Proof. Toward a contradiction, assume not. Fix a transitive $N'$ such that

1. (1.1) $\left |N'\right |<\Theta $ ,

2. (1.2) $\omega _1\subseteq N'$ and

3. (1.3) the set of $\alpha <\omega _1$ such that there is no elementary embedding $j: N'\rightarrow N'$ with the property that $\mathrm {crit }(j)=\alpha $ is stationary in $\omega _1$ .

Let $\phi (u)$ be the formula expressing (1.1)–(1.3). Thus, $\phi [N']$ holds.

Fix a real $x_0$ such that $N'$ is ordinal definable from $x_0$ . By minimizing the ordinal parameter, we can find N such that N is the ${\textsf {OD}}_{x_0}$ -leastFootnote ³ M such that $\phi [M]$ holds.

We observe that .Footnote ⁴ Indeed, because , we have some such that $L_\alpha ({\mathbb {R}})\vDash {\textsf {ZF-Powerset}}$ and such that for some $K\in L_\alpha ({\mathbb {R}})$ , $L_\alpha ({\mathbb {R}})\vDash \text {`}K$ is ${\textsf {OD}}_{x_0}$ and $\phi [K]$ ’. Since any function $k: K\rightarrow K$ is essentially a set of ordinals, Moschovakis’ Coding LemmaFootnote <sup>5</sup> implies that $L({\mathbb {R}})\vDash \text {`}K$ is ${\textsf {OD}}_{x_0}$ and $\phi [K]$ ’. Since N was the ${\textsf {OD}}_{x_0}$ -least, it follows that $N\in L_\alpha ({\mathbb {R}})$ .

Now let be such that

1. (2.1) $L_\alpha ({\mathbb {R}})\vDash {\textsf {ZF-Replacement}}+\text {`}N$ is the ${\textsf {OD}}_{x_0}$ -least K such that ’.

2. (2.2) $L_\alpha ({\mathbb {R}})$ is the derived model of some pair $({\mathcal {P} }, \Sigma )$ Footnote ⁶ such that ${\mathcal {P} }$ is an $x_0$ -mouse and

   $$ \begin{align*} ({\textsf{HOD}}_{x_0}|\Theta)^{L_\alpha({\mathbb{R}})}={\mathcal{M}}_\infty({\mathcal{P} }, \Sigma)|\Theta^{L_\alpha({\mathbb{R}})}. \end{align*} $$

3. (2.3) LettingFootnote ⁷ $(\delta ^i_{\mathcal {P} }: i\leq \omega )$ be the Woodin cardinals of ${\mathcal {P} }$ and their limit, for some ${\mathcal {P} }$ -successor cutpoint cardinal $\nu <\delta ^0_{\mathcal {P} }$ , ${\textsf {Ord}}\cap N<\pi _{{\mathcal {P} }, \infty }^\Sigma (\nu )$ .

4. (2.4) For every ${\mathcal {P} }$ -successor cutpoint cardinal $\nu <\delta ^0_{\mathcal {P} }$ , $\Sigma _{{\mathcal {P} }|\nu }\in L_\alpha ({\mathbb {R}})$ .Footnote ⁸

5. (2.5) $\Sigma $ has full normalization.

To obtain $({\mathcal {P} }, \Sigma )$ as above, we first let $\alpha $ be the least satisfying clause (2.1) and then use [Reference Sargsyan and Steel17, Lemma 2.5] to build $({\mathcal {P} }, \Sigma )$ . (2.5) follows from the results of [Reference Steel32] and [Reference Steel31, Theorem 1.4]. Now let ${\mathcal {M}}={\mathcal {M}}_\infty ({\mathcal {P} }, \Sigma )$ . Let $\tau $ be the least measurable cardinal of ${\mathcal {P} }$ and let $E\in \vec {E}^{\mathcal {P} }$ be such that

1. (3.1) $\mathrm {crit }(E)=\tau $ and E is total, and

2. (3.2) $\mathrm {lh}(E)$ is the least among all extenders of $\vec {E}^{\mathcal {P} }$ that satisfy (3.1).

Set ${\mathcal { Q}}=Ult({\mathcal {P} }, E)$ . We let ${\mathcal {P} }_0={\mathcal {P} }|(\tau ^{++})^{\mathcal {P} }$ and ${\mathcal { Q}}_0=\pi _E({\mathcal {P} }_0)$ . Notice that we can view ${\mathcal {P} }$ as a premouse over ${\mathcal {P} }_0$ and ${\mathcal { Q}}$ as a premouse over ${\mathcal { Q}}_0$ . We then let $\Lambda $ be the fragment of $\Sigma $ that acts on iteration trees on ${\mathcal {P} }$ which are above $(\tau ^{++})^{\mathcal {P} }$ , and let $\Phi $ be the fragment of $\Sigma _{\mathcal { Q}}$ that acts on iteration trees on ${\mathcal { Q}}$ which are above $\pi _E((\tau ^{++})^{\mathcal {P} })$ . We then have that

1. (4.1) ${\mathcal {M}}_\infty ({\mathcal {P} }, \Lambda )|\Theta ^{L_\alpha ({\mathbb {R}})}=({\textsf {HOD}}_{{\mathcal {P} }_0}|\Theta )^{L_\alpha ({\mathbb {R}})}$ and ${\mathcal {M}}_\infty ({\mathcal { Q}}, \Phi )|\Theta ^{L_\alpha ({\mathbb {R}})}=({\textsf {HOD}}_{{\mathcal { Q}}_0}|\Theta )^{L_\alpha ({\mathbb {R}})}$ .

We now define an elementary embedding $j_{{\mathcal {P} }, \Sigma }=_{def}j:{\mathcal {M}}_\infty ({\mathcal {P} }, \Lambda )\rightarrow {\mathcal {M}}_\infty ({\mathcal { Q}}, \Phi )$ such that $j\restriction {\mathcal {P} }_0=\pi _E\restriction {\mathcal {P} }_0$ .

Given $x\in {\mathcal {M}}_\infty ({\mathcal {P} }, \Lambda )$ , fix some normal $\Lambda $ -iterate ${\mathcal R}$ of ${\mathcal {P} }$ such that for some $y\in {\mathcal R}$ , $\pi ^{\Lambda _{\mathcal R}}_{{\mathcal R}, \infty }(y)=x$ . Let ${\mathcal {T}}={\mathcal {T}}_{{\mathcal {P} }, {\mathcal R}}$ . Let ${\mathcal {U}}$ be the full normalization of ${\mathcal {T}}^{\frown } \{E\}$ .Footnote ⁹ Clearly, ${\mathcal {U}}$ starts with E and continues with the minimal $\pi _E$ -copy of ${\mathcal {T}}$ .Footnote ¹⁰ Thus, ${\mathcal {U}}$ can be written as $\{E\}^{\frown } {\mathcal {W} }$ , where ${\mathcal {W} }$ is a normal iteration tree on ${\mathcal { Q}}$ according to $\Phi $ . If ${\mathcal {S}}$ is the last model of ${\mathcal {W} }$ , then ${\mathcal {S}}=Ult({\mathcal R}, E)$ . We set $j(x)=\pi _{{\mathcal {S}}, \infty }^{\Phi _{\mathcal {S}}}(\pi _E^{\mathcal R}(y))$ .

Proof. Pick another normal $\Lambda $ -iterate ${\mathcal R}'$ of ${\mathcal {P} }$ such that for some $y'\in {\mathcal R}'$ , $\pi _{{\mathcal R}', \infty }^{\Lambda _{{\mathcal R}'}}(y')=x$ . It then follows from full normalization that we can compare $({\mathcal R}, \Lambda _{{\mathcal R}})$ and $({\mathcal R}', \Lambda _{{\mathcal R}'})$ via the least-extender-disagreement process and get some common iterate $({\mathcal R}", \Lambda _{{\mathcal R}"})$ .Footnote ¹² It then follows that $\pi _{{\mathcal R}, {\mathcal R}"}(y)=\pi _{{\mathcal R}', {\mathcal R}"}(y')$ . Set then $y"= \pi _{{\mathcal R}, {\mathcal R}"}(y)$ .

Next, let ${\mathcal {T}}={\mathcal {T}}_{{\mathcal {P} }, {\mathcal R}}$ , ${\mathcal {T}}'={\mathcal {T}}_{{\mathcal {P} }, {\mathcal R}'}$ , ${\mathcal {U}}={\mathcal {T}}_{{\mathcal R}, {\mathcal R}"}$ and ${\mathcal {U}}'={\mathcal {T}}_{{\mathcal R}', {\mathcal R}"}$ . Let ${\mathcal R}_E=Ult({\mathcal R}, E)$ , ${\mathcal R}^{\prime }_E=Ult({\mathcal R}', E)$ and ${\mathcal R}^{\prime \prime }_E=Ult({\mathcal R}", E)$ . Notice that

1. (5.1) ${\mathcal R}_E$ is the last model of the full normalization of ${\mathcal {T}}^{\frown } \{E\}$ ,

2. (5.2) ${\mathcal R}^{\prime }_E$ is the last model of the full normalization of ${\mathcal {T}}^{\prime \frown } \{E\}$ ,

3. (5.3) ${\mathcal R}^{\prime \prime }_E$ is the last model of the full normalization of ${\mathcal {U}}^{\frown } \{E\}$ and the full normalization of ${\mathcal {U}}^{\prime \frown } \{E\}$ ,

4. (5.4) ${\mathcal R}_E$ is a $\Phi $ -iterate of ${\mathcal { Q}}$ via some normal tree ${\mathcal {X}}$ such that the full normalization of ${\mathcal {T}}^{\frown } \{E\}$ is $\{E\}^{\frown } {\mathcal {X}}$ ,

5. (5.5) ${\mathcal R}^{\prime }_E$ is a $\Phi $ -iterate of ${\mathcal { Q}}$ via some normal tree ${\mathcal {X}}'$ such that the full normalization of ${\mathcal {T}}^{\prime \frown } \{E\}$ is $\{E\}^{\frown } {\mathcal {X}}'$ ,

6. (5.6) letting ${\mathcal {Y}}$ and ${\mathcal {Y}}'$ be the iteration trees according to $\Phi _{{\mathcal R}_E}$ and $\Phi _{{\mathcal R}_E'}$ , respectively, such that $\{E\}^{\frown } {\mathcal {Y}}$ is the full normalization of ${\mathcal {U}}^{\frown }\{E\}$ and $\{E\}^{\frown } {\mathcal {Y}}'$ is the full normalization of ${\mathcal {U}}^{\prime \frown }\{E\}$ , then ${\mathcal R}^{\prime \prime }_E$ is the last model of both ${\mathcal {Y}}$ and ${\mathcal {Y}}'$ , and hence, ${\mathcal R}^{\prime \prime }_E$ is a $\Phi $ -iterate of ${\mathcal { Q}}$ via both ${\mathcal {X}}^{\frown } {\mathcal {Y}}$ and ${\mathcal {X}}^{\prime \frown } {\mathcal {Y}}'$ .

We want to see that

$$\begin{align*}(*)\quad \pi _{{\mathcal R}_E, \infty }^{\Phi _{{\mathcal R}_E}}(\pi _E^{\mathcal R}(y))=\pi _{{\mathcal R}^{\prime }_E, \infty }^{\Phi _{{\mathcal R}^{\prime }_E}}(\pi _E^{{\mathcal R}'}(y')). \end{align*}$$

It follows from (5.1)–(5.6) that (here, we drop script $\Phi $ to make the formulas readable; all iteration embeddings appearing below are defined using $\Phi $ )

1. (6.1) $\pi _{{\mathcal R}_E, \infty }(\pi _E^{\mathcal R}(y))=\pi _{{\mathcal R}^{\prime \prime }_E, \infty }(\pi _{{\mathcal R}_E,{\mathcal R}_E^{\prime \prime }} (\pi _E^{\mathcal R}(y)))$ ,

2. (6.2) $\pi _{{\mathcal R}^{\prime }_E, \infty }(\pi _E^{{\mathcal R}'}(y))=\pi _{{\mathcal R}^{\prime \prime }_E, \infty }(\pi _{{\mathcal R}^{\prime }_E,{\mathcal R}_E^{\prime \prime }} (\pi _E^{{\mathcal R}'}(y')))$ ,

3. (6.3) $\pi _{{\mathcal R}_E,{\mathcal R}_E^{\prime \prime }} (\pi _E^{\mathcal R}(y))=\pi _E^{{\mathcal R}"}(\pi _{{\mathcal R}, {\mathcal R}"}(y))$ ,

4. (6.4) $\pi _{{\mathcal R}^{\prime }_E,{\mathcal R}_E^{\prime \prime }} (\pi _E^{{\mathcal R}'}(y'))=\pi _E^{{\mathcal R}"}(\pi _{{\mathcal R}', {\mathcal R}"}(y'))$ ,

5. (6.5) $\pi _{{\mathcal R}, {\mathcal R}"}(y)=\pi _{{\mathcal R}', {\mathcal R}"}(y')$ Footnote ¹³ .

(*) now easily follows from (6.1)–(6.5). Indeed,

$$ \begin{align*} \pi_{{\mathcal R}_E, \infty}(\pi_E^{\mathcal R}(y))&=\pi_{{\mathcal R}^{\prime\prime}_E, \infty}(\pi_{{\mathcal R}_E,{\mathcal R}_E^{\prime\prime}} (\pi_E^{\mathcal R}(y)))\\ &=\pi_{{\mathcal R}^{\prime\prime}_E, \infty}(\pi_E^{{\mathcal R}"}(\pi_{{\mathcal R}, {\mathcal R}"}(y)))\\ &=\pi_{{\mathcal R}^{\prime\prime}_E, \infty}(\pi_E^{{\mathcal R}"}(\pi_{{\mathcal R}', {\mathcal R}"}(y')))\\ &=\pi_{{\mathcal R}^{\prime}_E, \infty}(\pi_E^{{\mathcal R}'}(y')).\\[-34pt] \end{align*} $$

The next claim essentially finishes the proof of Theorem 3.1.

Proof. Working in ${\mathcal {P} }$ , let $N_{\mathcal {P} }$ be the set of $x\in {\mathcal {P} }|\delta ^0_{{\mathcal {P} }}$ such that $\pi _{{\mathcal {P} }, \infty }^\Lambda (x)\in N$ . $N_{\mathcal {P} }$ is definable in ${\mathcal {P} }$ by the following formula. Let $(\delta ^i_{\mathcal {P} }: i\leq \omega )$ be the Woodin cardinals of ${\mathcal {P} }$ and their limit. Let $\sigma [u, v, y, c]$ be a formula in the language containing $\{\in , c\}$ , where c is a constant for $x_0$ , expressing the following:

1. (7.1) v is a premouse and u is an $\omega _1$ -iteration strategy for v,

2. (7.2) $y\in v$ ,

3. (7.3) if w is the ${\textsf {OD}}_c$ -least $w'$ such that $\phi [w']$ , then $\pi _{v, \infty }^u(x)\in w$ .

We then have that

1. (a) $x\in N_{\mathcal {P} }$ if and only if

2. (a.1) there is a ${\mathcal {P} }$ -successor cutpoint cardinal $\beta <\delta ^0_{\mathcal {P} }$ such that $x\in N_{\mathcal {P} }|\beta $ , and

3. (a.2) whenever $\beta>(\tau ^{++})^{\mathcal {P} }$ is a successor cutpoint cardinal of ${\mathcal {P} }$ such that $x\in {\mathcal {P} }|\beta $ ,

   $$ \begin{align*} {\mathcal{P} }\vDash (\exists \Psi \sigma[\Psi, {\mathcal{P} }|\beta, x, x_0])^{D(\delta^\omega_{\mathcal{P} })}. \end{align*} $$

It is important to note that the strategy $\Psi $ is just $\Lambda _{{\mathcal {P} }|\beta }$ , as ${\mathcal {P} }|\beta $ has a unique iteration strategy. Moreover, since $\beta $ is a successor cutpoint cardinal of ${\mathcal {P} }$ , $\pi _{{\mathcal {P} }, \infty }^\Lambda (x)=\pi ^\Psi _{{\mathcal {P} }|\beta , \infty }(x)$ .Footnote ¹⁴

Now let $\psi (u, v, w)$ be the formula on the right side of the above equivalence. Then $x\in N_{\mathcal {P} }$ if and only if ${\mathcal {P} }\vDash \psi [x]$ .

Notice that $\pi _E(N_{\mathcal {P} })=N_{\mathcal { Q}}$ , where $N_{\mathcal { Q}}$ is such that $x\in N_{\mathcal { Q}}$ if and only if ${\mathcal { Q}}\vDash \psi [x]$ . To finish the proof of the claim, we need to show that

$$\begin{align*}(*)\quad\pi _{{\mathcal {P} }, \infty }^\Lambda (N_{\mathcal {P} })=N \text { and } \pi _{{\mathcal { Q}}, \infty }^\Phi (N_{\mathcal { Q}})=N . \end{align*}$$

We only establish the first equality, as the second is very similar.Footnote ¹⁵

Suppose $x\in \pi _{{\mathcal {P} },\infty }^\Lambda (N_{\mathcal {P} })$ . We want to see that $x\in N$ . Let ${\mathcal {S}}$ be a $\Lambda $ -iterate of ${\mathcal {P} }$ such that $x=\pi _{{\mathcal {S}}, \infty }^{\Lambda _{\mathcal {S}}}(y)$ for some $y\in {\mathcal {S}}$ . We then have that ${\mathcal {S}}\vDash \psi [y]$ . Since we can realize $L_\alpha ({\mathbb {R}})$ as the derived model of ${\mathcal {S}}$ , we have that $\pi _{{\mathcal {S}}, \infty }^{\Lambda _{\mathcal {S}}}(y)\in N$ .

Conversely, suppose $x\in N$ . Let $(y, {\mathcal {S}})$ be such that ${\mathcal {S}}$ is a $\Lambda $ -iterate of ${\mathcal {P} }$ , $y\in {\mathcal {S}}$ , and $\pi _{{\mathcal {S}}, \infty }^{\Lambda _{\mathcal {S}}}(y)=x$ . Then ${\mathcal {S}}\vDash \psi [y]$ , which implies that $y\in \pi _{{\mathcal {P} }, {\mathcal {S}}}(N_{\mathcal {P} })$ . Therefore, $x=\pi _{{\mathcal {S}}, \infty }^{\Lambda _{\mathcal {S}}}(y)\in \pi _{{\mathcal {P} }, \infty }^\Lambda (N_{\mathcal {P} })$ .

To finish the proof of Theorem 3.1, we need to produce a club $C\subseteq \omega _1$ such that for each $\alpha \in C$ , there is an embedding $k: N\rightarrow N$ with $\mathrm {crit }(k)=\alpha $ . Above, we have produced an elementary embedding $j_{{\mathcal {P} }, \Sigma }:N\rightarrow N$ such that $\mathrm { crit }(j_{{\mathcal {P} }, \Sigma })=\tau $ , for $\tau $ the least measurable cardinal of ${\mathcal {P} }$ . Let $({\mathcal {P} }_\alpha : \alpha <\omega _1)$ be the sequence of linear iterates of ${\mathcal {P} }$ by E and its images, and let $\tau _\alpha $ be the least measurable cardinal of ${\mathcal {P} }_\alpha $ . Then $\mathrm {crit }(j_{{\mathcal {P} }_\alpha , \Sigma _{{\mathcal {P} }_\alpha }})=\tau _\alpha $ , and since $C=\{\tau _\alpha : \alpha <\omega _1\}$ is a club, we get a contradiction to the fact that $\phi [N]$ is true.

Proof. For each $x\in \mathrm {dom}(H)$ , set ${\mathcal {M}}_x={\mathcal {M}}_\infty ({\mathcal {P} }_x, \Sigma _x)$ . We assume toward a contradiction that

1. (*) for a Turing cone of x, there is $\kappa <\delta $ such that $o^{{\mathcal {M}}_x}(\kappa )\geq \delta $ .

Because $\delta $ is a regular cardinal, we have that for every $x\in {\mathbb {R}}$ , ${\mathcal {M}}_x\vDash \text {`}\delta $ is a measurable cardinal’. To see this, assume not. Let ${\mathcal { Q}}$ be a $\Sigma _x$ -iterate of ${\mathcal {P} }_x$ such that $\delta \in \mathrm {rge}(\pi _{{\mathcal { Q}}, \infty })$ , and set $\delta _{\mathcal { Q}}=\pi _{{\mathcal { Q}}, \infty }^{-1}(\delta )$ . We then have that ${\mathcal { Q}}\vDash \text {`}\delta _{\mathcal { Q}}$ is a regular non-measurable cardinal’. But then $\pi _{{\mathcal { Q}}, \infty }[\delta _{\mathcal { Q}}]$ is cofinal in $\delta $ , implying that $\mathrm {cf}(\delta )=\omega $ . It then follows from (*) that

1. (**) for a Turing cone of x, there is $\kappa <\delta $ such that $o^{{\mathcal {M}}_x}(\kappa )\geq (\delta ^+)^{{\mathcal {M}}_x}$ .

Because $\delta $ is a Suslin cardinal, we have a tree T on $\omega \times \delta $ such that $p[T]$ is not $\alpha $ -Suslin for any $\alpha <\delta $ . Because $(R, \Psi , H, \alpha )$ absorbs $\delta $ , we have that for a Turing cone of x, $T\in {\mathcal {M}}_x$ . Thus, we have an $x\in \mathrm {dom}(H)$ such that

1. (***) there is $\kappa <\delta $ such that $o^{{\mathcal {M}}_x}(\kappa )\geq (\delta ^+)^{{\mathcal {M}}_x}$ and $T\in {\mathcal {M}}_x$ .

Let $(\kappa , \iota )$ be the lexicographically least pair $(\nu , \zeta )$ such that $\nu $ witnesses (***) and, letting $F=\vec {E}^{{\mathcal {M}}_x}(\zeta )$ , $\mathrm {crit }(F)=\nu $ and $T\in Ult({\mathcal {M}}_x, F)$ . Thus, $\kappa $ is a limit of cutpoints of ${\mathcal {M}}_x$ . Let ${\mathcal { Q}}$ be a $\Sigma _x$ -iterate of ${\mathcal {P} }_x$ such that $(\kappa , \delta , T, E)\in \mathrm {rge}(\pi _{{\mathcal { Q}}, \infty })$ where $E=\vec {E}^{{\mathcal {M}}_x}(\iota )$ . Set $\Lambda =(\Sigma _x)_{\mathcal { Q}}$ . Given a complete $\Lambda $ -iterate ${\mathcal R}$ of ${\mathcal { Q}}$ , let $s_{\mathcal R}=_{def}(\kappa _{\mathcal R}, \delta _{\mathcal R}, T_{\mathcal R}, E_{\mathcal R})\in {\mathcal R}$ be such that

$$ \begin{align*} \pi_{{\mathcal R}, \infty}(s_{\mathcal R})=(\kappa, \delta, T, E). \end{align*} $$

If ${\mathcal R}$ is a complete $\Lambda $ -iterate of ${\mathcal { Q}}$ , let ${\mathcal R}_E=Ult({\mathcal R}, E_{\mathcal R})$ . Let $(f_{\mathcal { Q}}, s_{\mathcal { Q}})\in {\mathcal { Q}}$ be such that $s_{\mathcal { Q}}\in \nu (E_{\mathcal { Q}})^{<\omega }$ , $f_{\mathcal { Q}}:[\kappa _{\mathcal { Q}}]^{\left |s_{\mathcal { Q}}\right |}\rightarrow {\mathcal { Q}}|\kappa _{\mathcal { Q}}$ , and $\pi _{E_{\mathcal { Q}}}(f_{\mathcal { Q}})(s_{\mathcal { Q}})=T_{\mathcal { Q}}$ . Thus, if ${\mathcal R}$ is a complete $\Lambda $ -iterate of ${\mathcal { Q}}$ , then $\pi _{E_{\mathcal R}}(f_{\mathcal R})(s_{\mathcal R})=T_{\mathcal R}$ .

Say $(\lambda , s)\in \kappa \times \kappa ^{<\omega }$ is good if there is a complete $\Lambda $ -iterate ${\mathcal R}$ of ${\mathcal { Q}}$ such that $(\lambda , s)=\pi _{{\mathcal R}_E, \infty }(\delta _{\mathcal R}, s_{\mathcal R})$ . Suppose $(\lambda , s)$ is good and ${\mathcal R}$ witnesses it. Then let

$$ \begin{align*} T_{{\mathcal R}, \lambda, s}=\pi_{{\mathcal R}_E, \infty}(T_{\mathcal R}). \end{align*} $$

Thus, $T_{{\mathcal R}, \lambda , s}$ is independent of ${\mathcal R}$ , and so we let $T_{\lambda , s}=T_{{\mathcal R}, \lambda , s}$ .

We thus have that $p[T]=\bigcup \{ p[T_{\lambda , s}]: (\lambda , s)\in \kappa \times \kappa ^{<\omega }$ and $(\lambda , s)$ is good $\}$ . Consider now the tree U give by: $(u, h)\in U$ if and only if

1. 1. if $0\in \mathrm {dom}(h)$ , then $h(0)=(\lambda _0, s_0)$ is good, and

2. 2. if $\mathrm {dom}(u)=\mathrm {dom}(h)=_{def}m>1$ , then

   $$ \begin{align*} (\langle u(0),..., u(m-2)\rangle, \langle h(1), h(2),..., h(m-1) \rangle)\in T_{\lambda_0, s_0}. \end{align*} $$

Then if $(x, h)\in [U]$ , then $(x, h')\in [T]$ , where $h'(n)=h(n+1)$ . Also, if $(x, h)\in [T]$ , then for some good $(\lambda , s)$ and $g\in \lambda ^\omega $ , $(x, g)\in [T_{\lambda , s}]$ . Consequently, $(x, g')\in [U]$ , where $g'(0)=(\lambda , s)$ and for $n\geq 1$ , $g'(n)=h(n-1)$ . We thus have that $p[T]=p[U]$ , and as U can be represented as a tree on $\omega \times \kappa $ and $\kappa <\delta $ , we get a contradiction to our assumption that $p[T]$ is not $\alpha $ -Suslin for any $\alpha <\delta $ .

Proof. We will use the following straightforward lemma.

Fix ${\mathcal { Q}}, {\mathcal R}\in \mathcal {F}({\mathcal {P} }, \Sigma , F)$ , and let ${\mathcal {T}}={\mathcal {T}}^\Sigma _{{\mathcal {P} }, {\mathcal { Q}}}$ and ${\mathcal {U}}={\mathcal {T}}^\Sigma _{{\mathcal {P} }, {\mathcal R}}$ . It follows from full normalization that the least-extender-disagreement comparison between $({\mathcal { Q}}, \Sigma _{\mathcal { Q}})$ and $({\mathcal R}, \Sigma _{\mathcal R})$ produces a common iterate $({\mathcal {S}}, \Sigma _{\mathcal {S}})$ . We want to see that ${\mathcal {S}}\in \mathcal {F}({\mathcal {P} }, \Sigma , F)$ , which amounts to showing that ${\mathcal {Z}}=_{def}{\mathcal {T}}^\Sigma _{{\mathcal {P} }, {\mathcal {S}}}$ omits F.Footnote ³⁵ Set ${\mathcal {X}}={\mathcal {T}}_{{\mathcal { Q}}, {\mathcal {S}}}$ and ${\mathcal {Y}}={\mathcal {T}}_{{\mathcal R}, {\mathcal {S}}}$ . Then ${\mathcal {Z}}$ is the full normalization of ${\mathcal {T}}^{\frown } {\mathcal {X}}$ .

We now assume that ${\mathcal {X}}$ and ${\mathcal {Y}}$ were built by allowing padding, so that $\mathrm { lh}({\mathcal {X}})=\mathrm {lh}({\mathcal {Y}})$ , and our strategy is to analyze the full normalization process that produces ${\mathcal {Z}}$ out of $({\mathcal {T}}, {\mathcal {X}})$ and $({\mathcal {T}}, {\mathcal {Y}})$ . We review some facts about the normalization process, and we do this for $({\mathcal {T}}, {\mathcal {X}})$ .

Recall that the full normalization process for ${\mathcal {T}}^{\frown } {\mathcal {X}}$ produces iteration trees $({\mathcal {Z}}_\alpha : \alpha <\mathrm {lh}({\mathcal {X}}))$ on ${\mathcal {P} }$ , and ${\mathcal {Z}}$ is simply ${\mathcal {Z}}_{\mathrm {lh}({\mathcal {X}})-1}$ (e.g., see [Reference Schlutzenberg19] or [Reference Siskind and Steel21]). The sequence satisfies the following conditions.

1. (1.1) The last model of each ${\mathcal {Z}}_\alpha $ is ${\mathcal {M}}_\alpha ^{\mathcal {X}}$ .

2. (1.2) For $\alpha +1<\mathrm {lh}({\mathcal {X}})$ , ${\mathcal {Z}}_{\alpha +1}$ is obtained by letting $\beta $ be the ${\mathcal {X}}$ -predecessor of $\alpha +1$ , and minimally inflating ${\mathcal {Z}}_\beta $ by $E_\alpha ^{\mathcal {X}}$ . More precisely, letting $\gamma _0$ be the least $\gamma $ such that $E_\alpha ^{\mathcal {X}}\in \vec {E}^{{\mathcal {M}}_{\gamma }^{{\mathcal {Z}}_\alpha }}$ and $\gamma _1$ be the least $\gamma $ such that $\mathrm {lh}(E_{\gamma }^{{\mathcal {Z}}_\beta })>\mathrm {crit }(E_\alpha ^{\mathcal {X}})$ , ${\mathcal {Z}}_{\alpha +1}\restriction \gamma _0+1={\mathcal {Z}}_\alpha \restriction \gamma _0+1$ , and for $\iota>0$ such that $\gamma _0+\iota <\mathrm {lh}({\mathcal {Z}}_{\alpha +1})$ , ${\mathcal {M}}_{\gamma _0+\iota }^{{\mathcal {Z}}_{\alpha +1}}=Ult({\mathcal {M}}_{\gamma _1+\iota -1}^{{\mathcal {Z}}_\beta }, E_\alpha ^{\mathcal {X}})$ . Also, for $\iota <\gamma _0$ , $E_\iota ^{{\mathcal {Z}}_{\alpha +1}}=E_\iota ^{{\mathcal {Z}}_\alpha }$ , $E_{\gamma _0}^{{\mathcal {Z}}_\alpha }=E_\alpha ^{\mathcal {X}}$ and for $\iota>0$ such that $\gamma _0+\iota <\mathrm {lh}({\mathcal {Z}}_{\alpha +1})$ , $E_{\gamma _0+\iota }^{{\mathcal {Z}}_{\alpha +1}}$ is the last extender of $Ult({\mathcal {M}}_\iota ^{{\mathcal {Z}}_\beta }||\mathrm {lh}(E_\iota ^{{\mathcal {Z}}_\beta }), E_\alpha ^{\mathcal {X}})$ .

3. (1.3) Clause (1.2) above describes a natural embedding $\pi _{\beta , \alpha +1}: {\mathcal {Z}}_\beta \rightarrow {\mathcal {Z}}_{\alpha +1}$ , a tree embedding. If now $\alpha <\mathrm {lh}({\mathcal {X}})$ is a limit ordinal, then ${\mathcal {Z}}_\alpha $ is obtained as the direct limit of the system $({\mathcal {Z}}_\beta , \pi _{\beta , \beta '}: \beta <\beta ', \beta \in [0, \alpha )_{\mathcal {X}}, \beta '\in [0, \alpha )_{\mathcal {X}})$ .

Now set $p=({\mathcal {T}}, {\mathcal {X}})$ and $q=({\mathcal {U}}, {\mathcal {Y}})$ , and let $({\mathcal {Z}}_\alpha ^p, {\mathcal {Z}}_\alpha ^q: \alpha <\mathrm {lh}({\mathcal {X}}))$ be the two sequences produced by the respective normalization processes. To show that ${\mathcal {Z}}$ omits F, we inductively show that for $\alpha <\mathrm {lh}({\mathcal {X}})$ , ${\mathcal {Z}}_\alpha ^p$ and ${\mathcal {Z}}_\alpha ^q$ omit F, and a close examination shows that the limit case is trivial.

We now examine the successor stage of the induction. Suppose $\alpha +1<\mathrm {lh}({\mathcal {X}})$ is such that ${\mathcal {Z}}_\alpha ^p$ and ${\mathcal {Z}}_\alpha ^q$ omit F. We want to see that ${\mathcal {Z}}_{\alpha +1}^p$ and ${\mathcal {Z}}_{\alpha +1}^q$ also omit F. Let $\gamma $ be such that ${\mathcal {Z}}_\alpha ^p\restriction \gamma ={\mathcal {Z}}_\alpha ^q\restriction \gamma $ , $\gamma +1\leq \max (\mathrm {lh}({\mathcal {Z}}_\alpha ^p), \mathrm {lh}({\mathcal {Z}}_\alpha ^q))$ and $E_{\gamma }^{{\mathcal {Z}}_\alpha ^p}\not = E_{\gamma }^{{\mathcal {Z}}_\alpha ^q}$ . Assume without loss of generality that $\mathrm {lh}(E_{\gamma }^{{\mathcal {Z}}_\alpha ^q})<\mathrm { lh}(E_{\gamma }^{{\mathcal {Z}}_\alpha ^p})$ . In this case, setting $G=E_{\gamma }^{{\mathcal {Z}}_\alpha ^q}$ , we have that ${\mathcal {Z}}_{\alpha +1}^q={\mathcal {Z}}_{\alpha }^q$ and ${\mathcal {Z}}_{\alpha +1}^p$ is the full normalization of $({\mathcal {Z}}_{\alpha }^p)^{\frown }\{G\}$ .

Notice that since $G=E_{\gamma }^{{\mathcal {Z}}_\alpha ^q}$ and ${\mathcal {Z}}_\alpha ^q$ omits F, G cannot witness that ${\mathcal {Z}}_{\alpha +1}^p$ does not omit F. Also, because ${\mathcal {Z}}_{\alpha +1}^p\restriction \gamma +1={\mathcal {Z}}_{\alpha }^p\restriction \gamma +1$ , we have that ${\mathcal {Z}}_{\alpha +1}^p\restriction \gamma +2$ omits F.

Fix some $\iota>0$ , and let $\xi \leq \gamma $ be the predecessor of $\gamma +1$ in ${\mathcal {Z}}_{\alpha +1}^p$ . Then ${\mathcal {M}}_{\gamma +\iota }^{{\mathcal {Z}}_{\alpha +1}^p}=Ult({\mathcal {M}}_{\xi +\iota -1}^{{\mathcal {Z}}_\alpha ^p}, G)$ , and $E_{\gamma +\iota }^{{\mathcal {Z}}_{\alpha +1}^p}$ is the last extender of $Ult({\mathcal {M}}_{\xi +\iota -1}||\mathrm {lh}(E_{\xi +\iota -1}^{{\mathcal {Z}}_\alpha ^p}), G)$ . Because ${\mathcal {Z}}_\alpha ^p$ omits F, it is now straightforward to verify that ${\mathcal {Z}}_{\alpha +1}^p\restriction \gamma +\iota +1$ omits F.

Proof. TheFootnote ³⁷ argument is somewhat standard, so we will only give an outline. The key fact to keep in mind is that if ${\mathcal {T}}$ is an iteration tree on ${\mathcal {P} }(F)$ which omits F, then ${\mathcal {T}}$ can be split into two components ${\mathcal {T}}_l^{\frown } {\mathcal {T}}_r$ such that ${\mathcal {T}}_l$ is based on ${\mathcal {P} }|\nu (F)$ , and if ${\mathcal {T}}_l\not ={\mathcal {T}}$ , then $\pi ^{{\mathcal {T}}_l}$ is defined and ${\mathcal {T}}_r$ is strictly above $\mathrm {lh}(\pi ^{{\mathcal {T}}_l}(F))$ . Below and elsewhere, $F_{\mathcal { Q}}=\pi ^{{\mathcal {T}}_{{\mathcal {P} }(F), {\mathcal { Q}}}}(F)$ .

We merely sketch Claim 6.8, as it follows from standard facts from the HOD analysis (see [Reference Steel and Woodin33]). The key point is that if $\eta $ is the least Woodin cardinal of ${\mathcal { Q}}$ , ${\mathcal {U}}$ is ${\mathcal {P} }(F)$ -to- ${\mathcal { Q}}$ iteration tree according to $\Sigma $ , and $\alpha <\mathrm { lh}({\mathcal {U}})$ is the least such that the generators of ${\mathcal {U}}\restriction \alpha +1$ are contained in $\eta $ , then letting ${\mathcal {S}}={\mathcal {M}}_\alpha ^{\mathcal {U}}$ , we have ${\mathcal {S}}\in \mathcal {F}({\mathcal {P} }(F), \Sigma , F)$ . Then if $i: {\mathcal {P} }\rightarrow {\mathcal {S}}$ and $k:{\mathcal {S}}\rightarrow {\mathcal { Q}}$ are the iteration embeddings,Footnote ³⁸ then $j=k\circ i$ and ${\mathcal {M}}_\infty ({\mathcal {P} }(F), \Sigma , F)={\mathcal {M}}_\infty ({\mathcal {S}}, \Sigma _{\mathcal {S}}, i(F))$ . In addition, $X_{{\mathcal {P} }, \Sigma , F}\triangleleft {\mathcal {M}}_\infty ({\mathcal {S}}|\eta , \Sigma _{{\mathcal {S}}|\eta }, i(F))$ . Now let $\zeta $ be the least inaccessible cardinal of ${\mathcal {S}}$ such that $i(F)\in {\mathcal {S}}|\zeta $ . It follows that $X_{{\mathcal {P} }, \Sigma , F}\triangleleft {\mathcal {M}}_\infty ({\mathcal {S}}|\eta , \Sigma _{{\mathcal {S}}|\eta }, i(F))$ , and we have that ${\mathcal { Q}}|\zeta ={\mathcal {S}}|\zeta $ . The last missing piece is that $\Sigma _{{\mathcal { Q}}|\zeta }$ is captured by ${\mathcal { Q}}$ , and ${\mathcal { Q}}$ can compute ${\mathcal {M}}_\infty ({\mathcal {S}}|\eta , \Sigma _{{\mathcal {S}}|\eta }, i(F))$ .Footnote ³⁹

We can then develop the concept of suitable premouse, A-iterable suitable premouse, and other concepts used in the HOD analysis for iterations that omit F. For example, we define ${\mathcal {S}}$ to be suitable if, in addition to the usual properties of suitability (see [Reference Steel and Woodin33]), for some $G\in \vec {{\mathcal {S}}}$ , $X_{{\mathcal {P} }, \Sigma , F}$ is a complete iterate of ${\mathcal {S}}||\mathrm {lh}(G)$ . A-iterability is defined for those $A\subseteq {\mathbb {R}}$ which are ordinal definable from $X_{{\mathcal {P} }, \Sigma , F}$ , and given a suitable ${\mathcal {S}}$ we define the concept of A-iterability only for those iterations of ${\mathcal {S}}$ that omit G, where G is as above. Claim 6.8 can now be used to show that for every $A\subseteq {\mathbb {R}}$ that is ordinal definable from $X_{{\mathcal {P} }, \Sigma , F}$ , there is a strongly A-iterable pair. The rest is just like in the ordinary HOD analysis, and we leave it to the reader.

Proof. Fix a regular Suslin cardinal $\delta $ , and toward a contradiction, assume that $\delta $ is not an $\omega $ -club $\Theta $ -Berkeley cardinal. Fix a transitive $N'$ such that

1. (1.1) $\left |N'\right |<\Theta $ ,

2. (1.2) $\delta \subseteq N'$ , and

3. (1.3) the set of $\alpha <\delta $ such that there is no elementary embedding $j: N'\rightarrow N'$ with the property that $\mathrm {crit }(j)=\alpha $ is $\omega $ -stationary in $\delta $ .Footnote ⁴⁰

Let $\phi (u, v)$ be the formula expressing (1.1)–(1.3). Thus, $\phi [\delta , N']$ holds. Since $L({\wp }({\mathbb {R}}))\vDash \phi [\delta , N']$ , we can assume that $V=L({\wp }({\mathbb {R}}))$ . Without loss of generality, assume that $\delta $ is the least regular Suslin cardinal $\kappa $ such that $\phi [\kappa , M]$ holds for some M. It then immediately follows that $\delta $ cannot be the largest Suslin cardinal, as if $\delta $ were the largest Suslin cardinal, then for some $\alpha <\delta $ and some $\beta <\delta $ , letting $\Delta =\{A\subseteq {\mathbb {R}}: w(A)<\beta \},$ Footnote ⁴¹ $L_\alpha (\Delta )\vDash {\textsf {ZF-Replacement}}+\exists \kappa \exists M\phi [\kappa , M]$ .Footnote ⁴² A similar reflection argument shows that we can assume without losing generality that $\left |N'\right |$ is less than some Suslin cardinal $\delta '$ such that $\delta <\delta '$ and $\delta '$ is not the largest Suslin cardinal. Fix now such a $\delta '$ so that $\left |N'\right |<\delta '$ .

Let $(R, \Psi , H, \alpha )$ be as in Theorem 4.7 absorbing $\delta '$ . Notice that our discussion above implies that $W\vDash \phi [\delta , N']$ , for $W=L_\alpha ^\Psi ({\mathbb {R}})$ . We can then find $x_0$ such that $N'$ is ${\textsf {OD}}^W_{\Psi , x_0}$ , and then by minimizing, we can let N be the ${\textsf {OD}}^W_{\Psi , x_0}$ -least M such that $W\vDash \phi [\delta , M]$ . We now fix $x_1\in \mathrm {dom}(H)$ such that

1. (2.1) $x_1$ is Turing above $x_0$ and $N\in {\mathcal {M}}_\infty ({\mathcal {P} }_{x_1}, \Sigma _{x_1})$ , where $({\mathcal {P} }_{x_1}, \Sigma _{x_1})=H(x_1)$ , and

2. (2.2) $\delta $ is a limit of cutpoints of ${\mathcal {M}}_\infty ({\mathcal {P} }, \Sigma )$ , where $({\mathcal {P} }, \Sigma )=({\mathcal {P} }_{x_1}, \Sigma _{x_1})$ .

Without loss of generality, we assume that $\delta \in \mathrm {rge}(\pi ^\Sigma _{{\mathcal {P} }, \infty })$ , and for ${\mathcal { Q}}$ a complete $\Sigma $ -iterate of ${\mathcal {P} }$ , we let $\delta _{\mathcal { Q}}=\pi _{{\mathcal { Q}}, \infty }^{-1}(\delta )$ .

Fix now any completely total extender $F\in \vec {E}^{\mathcal {P} }$ such that $\mathrm {crit }(F)=\delta _{\mathcal {P} }$ , and set $X_{\mathcal {P} }=X_{{\mathcal {P} }, \Sigma , F}$ . Let ${\mathcal { Q}}=Ult({\mathcal {P} }(F), F)$ , $F_{\mathcal { Q}}=\pi _F(F)$ and $X_{\mathcal { Q}}=X_{{\mathcal { Q}}, \Sigma _{\mathcal { Q}}, F_{\mathcal { Q}}}$ . We set $\Lambda =\Sigma ^F$ and $\Phi =\Sigma ^{F_{\mathcal { Q}}}_{\mathcal { Q}}$ . As in the proof of Theorem 3.1, we define $j: {\mathcal {M}}_\infty ({\mathcal {P} }, \Sigma , F)\rightarrow {\mathcal {M}}_\infty ({\mathcal { Q}}, \Sigma _{\mathcal { Q}}, F_{\mathcal { Q}})$ . It follows from Theorem 6.7 and the proof of Claim 3.3 that $N\in {\mathcal {M}}_\infty ({\mathcal {P} }, \Sigma , F)\cap {\mathcal {M}}_\infty ({\mathcal { Q}}, \Sigma _{\mathcal { Q}}, F_{\mathcal { Q}})$ and $j(N)=N$ , and therefore, defining j is all that we will do.

Fix $u\in {\mathcal {M}}_\infty ({\mathcal {P} }, \Sigma , F)$ , and let ${\mathcal {S}}$ be a complete $\Lambda $ -iterate of ${\mathcal {P} }(F)$ such that $u=\pi _{{\mathcal {S}}, \infty }^{\Lambda }(u_{\mathcal {S}})$ for some $u_{\mathcal {S}}\in {\mathcal {S}}$ . Let $F_{\mathcal {S}}=\pi _{{\mathcal {P} }(F), {\mathcal {S}}}(F)$ Footnote ⁴³ and ${\mathcal {S}}_F=Ult({\mathcal {S}}, F_{\mathcal {S}})$ . We then letFootnote ⁴⁴

$$ \begin{align*} j(u)=\pi_{{\mathcal{S}}_F, \infty}^{\Phi}(\pi_{F_{\mathcal{S}}}(u_{\mathcal{S}})). \end{align*} $$

The definition of $j(u)$ makes sense, as full normalization implies that ${\mathcal {S}}_F$ is a complete $\Phi $ -iterate of ${\mathcal { Q}}$ . To prove this and other claims in this section, we set ${\mathcal {P} }={\mathcal {P} }(F)$ to simplify the notation.