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

$$\begin{align*}q=(q_0,q_1,\ldots,q_{m-1}),\end{align*}$$

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

$$\begin{align*}u_i,r\in\mathrm{Hull}_{k+1}^H(\gamma_i\cup\{\vec{p}_k^H,q\!\upharpoonright\! i\}), \end{align*}$$

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

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

and

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

Let $q_i=\min (\mathrm {Ord}\backslash H_i)=\gamma _i^{+H}\cap H_i$ and let $\gamma _i=\mathrm {card}^H(q_i)=\mathrm {card}^H(\eta )$ .

Clearly

$$\begin{align*}H_i\subsetneq H_i'=\mathrm{Hull}_{k+1}^H(\gamma_i\cup\{\vec{p}_k^H,q\!\upharpoonright\!(i+1)\}), \end{align*}$$

so it suffices to see that $u_{i+1}\in H_i'$ . 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 $W_i\in H_i'$ . For this, noting that $q_i=\gamma _i^{+W_i}$ , it suffices to see that $W_i\triangleleft H$ , since then $W_i$ is the least segment W of H such that $\mathrm {Ord}^W\geq q_i$ 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

$$\begin{align*}\rho>\rho_{k+1}^{W}=\gamma=\mathrm{lgcd}(W|\rho) \end{align*}$$

( $\rho _{k+1}^{W}\geq \gamma $ because $W\in H$ , and $\rho _{k+1}^W<\rho $ because $\eta <\rho $ and by line (1)). So by condensation as stated in [Reference Schlutzenberg9, Theorem 5.2], either (a) $W\triangleleft H$ or (b) letting $J\triangleleft H$ be least such that $q_i\leq \mathrm {Ord}^J$ 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 $W=\mathrm {Ult}_k(J,F)$ . But since $\eta>\gamma $ and because of line (2), we have

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

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.

Proof For $\gamma \in (\kappa ,\rho )$ , 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 $\gamma =\kappa ^{+W_\gamma }$ so $\rho _{k+1}^{W_\gamma }\in \{\kappa ,\gamma \}$ ; moreover, if $\gamma $ is a successor generator then $\rho _{k+1}^{W_\gamma }=\kappa $ .

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 '<\gamma $ , we have $W_{\gamma '}\triangleleft H$ and $\gamma '=\kappa ^{+W_{\gamma '}}$ and $W_{\gamma '}$ projects to $\kappa $ . It follows that $W_{\gamma '}\in H_\gamma $ , so $W_{\gamma '}\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 [Reference Schlutzenberg9, Theorem 5.2]) as in the proof of 2.3, we get $W_\gamma \triangleleft H$ .

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

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 $E=(\pi _{MR}^{-1})"F$ . So E is a standard M-extender with support $\tau +\ell $ , where either $\tau <\rho _0^M$ and $\pi _{MR}(\tau )=\sigma $ , or $\tau =\rho _0^M$ and $\rho _0^R=\sigma $ . And E is -definable, M is $(m+1)$ -sound, $n+10< m$ and

$$\begin{align*}\rho_{m+1}^M=\omega<\kappa=\mathrm{cr}(E)<\tau=\rho_{m}^M=\rho_{n+1}^M. \end{align*}$$

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

(3) $$ \begin{align} U=\mathrm{Ult}_m(M,E)\text{ is wellfounded and }(m,\omega_1+1)\text{-iterable},\end{align} $$

by the countable completeness of F in N and the $(\omega ,\omega _1+1)$ -iterability of R.

Now $\tau <\mathrm {Ord}^M$ . For suppose $\tau =\mathrm {Ord}^M$ . Since $\rho _{n+10}^M=\tau $ , therefore M is passive. If $\tau =\kappa ^{+M}$ (that is, $\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 $\theta $ be the largest M-cardinal $\leq \tau $ such that $M|\theta =U|\theta $ .Footnote ⁸ Let t be $(m,\omega )$ -self-solid for M, and such that letting

$$\begin{align*}\bar{M}=\mathrm{cHull}_m^M(\{t\}) \end{align*}$$

and $\pi :\bar {M}\to M$ be the uncollapse, then $r,s\in \mathrm {rg}(\pi _{MR}\circ \pi )$ , and $\tau \in \mathrm {rg}(\pi )$ if $\tau <\rho _0^M$ , and $\theta \in \mathrm {rg}(\pi )$ if $\theta <\rho _0^M$ . 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 $\bar {U}=\mathrm {Ult}_{m-1}(\bar {M},\bar {E})$ .

Let $\pi (\bar {\theta })=\theta $ if $\theta <\rho _0^M$ , and otherwise $\bar {\theta }=\rho _0^{\bar {M}}$ . Likewise for $\bar {\tau }$ . So $\bar {\theta },\bar {\tau }$ have the same defining properties with respect to $\bar {M},\bar {U}$ . Define the phalanx (see [Reference Schlutzenberg11, Section 1.1] for the notation)

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

Proof We will lift trees on $\mathfrak {P}$ to essentially m-maximal trees on U, which by 2.7 and line (3) suffices. Let $\psi :\bar {U}\to U$ be the Shift Lemma map. Let $\theta '=\sup \pi "\bar {\theta }$ .

Case 1. $\theta '<\theta $ .

Let $\gamma =\mathrm {card}^{M}(\theta ')$ , so $\gamma <\theta $ . Let $t'$ be such that $(\gamma ,t')$ is m-self-solid for M, with

$$\begin{align*}M'=\mathrm{cHull}^{M}_m(\gamma\cup\{t'\})\triangleleft M, \end{align*}$$

and $\widetilde {\pi }:M'\to M$ be the uncollapse, such that $t\in \mathrm {rg}(\widetilde {\pi })$ . Let $\pi ':\bar {M}\to M'$ be ${\pi '=\widetilde {\pi }^{-1}\circ \pi }$ . So

$$\begin{align*}\pi'\!\upharpoonright\!\bar{\theta}=\pi\!\upharpoonright\!\bar{\theta}=\psi\!\upharpoonright\!\bar{\theta}.\end{align*}$$

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

We can use $(\pi ',\psi )$ 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 $\alpha>0$ , let $\mathrm {root}^{\mathcal {T}}_\alpha =0$ 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

$$\begin{align*}\pi_\alpha:M^{\mathcal{T}}_\alpha\to M^{\mathcal{U}}_\alpha \end{align*}$$

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

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

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

$$\begin{align*}\pi_\alpha:M^{\mathcal{T}}_\alpha\to Q_\alpha=i^{\mathcal{U}}_{0\alpha}(M')\triangleleft M^{\mathcal{U}}_\alpha,\end{align*}$$

and $\pi _\alpha $ is an $(m-1)$ -lifting embedding which is produced in the obvious manner via the Shift Lemma. Otherwise, $(0,\alpha ]_{\mathcal {U}}$ drops in model, and $\pi _\alpha :M^{\mathcal {T}}_\alpha \to M^{\mathcal {U}}_\alpha $ , which is 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

$$\begin{align*}\bar{\gamma}^{+\bar{U}}=\bar{\gamma}^{+\bar{M}}=\bar{\theta}<\mathrm{lh}(E^{\mathcal{T}}_0)\end{align*}$$

and

$$\begin{align*}\gamma^{+U}=\gamma^{+M}=\theta<\mathrm{lh}(E^{\mathcal{U}}_0),\end{align*}$$

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

$$\begin{align*}\mathrm{lh}(E^{\mathcal{U}}_\alpha)<i^{\mathcal{U}}_{0,\alpha+1}(\gamma)<\mathrm{Ord}^{Q_{\alpha+1}} \end{align*}$$

and

$$\begin{align*}\pi_{\alpha+1}(\mathrm{lh}(E^{\mathcal{T}}_{\alpha}))=\mathrm{lh}(E^{\mathcal{U}}_{\alpha}) \end{align*}$$

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

$$\begin{align*}Q_{\alpha+1}=i^{\mathcal{U}}_{0,\alpha+1}(M')\triangleleft M^{\mathcal{U}}_{\alpha+1}||\mathrm{lh}(E^{\mathcal{U}}_\alpha),\end{align*}$$

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,

$$\begin{align*}\pi_{\alpha+1}(\lambda(E^{\mathcal{T}}_\alpha))=\lambda(E^{\mathcal{U}}_\alpha).\end{align*}$$

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

$$\begin{align*}\nu(E^{\mathcal{U}}_\alpha)\leq\nu(E^{\mathcal{U}}_{\alpha+1}),\end{align*}$$

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

$$\begin{align*}\pi_{\alpha+1}(\nu(E^{\mathcal{T}}_\alpha))=\psi_{\pi_\alpha}(\nu(E^{\mathcal{T}}_\alpha))<\lambda(E^{\mathcal{U}}_\alpha)=\nu(E^{\mathcal{U}}_\alpha),\end{align*}$$

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 $0$ -embedding. Let $j=\mathrm {root}^{\mathcal {T}}_\alpha \in \{-1,0\}$ . By the proof that the copying construction propagates near embeddings (see [Reference Schimmerling and Steel4]), $(j,\alpha ]_{\mathcal {T}}$ 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

$$\begin{align*}\mathrm{cr}(\bar{E})\leq\bar{\gamma}<\bar{\theta}\leq\lambda(\bar{E}),\end{align*}$$

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 -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 $\sigma _\gamma :\bar {M}\to M^{\prime }_\gamma $ is a near $(m-1)$ -embedding with

$$\begin{align*}\sigma_\gamma\!\upharpoonright\!\gamma^{+\bar{M}}=\pi\!\upharpoonright\!\gamma^{+\bar{M}}=\psi\!\upharpoonright\!\gamma^{+\bar{M}}\end{align*}$$

and $\rho _m^{M^{\prime }_\gamma }=\sigma _\gamma (\gamma )^{+M}$ ; 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 $\sigma _\gamma (\gamma )^{+}<\mathrm {Ord}^{M^{\prime }_\gamma }$ . 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 $(\bar {M},\mathfrak {P})$ , with $\mathcal {T}$ being $(m-1)$ -maximal. 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 $b^{\mathcal {T}},b^{\mathcal {U}}$ are non-model-dropping, $M^{\mathcal {T}}_\infty =Q=M^{\mathcal {U}}_\infty $ and $\deg ^{\mathcal {T}}_\infty =m-1=\deg ^{\mathcal {U}}_\infty $ . So $\bar {\theta }\leq \mathrm {cr}(i^{\mathcal {U}})$ , so $\bar {U}|\bar {\theta }^{+\bar {U}}=Q|\bar {\theta }^{+Q}$ , and since $\mathrm {lh}(E^{\mathcal {T}}_0)>\bar {\theta }$ , therefore

(4) $$ \begin{align} \bar{U}|\bar{\theta}^{+\bar{U}}=\bar{M}||\bar{\theta}^{+\bar{U}}.\end{align} $$

But if $\bar {\theta }<\bar {\tau }$ then because $\mathcal {H}_{\bar {\tau }}^{\bar {M}}\subseteq \bar {U}$ , it follows that $\bar {U}|\bar {\theta }^{+\bar {U}}=\bar {M}|\bar {\theta }^{+\bar {M}}$ , which contradicts the choice of $\bar {\theta }$ . So $\bar {\theta }=\bar {\tau }$ , which with line (4) 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 $\ell =0$ , so $\bar {E}$ is generated by $\bar {\tau }$ , then standard arguments show that $\bar {E}$ is just the $(\bar {\kappa },\bar {\tau })$ -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 1.2 and 1.4

Theorem 2.12 directly implies 1.2. For 1.4 note that we may replace the given extender with a sub-extender derived from a set of form $\tau \cup t$ , where t is a finite set of ordinals, and then appeal to 2.12.

Fact 2.14 (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.

Proof Let $j:M\to U$ be the ultrapower map. Let t be as in Definition 2.16. So $E\equiv E_j\!\upharpoonright \!\tau \cup t$ , and we will identify these two extenders in what follows.

If $U||\tau =M||\tau $ then let $\theta =\tau $ , and otherwise let $\lambda $ be least such that $U|\lambda \neq M|\lambda $ and let $\theta =\mathrm {card}^M(\lambda )$ . So $\theta $ is an M-cardinal and $\theta \leq \tau $ , and by condensation and weak amenability, $\kappa ^{+M}=\kappa ^{+U} \leq \theta $ . Note that if E is explicitly Dodd-solid above $\tau $ then E is Dodd-sound. (For suppose $\kappa ^{+M}<\tau $ . As $E\!\upharpoonright \!(\tau \cup t)$ is amenably 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 $e\in [\mathfrak {C}_0(M)]^{<\omega }$ be such that:

1. 1. $\theta ,\tau \in e$ (recall that $\tau <\rho _0^M$ by 2.16).

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 2.16) is $\mathrm {r}\Sigma _{m+1}^M(\{e\})$ .

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

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 $\lambda \in (\tau ,\tau ^{+U}]$ is least such that $U|\lambda \neq M|\lambda $ (note that in fact, $\lambda <\tau ^{+U}$ , because $\tau ^{+U}<\mathrm {Ord}^U$ , because $\kappa ^{+M}\leq \tau <\rho _0^M$ ).

7. 7. If E is explicitly Dodd-solid above $\tau $ 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. Footnote ¹⁰ If E is explicitly Dodd-solid above $\tau $ and $\theta =\tau $ and

   $$\begin{align*}\varsigma=\tau^{+U}<\tau^{+M} \end{align*}$$
   and $U|\varsigma =M||\varsigma $ but $M|\varsigma $ 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
   $$\begin{align*}[a\cup t,f]^{M,m}_E=E\!\upharpoonright\!(\mathrm{cr}(F)\cup t).\end{align*}$$

Let q be such that $(\omega ,q)$ is $(m+1)$ -self-solid for M and $e\in \mathrm {rg}(\pi ),$ where ${{\bar {M}}=\mathrm {cHull}_{m+1}^M(\{\vec {p}_m^M,q\})}$ and $\pi :{\bar {M}}\to M$ is the uncollapse (q exists by 2.3).

Let $\pi ({\bar {q}})=q$ , $\pi ({\bar {\theta }})=\theta $ , etc. Also write ${\bar {t}}$ for the preimage of t “in the codes”; but note we did not demand that $t\subseteq \mathrm {Ord}^M$ . 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 premiceFootnote ¹¹ 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 short ${\bar {M}}$ -extender with ${\bar {\kappa }}=\mathrm {cr}({\bar {E}})<\rho _m^{\bar {M}}$ .

Let ${\bar {E}}$ be the short $\bar {M}$ -extender generated by ${\bar {E}}\!\upharpoonright \!{\bar {\tau }}\cup {\bar {t}}$ and let

$$\begin{align*}{\bar{U}}=\mathrm{Ult}_m({\bar{M}},{\bar{E}}).\end{align*}$$

1. 2’. If M has largest cardinal $\Omega $ then ${\bar {M}}$ has largest cardinal $\bar {\Omega }$ .

2. 3’. ${\bar {\theta }},{\bar {\tau }}$ are ${\bar {M}}$ -cardinals, $\mathcal {H}^{\bar {M}}_{{\bar {\tau }}}\subseteq {\bar {U}}$ and ${\bar {M}}|{\bar {\theta }}={\bar {U}}|{\bar {\theta }}$ .

3. 4’. If $\theta <\tau $ then ${\bar {M}}|{\bar {\theta }}^{+{\bar {M}}}\neq {\bar {U}}|{\bar {\theta }}^{+{\bar {U}}}$ , and $\bar {\chi }$ is least such that ${\bar {M}}|\bar {\chi }\neq {\bar {U}}|\bar {\chi }$ .

4. 5’. If $\theta =\tau $ then:

   1. – If $\tau ^{+U}<\tau ^{+M}$ then ${\bar {\tau }}^{+{\bar {U}}}<{\bar {\tau }}^{+{\bar {M}}}$ and $\pi ({\bar {\tau }}^{+{\bar {U}}})=\tau ^{+U}$ .

   2. – If $\tau ^{+U}=\tau ^{+M}$ then ${\bar {\tau }}^{+{\bar {U}}}={\bar {\tau }}^{+{\bar {M}}}$ .

5. 6’. If $\theta =\tau $ then:

   1. – If $U|\tau ^{+U}=M||\tau ^{+U}$ then ${\bar {U}}|{\bar {\tau }}^{+{\bar {U}}}={\bar {M}}||{\bar {\tau }}^{+{\bar {U}}}$ .

   2. – If $U|\tau ^{+U}\neq M||\tau ^{+U}$ then ${\bar {\tau }}<\bar {\lambda }<{\bar {\tau }}^{+{\bar {U}}}$ and $\bar {\lambda }$ is least such that ${\bar {U}}|\bar {\lambda }\neq {\bar {M}}|\bar {\lambda }$ .

6. 7’. If E is explicitly Dodd-solid above $\tau $ then ${\bar {E}}$ is Dodd-solid with respect to ${\bar {t}}$ . That is, for each $\alpha \in {\bar {t}}$ , we have ${\bar {E}}\!\upharpoonright \!(\alpha \cup ({\bar {t}}\backslash (\alpha +1)))\in {\bar {U}}$ .

7. 8’. If E is explicitly Dodd-solid above $\tau $ and $U,M,\varsigma ,F$ are as in condition 8, then $\bar {\varsigma }={\bar {\tau }}^{+{\bar {U}}}$ and ${\bar {M}}|\bar {\varsigma }$ is active with ${\bar {F}}$ and the Dodd-soundness witness ${\bar {E}}\!\upharpoonright \!(\mathrm {cr}({\bar {F}})\cup {\bar {t}})$ is in ${\bar {U}}$ .

(We do not yet know that ${\bar {U}}$ is wellfounded.) Let ${\bar {j}}:{\bar {M}}\to {\bar {U}}$ be the ultrapower map. Let $\psi :{\bar {U}}\to U$ be the Shift Lemma map. Define the phalanx $\mathfrak {P}=(({\bar {M}},m,{\bar {\theta }}), ({\bar {U}},m), {\bar {\theta }})$ .

Proof The argument is mostly similar to that in the proof of 2.12. We will lift m-maximal trees $\mathcal {T}$ on $\mathfrak {P}$ to m-maximal trees on M.Footnote ¹² 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 to 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 $S\triangleleft \mathrm {Ult}_0(R|\alpha ,F^{R|\alpha })$ ”. 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

$$\begin{align*}U'=\mathrm{cHull}_{m+1}^{j(M_\kappa)}(\tau\cup\{r\}) \end{align*}$$

with $r\in [\mathrm {Ord}^{j(M_\kappa)}]^{<\omega }$ chosen such that $U\models $ “ $(\tau ,r)$ is $(m+1)$ -self-solid for $j(M_\kappa )$ and letting $\varrho _\tau :U'\to j(M_\kappa )$ be the uncollapse, then $j(q_\kappa ),t\in \mathrm {rg}(\varrho _\tau )$ ”. Such an r exists by the elementarity of j and by 2.3. (Note that $U\models $ “ $j(M_\kappa )$ is wellfounded”; the transitive collapse $U'$ is computed inside U, where it is well-defined.) Note that if $\tau =j(\kappa )$ then $t=\emptyset $ and $U'=j(M_\kappa )$ and $r=j(q_\kappa )$ . And if $\tau <j(\kappa )$ then $\rho _{m+1}^{U'}=\tau $ , so $U'\triangleleft j(M_\kappa )$ by condensation in U. In fact, $U'\triangleleft U|\tau ^{+U}$ , and we assumed that $U|\tau ^{+U}$ is wellfounded, so $U'$ is wellfounded.

Let $\psi :{\bar {U}}\to j(M_\kappa )$ be the Shift Lemma map induced by $\sigma _\kappa $ and $\pi $ . That is, givenFootnote ¹³

$$\begin{align*}x=[a\cup \bar{t},f^{{\bar{M}}}_{\tau,z}]^{\bar{M},m}_{\bar{E}}\in\mathfrak{C}_0({\bar{U}}),\end{align*}$$

where $\tau $ is an $\mathrm {r}\Sigma _m$ term, $z\in \mathfrak {C}_0({\bar {M}})$ and $a\in [\bar {\tau }]^{<\omega }$ , thenFootnote ¹⁴

$$ \begin{align*} \psi(x)&=[\pi(a)\cup t,f^{M_\kappa}_{\tau,\sigma_\kappa(z)}]^{M,m}_E\\ &=j(f^{M_\kappa}_{\tau,\sigma_\kappa(z)})(\pi(a)\cup t)\\ &=(\tau^{j(M_\kappa)}(j(\sigma_\kappa(z)),\pi(a)\cup t))^U.\end{align*} $$

Now $\mathrm {rg}(\psi )\subseteq \mathrm {rg}(\varrho _\tau )$ , for given x, etc. as above, we have $\pi (a)\subseteq \tau \subseteq \mathrm {rg}(\varrho _\tau )$ , ${t\in \mathrm {rg}(\varrho _\tau )}$ , and $j(\sigma _\kappa (z))\in \mathrm {rg}(\varrho _\tau )$ since

$$\begin{align*}j(\sigma_\kappa(z))\in\mathrm{rg}(j\circ\sigma_\kappa)=(\mathrm{Hull}_{m+1}^{j(M_\kappa)}(\{j(q_\kappa)\}))^U\end{align*}$$

and $j(q_\kappa )\in \mathrm {rg}(\varrho _\tau )$ . So $\psi (x)\in \mathrm {rg}(\varrho _\tau )$ .

So we can define $\psi ':{\bar {U}}\to U'$ by $\psi '=\varrho ^{-1}_\tau \circ \psi $ . Then $\psi '$ is m-lifting, because if $\varphi $ is $\mathrm {r}\Sigma _{m+1}$ and ${\bar {U}}\models \varphi (x)$ then easily

$$\begin{align*}U\models\text{"}j(M_\kappa)\models\varphi(\psi(x))\text{"},\end{align*}$$

so $U\models \text {"}U'\models \varphi (\psi '(x))\text {"}$ , so $U'\models \varphi (\psi '(x))$ . Also $\psi '\!\upharpoonright \!{\bar {\tau }}=\pi \!\upharpoonright \!{\bar {\tau }}$ . And $\psi '$ 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 $\xi $ , we have $M|\xi \preccurlyeq _1 M$ by condensation, and hence, $\kappa <\max (q)$ (as $\kappa \in \mathrm {rg}(\pi )$ ), and so ${\bar {M}},M_\kappa $ have largest cardinals $\Psi _\omega ,\Psi _\kappa $ respectively, with $\pi (\Psi _\omega )=\pi _\kappa (\Psi _\kappa )=\mathrm {card}^M(\max (q))$ .

For $\eta <\theta $ , let

$$\begin{align*}M_\eta=\mathrm{cHull}_{m+1}^{M}(\eta\cup\{q\}) \end{align*}$$

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 1. $\pi "{\bar {\theta }}$ 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

   $$\begin{align*}\eta\subseteq\mathrm{Hull}_{m+1}^M(\xi\cup\{q\})=\mathrm{Hull}_{m+1}^M(\eta\cup\{q\}), \end{align*}$$
   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 $\mathrm {cr}(\pi _\eta )\geq \eta $ . Since $\eta <\theta \leq \rho _{m+1}^M$ , condensation (see [Reference Schlutzenberg9, Theorem 5.2]) 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 $M_\eta \triangleleft \mathfrak {C}_0(U')$ as $\eta <\tau $ .

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

So $U\models \varphi ^+({\bar {M}},{\bar {U}},{\bar {\theta }})$ , where $\varphi ^+({\bar {M}},{\bar {U}},{\bar {\theta }})$ 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

$$\begin{align*}\pi^*:{\bar{M}}\to M^*\text{ and }\psi^*:{\bar{U}}\to U^*\end{align*}$$

such that $\pi ^*\!\upharpoonright \!{\bar {\theta }}=\psi ^*\!\upharpoonright \!{\bar {\theta }}$ and if $\bar {M}|\bar {\theta }$ has largest cardinal $\bar {\xi }$ then $\rho _{m+1}^{M^*}=\pi ^*(\bar {\xi })$ ”.

So by elementarity, $M\models \varphi ^+({\bar {M}},{\bar {U}},{\bar {\theta }})$ . Let $M^*,U^*,\pi ^*,\psi ^*$ witness this in M. These embeddings are enough to copy m-maximal trees on $\mathfrak {P}$ to m-maximal trees on M. Let us point out one detail of the copying process. Suppose $\bar {M}|{\bar {\theta }}$ has largest cardinal $\bar {\xi }$ and let $\xi ^*=\pi ^*(\bar {\xi })$ . Then $\rho _{m+1}^{M^*}=\xi ^*$ and

(5) $$ \begin{align}M|\xi^*\triangleleft M^*\triangleleft U^*|(\xi^*)^{+U^*}=U^*|\psi^*(\bar{\theta}). \end{align} $$

When iterating $\mathfrak {P}$ , extenders G with $\mathrm {cr}(G)=\bar {\xi }$ apply to ${\bar {M}}$ . Let $G^*$ be the lift of G. Then $\mathrm {cr}(G^*)=\xi ^*$ and $G^*$ is $U^*$ -total. From line (5), it follows that we can define a copy map

$$\begin{align*}\mathrm{Ult}_m({\bar{M}},G)\to i^{U^*}_{G^*}(M^*) \end{align*}$$

in the usual manner. Otherwise the copying is routine.Footnote ¹⁵

Case 2. $\pi "{\bar {\theta }}$ 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 -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 $M_\mu ,\sigma _\mu ,U'\in U|\tau ^{+U}$ .

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. – for each ${\bar {\mu }}\in C$ :

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

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

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

We have that $U\models $ “T is illfounded”, because $\psi ',U',\left <M_\mu ,\sigma _\mu \right>_\mu $ exist and $U|\tau ^{+U}$ is wellfounded and models $\mathrm {ZFC}^-$ .

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

$$\begin{align*}\varphi:\mathrm{Ult}_m({\bar{M}},G)\to i(\widetilde{M}_{\bar{\mu}}), \end{align*}$$

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. $\Box $ (Claim 1)

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

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,

$$\begin{align*}{\bar{U}}||{\bar{\theta}}^{+{\bar{U}}}=M^{\mathcal{U}}_\infty||{\bar{\theta}}^{+M^{\mathcal{U}}_\infty}= M^{\mathcal{T}}_\infty||{\bar{\theta}}^{+M^{\mathcal{T}}_\infty}={\bar{M}}||{\bar{\theta}}^{+{\bar{U}}}= {\bar{M}}||{\bar{\theta}}^{+{\bar{M}}}, \end{align*}$$

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 explicitly Dodd-solid above $\tau $ .

If $t=\emptyset $ then this follows from the ISC as in the proof of the ISC for pseudo-mice. Suppose instead that E is explicitly Dodd-solid above $\tau $ 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 2.14 and as in [Reference Schlutzenberg11, Section 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 largest generator $\gamma =i^{\mathcal {U}}(\max ({\bar {t}}))$ , and there is a unique $\beta \leq _{\mathcal {T}}\alpha $ such that the Dodd-core D of G is in $\mathbb {E}_+(M^{\mathcal {T}}_\beta )$ . Moreover, $\tau _D\leq {\bar {\tau }}$ , and if $\beta <_{\mathcal {T}}\alpha $ , then letting $\varepsilon +1=\mathrm {succ}^{\mathcal {T}}(\beta ,\alpha )$ , we have $M^{*\mathcal {T}}_{\varepsilon +1}=M^{\mathcal {T}}_\beta |\mathrm {lh}(D)$ and $\deg ^{\mathcal {T}}_{\varepsilon +1}=0$ , and letting $k=i^{*\mathcal {T}}_{\varepsilon +1,\alpha }$ , then $\mathrm {cr}(k)\geq \tau _D$ ,

$$\begin{align*}i^{\mathcal{U}}({\bar{t}})=k(t_D)\backslash{\bar{\tau}} \end{align*}$$

and

$$\begin{align*}{\bar{E}}\!\upharpoonright\!{\bar{\tau}}\cup{\bar{t}}\equiv G\!\upharpoonright\!{\bar{\tau}}\cup k(t_D). \end{align*}$$

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

$$\begin{align*}G\!\upharpoonright\!{\bar{\tau}}\cup k(t_D)\equiv D\!\upharpoonright\!\tau_D\cup t_D, \end{align*}$$

so $\bar {E}=D\in \mathbb {E}^{{\bar {M}}}$ , 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 [Reference Schlutzenberg11]) 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 ${\bar {\tau }}\leq \nu _{F^*}$ , as ${\bar {\tau }}<\mathrm {lh}(E^{\mathcal {T}}_0)$ and ${\bar {\tau }}$ is an ${\bar {M}}$ -cardinal. Note

$$\begin{align*}{\bar{U}}=\mathrm{Hull}_{m+1}^{M^{\mathcal{T}}_\infty}({\bar{\tau}}\cup k(t_D))=\mathrm{Ult}_m({\bar{M}},G'),\end{align*}$$

where $G'=F^{\mathrm {Ult}_0({\bar {M}},F^*\!\upharpoonright \!{\bar {\tau }})}$ . Therefore ${\bar {U}}$ is the iterate of ${\bar {M}}$ given by the tree $\mathcal {T}'$ which uses exactly two extenders, $E^{\mathcal {T}'}_0=F^*\!\upharpoonright \!{\bar {\tau }}$ and $E^{\mathcal {T}'}_1=G'$ . It follows that $\mathcal {T}=\mathcal {T}'$ , $\mathcal {U}$ is trivial, $E^{\mathcal {T}}_0=F^*$ , $\nu _{E^{\mathcal {T}}_0}={\bar {\tau }}$ , $\tau _D\leq \mathrm {cr}(E^{\mathcal {T}}_0)<{\bar {\tau }}$ and $E^{\mathcal {T}}_1=G=G'$ . 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 $\mathrm {lh}(E^{\mathcal {T}}_0)={\bar {\tau }}^{+{\bar {U}}}=\bar {\varsigma }<{\bar {\tau }}^{+{\bar {M}}}$ , ${\bar {M}}|\bar {\varsigma }$ 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 8’ above. But then by that property,

$$\begin{align*}{\bar{E}}\!\upharpoonright\!(\mathrm{cr}(E^{\mathcal{T}}_0)\cup{\bar{t}})\in{\bar{U}}\cap{\bar{M}}.\end{align*}$$

Also ${\bar {t}}=k(t_D\backslash \mathrm {cr}(E^{\mathcal {T}}_0))$ and

$$\begin{align*}D\equiv D\!\upharpoonright\!(\tau_D\cup t_D)\equiv D\!\upharpoonright\!(\mathrm{cr}(E^{\mathcal{T}}_0)\cup (t_D\backslash\mathrm{cr}(E^{\mathcal{T}}_0))) \equiv{\bar{E}}\!\upharpoonright\!(\mathrm{cr}(E^{\mathcal{T}}_0)\cup{\bar{t}}). \end{align*}$$

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

This completes the proof of the theorem.

Fact 3.1 (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

$$\begin{align*}\mathrm{cr}(\pi)=\bar{\kappa}=\mathrm{cr}(\sigma)=\rho_\omega^{\bar{P}}=\rho_\omega^{\bar{Q}}<\kappa \end{align*}$$

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

1. (i) $\bar {P}\trianglelefteq N$ and $\bar {Q}\trianglelefteq N$ , or

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

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

Fact 3.2. 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 $ .

Fact 3.3. 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 $P\triangleleft M$ .

Proof of Corollary 1.5

We have that M is $(0,\omega _1+1)$ -iterable, $\kappa $ is an uncountable cardinal in M and $\kappa ^{+M}<\mathrm {Ord}^M$ . We want to see that $M|\kappa ^{+M}$ is definable from parameters over $\mathcal {H}=(\mathcal {H}_{\kappa ^+})^M$ . There are two cases.

Proof Let $M\triangleleft P\triangleleft N$ be such that $\pi \in P$ . Let $q\in [\mathrm {Ord}^P]^{<\omega }$ be such that $(\omega ,q)$ is $1$ -self-solid for P and such that $\pi ,H,M\in \mathrm {Hull}_1^P(\{q\})$ . Let ${\bar {P}}=\mathrm {cHull}_1^P(\{q\})$ . Then by 2.3, ${\bar {P}}\triangleleft N|\omega _1^N$ . 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.

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 3.4, but at degree n instead of $1$ , with $n>k+m+5$ .

Proof of Theorem 3.11

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

Part (c): We split into cases corresponding to hypotheses (i)–(vii). 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 1.2 to determine, for example, whether or not $\theta $ is a cutpoint of M.