sigma_{1})$, or* 6. 6.

$\operatorname{cof}^{\utilde{\Sigma}_{2}^{\left\lfloor M\right\rfloor}}(\mathrm{OR}^{M})>\omega$*;*141414By $\operatorname{cof}^{\utilde{\Sigma}_{n}^{\left\lfloor M\right\rfloor}}(\mathrm{OR}^{M})$, we mean the least ordinal $\mu$ such that there is a total unbounded function $f:\mu\to\mathrm{OR}^{M}$ which is $\utilde{\Sigma}_{n}^{\left\lfloor M\right\rfloor}$-definable. Note that this is standard $\Sigma_{n}$, not $\mathrm{r}\Sigma_{n}$.let $(\Gamma,\Lambda)=(\Pi_{5},\Sigma_{4})$, or 7. 7.

$\operatorname{cof}^{\utilde{\Sigma}_{1}^{\left\lfloor M\right\rfloor}}(\mathrm{OR}^{M})>\omega$* and $M$ is ev. trivial; let $(\Gamma,\Lambda)=(\Pi_{3}\wedge\Sigma_{3},\Sigma_{3})$.*

Then $\mathbb{E}^{M}$ is $\Lambda^{\left\lfloor M\right\rfloor}(\{M|\theta\})$, and the case specification is $\Gamma^{\left\lfloor M\right\rfloor}(\{M|\theta\})$, both uniformly in such $M$.

Proof of Theorem 3.

Parts (a) and (b) follow immediately from part (c) by an easy induction on $M$-cardinals.

Part (c): We split into the cases given in the statement of this part. In each case we will give a characterization of $\mathbb{E}^{M}$ and leave to the reader the verification of the precise degree of definability. Note that for the definability of the case specification, we use 1 to determine, for example, whether or not $\theta$ is a cutpoint of $M$.

Case 1.

$\theta$* is regular in $M$.*

By 3, working in $M$, given any premouse $P$, we have $P\triangleleft M$ iff there is a sound premouse $Q$ and $n<\omega$ such that $P\triangleleft Q$ and $\rho_{n+1}^{Q}=\theta$ and $M|\theta\trianglelefteq Q$ and $Q$ satisfies $(n+1)$-condensation. And $\mathbb{E}^{M}$ is the stack of all structures of the form $\mathcal{S}_{m}(P)$ for such $P$ and $m<\omega$.

Case 2.

$\theta$* is a cutpoint of $M$.151515 The case specification is $\Pi_{2}$ because $\theta$ is a cutpoint of $M$ iff for all $E,\mathcal{H}\in M$, if $M\models$“$\mathcal{H}=\mathcal{H}_{\theta}^{M}$ and $E$ is a pre-extender with $\mathcal{H}\subseteq\mathrm{Ult}(M,E)$” then $M\models$“$E$ is not countably complete”; if $\left\lfloor M\right\rfloor$ is admissible then $\Pi_{1}$ suffices for the case specification, because we can replace the requirement that $M\models$“$E$ is not countably complete” with the requirement that “$M\models\mathrm{Ult}(M|(\kappa^{+})^{M},E)$ is illfounded”, where $\kappa=\mathrm{cr}(E)$.*

Use the proof of Corollary 1, or an obvious adaptation thereof if $M=\mathcal{J}(R)$, combined with 3 and 3.

Case 3.

$M$* has no cutpoint in $[\theta,\mathrm{OR}^{M})$.*

Use 1.

Case 4.

$\operatorname{cof}^{M}(\theta)>\omega$.

Let $P\in M$ and $n<\omega$ be such that $P$ is a sound premouse, $M|\theta\trianglelefteq P$, $\rho_{n+1}^{P}=\theta$, and $P$ satisfies $(n+1)$-condensation. We claim that $P\triangleleft M$; clearly this suffices. If $\theta$ is regular in $M$ we can use the proof of Case (c)1, so suppose otherwise; in particular, $\theta$ is a limit cardinal of $M$.

We prove that $P\triangleleft M$ using a phalanx comparison. Let $Q\triangleleft M$ and $x\in Q$ and $m<\omega$ be such that $P$ is $\mathrm{r}\Sigma_{m}^{Q}(\{x\})$; in particular, $\mathrm{OR}^{P}\leq\mathrm{OR}^{Q}$. We must show that $P\trianglelefteq Q$. Suppose not; note that the fact that $P\ntrianglelefteq Q$ is first-order over $Q$ (in the parameter $x$). So we may assume that $x=\emptyset$ (increasing $m$ if needed). Let $m+n+5<k<\omega$ and let $\bar{Q}=\mathrm{cHull}_{k+1}^{Q}(\emptyset)$. Then $\bar{Q}\triangleleft M$. Let $\bar{P}$ be defined over $\bar{Q}$ as $P$ is over $Q$. Let $\pi:\bar{Q}\to Q$ be the uncollapse, and $\pi(\bar{\theta})=\theta$. Then $\bar{P}$ is $(n+1)$-sound and $\rho_{n+1}^{\bar{P}}=\bar{\theta}$, $\bar{Q}$ is $\omega$-sound and $\rho_{k+1}^{\bar{Q}}=\omega$, $\bar{P}|\bar{\theta}=\bar{Q}|\bar{\theta}$, and $\bar{\theta}$ is a cardinal of both models.

Define the phalanx $\mathfrak{P}=((\bar{Q},k,\bar{\theta}),(\bar{P},n),\bar{\theta})$. By the following claim, a standard comparison argument (comparing $\mathfrak{P}$ with $\bar{Q}$) shows that $\bar{P}\trianglelefteq\bar{Q}$, so $P\trianglelefteq Q$, a contradiction, completing the proof.

Claim 3.

$\mathfrak{P}$* is $(\omega_{1}+1)$-iterable.*

Proof.

Let $\sigma:\bar{P}\to P$ be $\pi\!\upharpoonright\!\bar{P}$. Then $\sigma\!\upharpoonright\!\bar{\theta}=\pi\!\upharpoonright\!\bar{\theta}$. Let $\eta=\sup\pi``\bar{\theta}$. Then $\eta<\theta$ because $\operatorname{cof}^{M}(\theta)>\omega$. Because $\theta$ is a limit cardinal of $M$, so is $\eta$. Let

[TABLE]

and $\pi^{\prime}:P^{\prime}\to P$ be the uncollapse. Then $P^{\prime}$ is $(n+1)$-sound, $\rho_{n+1}^{P^{\prime}}=\eta$ and

[TABLE]

For clearly $\rho_{n+1}^{P^{\prime}}\leq\eta$. Using that $k>n+m+5$, note that $P^{\prime}$ is $(n+1,q)$-solid. We have $P^{\prime}|\eta=M|\eta$ and $P^{\prime}\in M$, and as $\eta$ is an $M$-cardinal, therefore $\rho_{n+1}^{P^{\prime}}=\eta$ and $q=p_{n+1}^{P^{\prime}}$. So we can apply $(n+1)$-condensation, and note then that $P^{\prime}\triangleleft M$.

Let $\sigma^{\prime}:\bar{P}\to P^{\prime}$ be the natural factor map. Then $\sigma^{\prime}$ is a near $n$-embedding, and $\sigma^{\prime}\!\upharpoonright\!\bar{\theta}=\pi\!\upharpoonright\!\bar{\theta}$. Using $(\pi,\sigma^{\prime})$, one can lift normal trees on $\mathfrak{P}$ to normal trees on $\bar{Q}$, completing the proof. ∎

Case 5.

$M\models$“$\theta$ is not a limit of cardinals which are $\mathcal{H}_{\theta}$-strong”.

This is almost the same as the previous case; we leave it to the reader.

The remaining cases are more subtle than the previous ones. We (may) now make the:

Assumption 1**.**

$\theta$ is a limit cardinal of $M$ and $M$ has a cutpoint in $[\theta,\mathrm{OR}^{M})$.

This must of course be incorporated appropriately into the $\Sigma_{4}(\{M|\theta\})$ (in case (c)6) and $\Sigma_{3}(\{M|\theta\})$ (in case (c)7) definitions one forms from the arguments to follow. But given the definability $(\Sigma,\Lambda)$ established for cases (c)1 and (c)3, this is no problem. (Note here that in case (c)7, $M$ does have a cutpoint $\geq\theta$, so the $\Pi_{3}$ complexity of asserting the non-existence of a cutpoint is not relevant in this case.)

Case 6.

$\operatorname{cof}^{\utilde{\Sigma}_{2}^{\left\lfloor M\right\rfloor}}(\mathrm{OR}^{M})>\omega$.

Work in $M$ and let $P$ be a premouse. Say that $P$ is good iff $P$ is sound, $M|\theta\trianglelefteq P$ and $\rho_{\omega}^{P}=\theta$. Say that $P$ is excellent iff

1. –

$P$* is good,* 2. –

$M$* and $\mathrm{Lp}^{M}_{\mathbbm{m}}(P)$ have the same universe, and* 3. –

$1$-condensation holds for every $Q\triangleleft\mathrm{Lp}^{M}_{\mathbbm{m}}(P)$.

By the case hypothesis, $M$ has no largest proper segment, so with Assumption 1, it follows that there are cofinally many excellent $N\triangleleft M$. Therefore it suffices to prove the following claim:

Claim 4.

Let $P,Q\in M$ be excellent. Then either $P\trianglelefteq Q$ or $Q\trianglelefteq P$.

Proof.

We may assume $Q\triangleleft M$ and $\mathrm{OR}^{Q}$ is a cutpoint of $M$, so $\mathrm{Lp}^{M}_{\mathbbm{m}}(Q)=M$. Define $\left<P_{n},Q_{n}\right>_{n<\omega}$ as follows. Let $P_{0}=P$ and $Q_{0}=Q$. Given $P_{n},Q_{n}$, let $Q_{n+1}$ be the least $N\triangleleft M$ such that $N$ is good, $Q_{n}\triangleleft N$ and $P_{n}\in N$. Given $P_{n},Q_{n+1}$, let $P_{n+1}$ be the least $R\triangleleft\mathrm{Lp}^{M}_{\mathbbm{m}}(P)$ such that $R$ is good, $P_{n}\triangleleft R$ and $Q_{n+1}\in R$.

Let $\widetilde{P}=\mathrm{stack}_{n<\omega}P_{n}$ and $\widetilde{Q}=\mathrm{stack}_{n<\omega}Q_{n}$. Note that $\widetilde{P}$ and $\widetilde{Q}$ have the same universe $U$ (but ostensibly may have different extender sequences). We have $\mathrm{OR}^{U}<\mathrm{OR}^{M}$ by our case hypothesis, as $\left<P_{n},Q_{n}\right>_{n<\omega}$ is $\Sigma_{2}^{\left\lfloor M\right\rfloor}(\{P,Q\})$.161616 It seems that $\Sigma_{1}$ is not in general enough, because to ensure that, for example, $P_{n}\triangleleft\mathrm{Lp}^{M}_{\mathbbm{m}}(P)$, requires a $\forall$-quantifier in order to deal with arbitrary countable substructures of $P_{n}$; note that if $\operatorname{cof}^{M}(\theta)>\omega$, one can dispense with this quantifier, however, as one can code the substructures via bounded subsets of $\theta$. Now $\widetilde{P}$ is definable over $U$ from the parameter $P$, and likewise $\widetilde{Q}$ over $U$ from $Q$; in fact,

[TABLE]

(Clearly cofinally many segments of $\widetilde{P}$ satisfy the requirements for premice in $\mathrm{Lp}^{U}_{\mathbbm{m}}(P)$; but if $R$ is some premouse satisfying these requirements then working in $U$, we can run the same proof as before to see that $R\triangleleft\mathrm{Lp}^{U}_{\mathbbm{m}}(P)$.) Also, $U$ has largest cardinal $\theta$, so $\mathrm{Lp}^{M}_{\mathbbm{m}}(P)|\mathrm{OR}^{U}$ and $M|\mathrm{OR}^{U}$ are both passive. So letting $P^{+}=\mathcal{J}(\widetilde{P})$ and $Q^{+}=\mathcal{J}(\widetilde{Q})$, we have $P^{+}\triangleleft\mathrm{Lp}^{M}_{\mathbbm{m}}(P)$ and $Q^{+}\triangleleft M$ and (because $\widetilde{P},\widetilde{Q}$ are definable from parameters over $U$),

[TABLE]

Also because $\mathrm{OR}^{U}$ has cofinality $\omega$, definably over $U$ from parameters, we have

[TABLE]

We claim that there is an $M$-cardinal $\gamma<\theta$ such that

[TABLE]

(“$\mathrm{Hull}$” denotes the uncollapsed hull), and the transitive collapses $\bar{P}^{+},\bar{Q}^{+}$ are $1$-sound and such that $\rho_{1}^{\bar{P}^{+}}=\gamma=\rho_{1}^{\bar{Q}^{+}}$. For recalling that $\theta$ is a limit cardinal of $M$, let $\gamma<\theta$ be an $M$-cardinal large enough that, defining $H,J$ as above, we have

[TABLE]

(recall $w_{1}^{P^{+}},w_{1}^{Q^{+}}$ are the $1$-solidity witnesses for $P^{+},Q^{+}$). Then because $\gamma$ is an $M$-cardinal and $w_{1}^{P^{+}}\in H$, we easily have that $\rho_{1}^{\bar{P}^{+}}=\gamma$ and $\bar{P}^{+}$ is $1$-sound, and likewise for $\bar{Q}^{+}$. And because

[TABLE]

and $P^{+},Q^{+}$ have the same universe, we have $J\subseteq H$. Similarly $H\subseteq J$, giving line (4).

By $1$-condensation for $P^{+},Q^{+}$ (a requirement of excellence), and because $\rho_{1}^{\bar{P}^{+}}=\gamma=\rho_{1}^{\bar{Q}^{+}}$ is an $M$-cardinal, we have $\bar{P}^{+}\triangleleft M$ and $\bar{Q}^{+}\triangleleft M$. By line (4), $\mathrm{OR}^{\bar{P}^{+}}=\mathrm{OR}^{\bar{Q}^{+}}$. Therefore $\bar{P}^{+}=\bar{Q}^{+}$. It easily follows that $\widetilde{P}=\widetilde{Q}$, giving the claim. ∎

Case 7.

$\operatorname{cof}^{\utilde{\Sigma}_{1}^{\left\lfloor M\right\rfloor}}(\mathrm{OR}^{M})>\omega$* and $M$ is eventually trivial.*

A simplification of the argument in the previous case shows that the collection of all $R\triangleleft M$ such that $M=\mathcal{J}_{\alpha}(R)$ for some $\alpha>0$, is $\Pi_{2}^{\left\lfloor M\right\rfloor}(\{M|\theta\})$. Regarding the complexity of the case specification, it is $\Sigma_{3}^{\left\lfloor M\right\rfloor}$ to assert “$M$ is eventually trivial”, as it is equivalent to

[TABLE]

(as if $M$ is not eventually trivial then $M$ is closed under sharps).

This completes all cases and hence, the proof of the theorem. ∎

Definition 3.5**.**

Let $M$ be a transitive structure. Let $\mathbbm{m}\in M$ be a premouse with $\left\lfloor\mathbbm{m}\right\rfloor=\mathrm{HC}^{M}$. The inductive condensation stack of $M$ above $\mathbbm{m}$ is the stack of premice in $M$, extending $\mathbbm{m}$, satisfying the inductive definition used in the proof of 3.

Of course, the inductive condensation stack $S$ could have $\mathrm{OR}^{S}<\mathrm{OR}^{M}$. But if $M$ is a $(0,\omega_{1}+1)$-iterable tractable premouse and $m=M|\omega_{1}^{M}$ then $M=S$.)

Remark 3.5**.**

In Case (c)3 of the preceding proof, it appeared that we used 1 for extenders $E$ generated by $\theta\cup t$ for some finite set $t$ of generators (in order that we can represent arbitrary segments $R\triangleleft M|(\theta^{+})^{M}$). Actually, it suffices to consider only extenders $E$ such that $\nu_{E}=\theta$ (and $\mathcal{H}_{\theta}^{M}\subseteq\mathrm{Ult}(M,E)$ etc). For we claim that (under the case hypothesis) there are unboundedly many $\beta<\mathrm{OR}^{M}$ such that $M|\beta$ is active with an extender $E$ such that $\nu_{E}=\theta$; clearly this suffices.

For let $Q\triangleleft M$ be such that $\rho_{\omega}^{Q}=\theta$ and let $\alpha$ be least such that $\alpha>\mathrm{OR}^{Q}$ and $M|\alpha$ is active with extender $F$ and $\kappa=\mathrm{cr}(F)<\theta$. We claim that $\nu_{F}=\theta$. So suppose that $\theta<\nu_{F}$. Easily by the ISC, $\theta$ is the largest cardinal of $M|\alpha$. So $F$ is type 2. Let $E=F\!\upharpoonright\!\theta$ and let

[TABLE]

be the standard factor map. So $\mathrm{cr}(\pi)$ is the least generator $\gamma$ of $F$ with $\gamma\geq\theta$.

Suppose $\gamma=\theta$. It follows easily that $\theta$ is a limit cardinal of $M$ and $\mathrm{Ult}(M,E)$, so

[TABLE]

By the ISC, $\kappa$ is then ${<\theta}$-strong in $M$, hence likewise in $\mathrm{Ult}(M,E)$. Therefore $\kappa$ is ${<\pi(\theta)}$-strong in $\mathrm{Ult}(M,F)$. But then again by the ISC, there are unboundedly many $\xi<(\theta^{+})^{\mathrm{Ult}(M,F)}$ indexing an extender $G$ with $\mathrm{cr}(G)=\kappa$, contradicting the minimality of $F$.

Now suppose $\gamma>\theta$. Then because

[TABLE]

we have $\gamma=\mathrm{lh}(E)$ and $\pi(\gamma)=\mathrm{lh}(F)$. But $E\in\mathbb{E}(\mathrm{Ult}(M,F))$, so by reflection, there are unboundedly many $\xi<\mathrm{lh}(E)$ such that $M|\xi$ is active with an extender $G$ with $\mathrm{cr}(G)=\kappa$, and so the same holds of $\pi(\mathrm{lh}(E))=\mathrm{lh}(F)$, again contradicting the minimality of $F$.

Remark 3.5**.**

Let $M$ be passive, $(0,\omega_{1}+1)$-iterable, satisfying “$\omega_{1}$ exists” and $\theta=\mathrm{lgcd}(M)$. We sketch, in a further case, the identification of $M$ from parameter $M|\theta$ over $\left\lfloor M\right\rfloor$. However, here we do not know whether the case specification itself is uniformly definable over $\left\lfloor M\right\rfloor$ as above. Say that $M$ is $\utilde{\mathrm{r}\Sigma}_{1}$-bounded iff $\mathrm{Hull}_{1}^{M}(\alpha\cup\{x\})$ is bounded in $\mathrm{OR}^{M}$ for every $\alpha<\rho_{1}^{M}$ and $x\in M$. Suppose that $M$ is $1$-sound and $\rho_{1}^{M}>\omega$, and $M$ is eventually trivial, or $M$ is not $\utilde{\mathrm{r}\Sigma}_{1}$-bounded. Then $M$ is definable from $M|\theta$ over $\left\lfloor M\right\rfloor$.

To see this, we argue much as in the last two cases of 3. We may make Assumption 1. If $M$ is eventually trivial things are easier (using then either the argument from Case (c)7 of 3 if $\operatorname{cof}^{\utilde{\Sigma}_{1}^{\left\lfloor M\right\rfloor}}(\mathrm{OR}^{M})>\omega$, or a variant of the argument to follow otherwise), so we leave this case to the reader, and suppose otherwise. So $M$ is closed under sharps and has no largest proper segment. The difference to Case (c)6 of 3 is that now, when we define $\widetilde{P},\widetilde{Q}$, we might have $\left\lfloor M\right\rfloor=\left\lfloor\widetilde{P}\right\rfloor=\left\lfloor\widetilde{Q}\right\rfloor$. Let $P\in M$ be good (good defined as before). Say that $P$ is excellent iff $P$ satisfies the conditions of excellence from before, and letting $P^{*}=\mathrm{Lp}_{\mathbbm{m}}^{M}(P)$, then $P^{*}$ is $1$-sound, $\rho_{1}^{P^{*}}>\omega$, $P^{*}$ is not $\utilde{\mathrm{r}\Sigma}_{1}$-bounded, $1$-condensation holds for $P^{*}$, and for all $R\trianglelefteq P^{*}$, if

[TABLE]

then for all sufficiently large $\gamma<\rho$,

[TABLE]

(so $1$-condensation applies to the uncollapse map). By 2, all sufficiently large $Q\triangleleft M$ are excellent; we take $Q$ such.

Let $P\in M$ be excellent. We claim that $\rho_{1}^{M}=\rho_{1}^{P^{*}}$. For suppose $\rho_{1}^{M}>\rho_{1}^{P^{*}}$. Let $\alpha\in[\rho_{1}^{P^{*}},\rho_{1}^{M})$ be large enough that

[TABLE]

is unbounded in $\mathrm{OR}^{M}$ (using non-$\utilde{\mathrm{r}\Sigma}_{1}$-boundedness) and $P,p_{1}^{P^{*}}\in H$. Then $P^{\prime}\in H$ for cofinally many $P^{\prime}\triangleleft P^{*}$. For given $\eta_{0},\eta_{1}\in H\cap\mathrm{OR}^{M}$ such that there is a good $P^{\prime}\triangleleft P^{*}$ with $\eta_{0}\leq\mathrm{OR}^{P^{\prime}}$ and $P^{\prime}\in M|\eta_{1}$, then the least good $P^{\prime\prime}\triangleleft P^{*}$ such that $\eta_{0}\leq\mathrm{OR}^{P^{\prime\prime}}$, is in $H$ (in order to ensure that the selected $P^{\prime\prime}\triangleleft P^{*}$, one can restrict their attention to all countable elementary substructures of $P^{\prime\prime}$ which appear in some reasonable segment of $M$; because $M$ is closed under sharps, there are plenty of very closed segments). It follows that

[TABLE]

But $P^{*}$ is $1$-sound and $\left\lfloor P^{*}\right\rfloor=\left\lfloor M\right\rfloor$, so $M=H$, contradicting the fact that $\alpha<\rho_{1}^{M}$. So $\rho_{1}^{M}\leq\rho_{1}^{P^{*}}$ and the converse is likewise.

The rest is much like the last part of the argument used in Case (c)6, but if $\widetilde{P}=P^{*}$ and $\widetilde{Q}=M$, there is a wrinkle. In this case, choose $\alpha<\rho_{1}^{M}=\rho_{1}^{P^{*}}$ such that

[TABLE]

by arguing as in the previous paragraph, and such that the transitive collapses $\bar{P},\bar{Q}$ of the hulls are $1$-sound (using 2 and excellence if $\rho_{1}^{M}=(\kappa^{+})^{M}$). Then by $1$-condensation we get $\bar{P}=\bar{Q}$, so $P=Q$.

Corollary 3.5**.**

Let $M$ be a $(0,\omega_{1}+1)$-iterable premouse satisfying either $\mathsf{PS}$ or $\mathsf{ZFC}^{-}+$“$\omega_{1}$ exists”. Suppose that either:

$\mathbbm{m}^{M}$* is $(\omega,\omega_{1}+1)$-iterable in $M$,171717If $\mathrm{OR}^{M}=\omega_{2}^{M}$ then this statement should be interpreted as “There is an $(\omega,\omega_{1})$-strategy $\Sigma$ for $M|\omega_{1}^{M}$ such that for every tree $\mathcal{T}$ via $\Sigma$ of length $\omega_{1}$, there is a $\mathcal{T}$-cofinal branch”. or* 2. 2.

$\mathbbm{m}^{M}$* is built by the181818Here one can naturally impose various other restrictions on the construction, but it should be uniquely specified somehow. maximal fully backgrounded $L[\mathbb{E}]$-construction of $M$ using background extenders $E\in\mathbb{E}^{M}$ such that $\nu_{E}$ is an $M$-cardinal.*

Then $\mathbb{E}^{M}$ is definable over $\left\lfloor M\right\rfloor$ without parameters, so if $\left\lfloor M\right\rfloor\models\mathsf{ZFC}$ then $\left\lfloor M\right\rfloor\models$“$V=\mathrm{HOD}$”.

Proof.

Part 1 follows from 3. For part 2, note that by 1, if $E\in M$ then we have that (i) $E\in\mathbb{E}^{M}$ and $\nu_{E}$ is an $M$-cardinal iff (ii) $M\models$“$E$ is a countably complete extender, $\nu_{E}$ is a cardinal and $\mathcal{H}_{\nu_{E}}\subseteq\mathrm{Ult}(V,E)$”. So the $L[\mathbb{E}]$-construction using these background extenders is definable over $\left\lfloor M\right\rfloor$ without parameters, so $\mathbbm{m}^{M}$ is likewise definable. ∎

Recall that $M_{\mathsf{wlim}}$ is the least proper class mouse with a Woodin limit of Woodins. Part 2 of the previous corollary gives:

Corollary 3.5**.**

$\left\lfloor M_{\mathsf{wlim}}\right\rfloor\models$“$V=\mathrm{HOD}$”.

There are of course many variants of this corollary. Using the background construction of [6] in place of the background construction used above, one gets that $\left\lfloor M\right\rfloor\models$“$V=\mathrm{HOD}$” where $M$ is, for example, the least proper class mouse with a $\lambda$ which is a limit of Woodins and strong cardinals.

4 Direct condensation stacks in $M[G]$

In this section we prove the following theorem, using a variant of the inductive condensation stack:

Theorem 4.0**.**

Let $M$ be a $(0,\omega_{1}+1)$-iterable premouse satisfying $\mathsf{PS}$. Let $\theta<\mathrm{OR}^{M}$ be a regular cardinal of $M$ and $\mathbb{P}\in M|\theta$ be a poset. Let $G$ be $(M,\mathbb{P})$-generic. Then $\mathbb{E}^{M}$ is definable over $M[G]$ from the parameter $M|\theta$.

Proof.

Work in $M[G]$. It suffices to give a definition of $M|(\eta^{+})^{M}$ from the parameter $M|\theta$, uniformly in $M$-regular cardinals $\eta\geq\theta$. Note that the Jensen stack over $M|\eta$ is exactly $M|\eta^{+}$, and this structure satisfies standard condensation facts.

Say that a premouse $P$ is excellent iff $M|\theta\trianglelefteq P$, $\mathrm{OR}^{P}=\eta$, the Jensen stack $P^{+}$ over $P$ has height $\eta^{+}$, $P^{+}$ satisfies standard condensation facts, and there is $\mathbb{Q}\in P|\theta$ and a $(P,\mathbb{Q})$-generic filter $h$ such that $P^{+}[h]$ has universe $\mathcal{H}_{\eta^{+}}$.

Clearly the following claim completes the proof:

Claim 5.

$M|\eta$* is the unique excellent premouse.*

Proof.

Clearly $N=M|\eta$ is excellent (with $N^{+}=M|\eta^{+}$), as witnessed by $\mathbb{P},g$.

So let $R$ also be excellent, as witnessed by $\mathbb{Q},h$. Define a sequence $\left<N_{n},R_{n}\right>_{n<\omega}$ as follows. Let $N_{0}=N$ and $R_{0}=R$. Given $N_{n},R_{n}$, let $N_{n+1}$ be the least $N^{\prime}$ such that $N_{n}\triangleleft N^{\prime}\triangleleft N^{+}$ and $\rho_{\omega}^{N^{\prime}}=\eta$ and $R_{n},h\in N^{\prime}[g]$; then let $R_{n+1}$ be the least $R^{\prime}$ such that $R_{n}\triangleleft R^{\prime}\triangleleft R^{+}$ and $\rho_{\omega}^{R^{\prime}}=\eta$ and $N_{n+1},g\in R^{\prime}[h]$. Let $N_{\omega}=\mathrm{stack}_{n<\omega}N_{n}$ and $\widetilde{N}=\mathcal{J}(N_{\omega})$, and $R_{\omega},\widetilde{R}$ likewise. Then $\widetilde{N}\triangleleft N^{+}$ and $\widetilde{R}\triangleleft R^{+}$. Note that $N_{\omega}[g]$ and $R_{\omega}[h]$ have the same universe $U$, and $N_{\omega},R_{\omega}$ are both definable from parameters over $U$ (via the Jensen stack). Hence, $\mathcal{J}(N_{\omega}[g])$ and $\mathcal{J}(R_{\omega}[h])$ and $\widetilde{N}[g]$ and $\widetilde{R}[h]$ all have the same universe $\widetilde{U}=\mathcal{J}(U)$.

Now $\widetilde{N},\widetilde{R}$ both satisfy standard $1$-condensation facts. Let $\gamma<\theta$ be a cardinal of the models $N,N[g],R[h],R$ such that $\mathbb{P},\mathbb{Q}$ have cardinality $\leq\gamma$ in $N,R$ respectively.

Subclaim 1.

For all $x\in\widetilde{U}$ there is $q\in[\mathrm{OR}^{\widetilde{U}}]^{<\omega}$ such that the hulls $H,H^{\prime},J,J^{\prime}$ all contain the same ordinals, where

[TABLE]

[TABLE]

and moreover, $\mathbb{P}\in H$, $\mathbb{Q}\in J$, $x\in H^{\prime}=J^{\prime}$, and the transitive collapses $C,D$ of $H,J$ respectively are sound.

Assuming the subclaim, let $\pi:C\to H$ and $\sigma:D\to J$ be the uncollapses. Then by $1$-condensation, $C\triangleleft N|\theta$ and $D\triangleleft R|\theta$, and hence $C=D$ (as $N|\theta=R|\theta$ and $\mathrm{OR}^{C}=\mathrm{OR}^{D}$), and $\pi\!\upharpoonright\!\mathrm{OR}=\sigma\!\upharpoonright\!\mathrm{OR}$. But then $\widetilde{N}=\widetilde{P}$ and $N=P$, as desired.

Proof of Subclaim.

Use a simple variant of the proof of 2 to choose $q$, running an algorithm much as there, but simultaneously for both models $\widetilde{N},\widetilde{R}$, and using the $\Sigma_{1}$-definability of the $\Sigma_{1}$-forcing relation to see that $H,H^{\prime}$ contain the same ordinals (and likewise $J,J^{\prime}$), and choosing elements of $q$ large enough to ensure that $H^{\prime}=J^{\prime}$ and $\mathbb{P}\in H$ etc. We leave the details to the reader. (Here is some more of a sketch: Given $q\!\upharpoonright\!i$ and $\gamma_{i}$ much as in the proof of 2, first select some $q^{\prime}_{i}$ satisfying the requirements much as before with respect to $\widetilde{N}$ (hence with $\gamma_{i}<q^{\prime}_{i}<(\gamma_{i}^{+})^{\widetilde{U}}$), and with $q^{\prime}_{i}$ large enough that $\mathbb{P}\cup\{\mathbb{P}\}\subseteq$ the relevant hulls of $\widetilde{N}$ (note this condition holds trivially unless $\gamma_{i}<\theta$) and $x,N_{\omega},R_{\omega},g,h$ are in the relevant hulls of $\widetilde{U}$. Then choose $q_{i}$ with $q^{\prime}_{i}<q_{i}<(\gamma_{i}^{+})^{\widetilde{U}}$ and much as before with respect to $\widetilde{R}$. In this manner it is easy to arrange that $q_{i}$ works.)

This completes the proof of the subclaim, claim and theorem. ∎

∎

∎

Definition 4.1**.**

Let $M$ be a transitive structure satisfying $\mathsf{PS}$. Work in $M$. Let $P$ be a premouse with $\mathrm{OR}^{P}$ regular. For a regular cardinal $\eta\geq\mathrm{OR}^{P}$, define $\eta$-excellent premice (relative to $P,\eta$) as in the proof above (there we have $P=M|\theta$). The direct condensation stack of $M$ above $P$ is the stack $S$ of all $\eta$-excellent premice, for all such $\eta$, as far as this is a well-defined stack.

Remark 4.1**.**

As a special case of the previous theorem, we get a shorter proof that if a mouse $M$ satisfies $\mathsf{PS}$, then $\mathbb{E}^{M}$ is definable over $\left\lfloor M\right\rfloor$ from the parameter $M|\omega_{1}^{M}$. Note that the proof also easily adapts to the case that $M$ has a largest cardinal $\lambda$, assuming that $\lambda$ is $M$-regular. However, for the singular case (most importantly $\operatorname{cof}^{M}(\lambda)=\omega$) we need the earlier methods.

5 A simplified fine structure

In [1], Mitchell-Steel fine structure is introduced, which makes use of the parameters $u_{n}$. We introduce a simplified fine structure here which avoids the parameters $u_{n}$, and show that in fact, the two fine structures are equivalent (we get the same notions of soundness etc).

Definition 5.1**.**

Let $N$ be a premouse. Given $X\subseteq N$, $\mathrm{Hull}_{k+1}^{N}(X)$ denotes the substructure of $N$ whose elements are those $z\in N$ such that there is $\vec{x}\in X^{<\omega}$ and an $\mathrm{r}\Sigma_{k+1}$ formula $\varphi$ such that $z$ is the unique $z^{\prime}\in N$ such that $N\models\varphi(\vec{x},z^{\prime})$. And $\mathrm{cHull}_{k+1}^{N}(X)$ denotes its transitive collapse. Also let $\operatorname{Th}_{k+1}^{N}(X)$ be the $\mathrm{r}\Sigma_{k+1}$ theory191919That is, the pure theory, in the language of [1]. of $N$ in parameters in $X$.

Definition 5.2** (Minimal Skolem terms).**

Let $\varphi$ be an $\mathrm{r}\Sigma_{k+1}$ formula of $n+1<\omega$ free variables. The minimal Skolem term associated to $\varphi$ is denoted $\mathrm{m}\tau_{\varphi}$, and has $n$ variables.

Let $R$ be a $k$-sound premouse with $\rho_{k}^{R}>\omega$. Let $q\in[\mathrm{OR}^{R}]^{<\omega}$ with

[TABLE]

and if $q\neq\emptyset$ then $\rho_{k}\leq\min(q)$. We define the partial function

[TABLE]

If $k=0$ then $\mathrm{m}\tau_{\varphi,q}^{R}$ is just the usual Skolem function associated to $\varphi$ (such that the graph of $\mathrm{m}\tau_{\varphi}^{R}$ is uniformly $\mathrm{r}\Sigma_{1}^{R}$). (Note $q=\emptyset$ in this case.)

Suppose $k>0$. Let $\vec{x}\in\mathfrak{C}_{0}(R)^{n}$. If $\mathfrak{C}_{0}(R)\models\neg\exists y\varphi(\vec{x},y)$, then $\mathrm{m}\tau_{\varphi,q}^{R}(\vec{x})$ is undefined.

Suppose $\mathfrak{C}_{0}(R)\models\exists y\varphi(\vec{x},y)$. Let $\tau_{\varphi}$ be the basic Skolem term associated to $\varphi$ (see [1, 2.3.3]). For $\beta<\rho_{k}^{R}$, let $(\tau_{\varphi})^{\beta}$ be defined over $R$ as in the proof of [1, 2.10], with $q$ as chosen above. Let $\beta_{0}$ be the least $\beta$ such that $(\tau_{\varphi})^{\beta}(\vec{x})$ is defined. Define

[TABLE]

Lemma 5.2**.**

The graph of $\mathrm{m}\tau_{\varphi,q}^{R}$ is $\mathrm{r}\Sigma_{k+1}^{R}(\{q\})$, recursively uniformly in $\varphi,R,q$ (for $R,q$ as in 5.2).

Given $\mathrm{r}\Sigma_{k+1}$ formulas $\varphi,\psi_{0},\ldots,\psi_{n-1}$, with $\varphi$ of $n$ free variables and $\psi_{i}$ of $n_{i}+1$ free variables, the relation over $\mathfrak{C}_{0}(R)$,

[TABLE]

is $\mathrm{r}\Sigma_{k+1}(\{q\})$, uniformly in $R,q$ as in 5.2, and moreover, there is a recursive function passing from $\varphi,\psi_{0},\ldots,\psi_{n-1}$ to an $\mathrm{r}\Sigma_{k+1}$ formula for $\varrho$.

Therefore minimal Skolem terms are effectively closed under composition. That is, there is a recursive function passing from $\varphi,\psi$ to $\varrho$, such that for all relevant $R,q$, we have

[TABLE]

In the following lemma, the standard Skolem terms are as in [1]; for example, $t_{\varphi}^{R}(x)$ is the $R$-least $y$ such that $\mathfrak{C}_{0}(R)\models\varphi(x,y)$.

Lemma 5.2**.**

Let $R,q$ be as in 5.2, $X\subseteq\mathfrak{C}_{0}(R)$ and

[TABLE]

Then $H_{1}=H_{2}=H_{3}$.

Proof.

The main thing is to see that $H_{3}\subseteq H_{2}$. For this, see the proof of [1, 2.10], combined with (for example) the observation that if $\vec{x}\in X^{<\omega}$ and $y\in H_{2}$ and $R\models\exists z<_{R}y\varphi(q,\vec{x},y)$, then there is $z\in H_{2}$ such that $z<_{R}y$ and $R\models\varphi(q,\vec{x},z)$; this is by 5. Applying this observation finitely many times shows that $t^{R}_{\varphi}(q,\vec{x})\in H_{2}$.∎

Definition 5.3**.**

For $k<\omega$, the terminology $k$-u-sound, $k$-u-solid, etc, mean just what $k$-sound, $k$-solid, etc, mean in [1].202020In this notation, “$k$” is a variable but “u” is just a symbol. The symbol “u” indicates that the $u_{n}$’s are being used in the definition.

Definition 5.4**.**

For $N$ a premouse, define $q_{k}=q_{k}^{N}$ and $k$-q-solidity and $k$-q-soundness for $k\in[0,\omega)$, recursively as follows. We also define $k$-q-universality in the obvious manner.

Suppose $q_{0},\ldots,q_{k}$ have been defined and $N$ is $k$-q-sound and $k$-u-sound.

Now if $k\geq 1$ then suppose by induction that

[TABLE]

Let $q_{k+1}$ be the $<_{\text{lex}}$-least $q\in[\mathrm{OR}]^{<\omega}$ such that212121In our notation, $\operatorname{Th}_{k+1}$ refers to pure $\mathrm{r}\Sigma_{k+1}$ theories, but by [1, §2], it would make no difference in the definition of $q_{k+1}$ (or $\rho_{k+1}$) whether we use pure or generalized theories.

[TABLE]

For $q,v\in N$ with $q\in[\mathrm{OR}]^{<\omega}$, and for $\alpha\in q$, the $(k+1)$-solidity witness for $((q,v),N)$ at $\alpha$, is

[TABLE]

We say $(q,v)$ is $(k+1)$-solid for $N$ iff $w_{\alpha}^{N}(q,v)\in N$ for each $\alpha\in q$. We say $N$ is $(k+1)$-q-solid iff $(q_{k+1},q_{k})$ is $(k+1)$-solid for $N$. We say $N$ is $(k+1)$-q-sound iff $N$ is $(k+1)$-q-solid and

[TABLE]

In general we define $\mathfrak{C}^{\text{q}}_{k+1}(N)=\mathrm{cHull}_{k+1}^{N}(\rho_{k+1}^{N}\cup\{q_{k+1},q_{k}\})$.

The theorem below establishes the equivalence between standard Mitchell-Steel fine structure (u-soundness, etc) and the fine structure introduced here (q-soundness, etc). In part 3 we show that the parameters provided by $u^{N}_{k}$ automatically get into the relevant hulls.

Theorem 5.4**.**

Let $k<\omega$. Let $N$ be a premouse. Then:

1. (a)

$N$* is $k$-q-sound iff $N$ is $k$-u-sound.*

If $N$ is $k$-u-sound and $\omega<\rho_{k}^{N}$ then:

$p_{i}^{N}=q_{i}^{N}$* for all $i\leq k$.* 2. 3.

Let $X\subseteq N$, let $M=\mathrm{cHull}_{k+1}^{N}(X\cup\{q_{k}^{N}\})$ and $\pi:M\to N$ the uncollapse. Then $M$ is a $k$-u-sound premouse and $\pi$ is a near $k$-u-embedding such that

1. (i)

if $\rho_{k}^{M}<\rho_{0}^{M}$ then $\pi(\rho_{k}^{M})\geq\rho_{k}^{N}$, 2. (ii)

for all $i\leq k$ and all $\alpha\in q_{i}^{M}$,

[TABLE] 3. 4.

$p_{k+1}^{N}=q_{k+1}^{N}$* and $N$ is $(k+1)$-u-solid iff $(k+1)$-q-solid, $(k+1)$-u-universal iff $(k+1)$-q-universal, $(k+1)$-u-sound iff $(k+1)$-q-sound.*

Proof.

We prove the proposition by induction on $k$. For $k=0$ it is easy. Assume $k>0$ and the lemma holds at all $k^{\prime}<k$. Parts (a) and 2 are trivial by induction (by part 4). So consider part 3. Let $H=\mathrm{rg}(\pi)$. Note that $X\cup\{q_{k}^{N}\}\subseteq H$ and if $\vec{x}\in H$ and $y\in\mathrm{Hull}_{k+1}^{N}(\{\vec{x}\})$ then $y\in H$. Now we prove:

Claim 6.

Let $i\leq k$. Then $q_{i}^{N}\in H$ and if $i<k$ and $\rho_{i}^{N}<\rho_{0}^{N}$ then $\rho_{i}^{N}\in H$.

Proof.

We prove the claim by induction on $i$. It is trivial for $i=0$ and $i=k$ (since $q_{0}^{N}=\emptyset$ and we put $q_{k}^{N}\in H$ directly). Suppose $0<i<k$ and the claim holds for all $i^{\prime}<i$.

We show $q_{i}^{N}\in H$. So assume $n=\mathrm{lh}(q_{i}^{N})>0$. Note that $q_{i}^{N}$ is the unique $q\in[\mathrm{OR}]^{<\omega}$ such that

1. (i)

$N=\mathrm{Hull}^{N}_{i}((\min(q)+1)\cup\{q_{i-1}^{N}\})$ and 2. (ii)

$(q,q_{i-1}^{N})$ is $i$-solid for $N$ and 3. (iii)

$\mathrm{lh}(q)=n$.

Now each of these statements are $\mathrm{r}\Sigma_{k+1}(\{q_{k}^{N}\})$. For statement (i), if $i>1$, this is because by induction,

[TABLE]

so by 5, (i) is equivalent to “$\forall x\in N$ there is $\vec{\beta}\in(\min(q)+1)^{<\omega}$ and an $\mathrm{r}\Sigma_{i}$ formula $\varphi$ such that $x=\mathrm{m}\tau_{\varphi,(q_{i-1}^{N},q_{i-2}^{N})}(\vec{\beta},q)$”. If $i=1$ then it is similar. So $q_{i}^{N}\in H$, as required.

Now if $i<k$ and $\rho_{i}^{N}<\rho_{0}^{N}$ then by induction, $\rho^{N}_{i}$ is the least $\rho$ with

[TABLE]

So as above and in the proof of 5, if $\rho_{i}^{N}<\mathrm{OR}^{N}$ then $\rho^{N}_{i}\in H$.222222That is, it seems we don’t get a uniform definition of $\rho^{N}_{k-1}$ here, but we do get some $\rho\in H$ such that $N=\mathrm{Hull}^{N}_{i}(\rho\cup\{q_{i}^{N},q_{i-1}^{N}\})$, and then if $\rho\neq\rho_{i}^{N}$, we can get a smaller such $\rho^{\prime}\in H$, and so on. ∎

In the next claim it is more important that $q_{k}^{N}\in H$.

Claim 7.

If $\varphi$ is $\mathrm{r}\Sigma_{k+1}$ and $\vec{x}\in H$ and $N\models\exists y\varphi(\vec{x},y)$ then $\exists y\in H$ such that $N\models\varphi(\vec{x},y)$.

Proof.

Let $q=\{q_{k}^{N},q_{k-1}^{N}\}$. Since $q\in H$ and $N=\mathrm{Hull}_{k}^{N}(\rho_{k}^{N}\cup\{q\})$, 5 applies and yields the claim. ∎

We have $H=\mathrm{Hull}^{N}_{i}(H)$ for each $i\leq k+1$. Therefore by induction, $M$ is $(k-1)$-sound, $\pi$ is a near $(k-1)$-embedding, and so on. Combined with Claim 6, this also gives that if $\rho_{k-1}^{M}<\rho_{0}^{M}$ then $\pi(\rho_{k-1}^{M})=\rho_{k-1}^{N}$, and if $\rho_{k-1}^{M}=\rho_{0}^{M}$ then $\rho_{k-1}^{N}=\rho_{0}^{N}$. By this and Claim 7 it is easy enough to see that: $\rho_{k}^{M}$ is the least $\rho$ such that either $\rho=\rho_{0}^{M}$ or $\pi(\rho)\geq\rho_{k}^{N}$; and

[TABLE]

and $\pi$ is $\mathrm{r}\Sigma_{k+1}$ elementary. To see that $q_{k}^{M}=\pi^{-1}(q_{k}^{N})$, we therefore just need that $(\pi^{-1}(q_{k}^{N}),q_{k-1}^{M})$ is $k$-solid for $M$. For this it suffices to know that $M$ has the appropriate generalized solidity witnesses; see [10]. But this follows from the fact that $N$ has generalized solidity witnesses for $(q_{k}^{N},q_{k-1}^{N})$ in $\mathrm{rg}(\pi)$, which follows from Claim 7. Also, the elementarity of $\pi$ then guarantees that 33(ii) holds for $i=k$.

Part 4 follows easily from part 3 (by part 3, the parameters provided by $u^{N}_{k}$ already get into the relevant hulls). ∎