Proof. See [Reference Steel16, Theorem 6.12].

Proof. Let us assume otherwise and let $Q_0\not =Q_1$ be two witnesses. Since we can always take a countable hull of some $H_\theta $ , for a large enough $\theta $ , we may assume w.l.o.g. that P, $Q_0$ , and $Q_1$ are countable. By the argument of [Reference Steel16, Corollary 3.12], we have that either $Q_0\unlhd Q_1$ or $Q_1\unlhd Q_0$ . However, the minimality which is a part of the definition of a Q-structure would then imply $Q_0=Q_1$ , which is a contradiction.

Proof. Let us assume otherwise and let $b_0\neq b_1$ be two such branches. Since M is tame, $\delta (\mathcal {T})$ is a strong cutpoint in both $\mathcal {Q}(\mathcal {T},b_0)$ and $\mathcal {Q}(\mathcal {T},b_1)$ . By the uniqueness of Q-structures (cf. Lemma 2.7), we have that $\mathcal {Q}(\mathcal {T},b_0)=\mathcal {Q}(\mathcal {T},b_1)$ , which contradicts Proposition 2.5.

Proof. The Q-structure $\mathcal {Q}(\mathcal {T},b)$ is countably iterable since $\mathcal {M}^{\mathcal {T}}_b$ is countably iterable. The ordinal $\delta (\mathcal {T})$ is a strong cutpoint in $\mathcal {Q}(T,b)$ because M is tame. This suffices for the conclusion.

Proof. Since M is $(\omega _1+1)$ -iterable, we have that it is weakly $\omega $ -iterable.Footnote ¹³ By the proof of [Reference Steel16, Theorem 7.10], we have that M has an $(\omega _1+1)$ -iteration strategy in $L(\mathbb {R})$ . The moreover part follows from Definition 2.13.

Proof. This is a straightforward generalization of [Reference Steel16, Corollary 3.12].

Proof. Since M projects to $1$ and is sound, there exists a surjection $f:J_1(a)\twoheadrightarrow M$ which is definable over M. Now, if we look at $J_1(M)$ , organized as an a-premouse, we see that it is obtained as the rudimentary closure of $M\cup \{M\}$ . However, the rudimentary functions can be listed recursively, so the fact that $V_\omega \cup \{f\}\subseteq J_1(a)$ allows us to define a surjection

$$ \begin{align*}g:J_1(a)\twoheadrightarrow J_1(M)\end{align*} $$

over $J_1(M)$ . This means that $J_1(M)$ projects to $1$ and is consequently $\mathsf {Lp}$ -good.

Proof.

1. 1^° If it were the case that $k(\mathsf {Lp}(a))>0$ , then the a-premouse obtained from $\mathsf {Lp}(a)$ by decreasing $k(\mathsf {Lp}(a))$ by 1 would also satisfy condition a of Definition 3.4, while being a strict initial segment of $\mathsf {Lp}(a)$ . This shows that a must hold.

2. 2^° To establish b, let us assume otherwise. Then the set C of all $\alpha \in (\mathsf {rank}(a),\omega _1)$ such that there exists an $\mathsf {Lp}$ -good a-premouse M satisfying $\hat o(M)=\alpha $ is cofinal in $\omega _1$ .

3. 3^° For all $\alpha \in C$ , an M witnessing this fact is unique and we denote it by $M_\alpha $ .

4. 4^° Let $(f_\alpha : \alpha \in C),g$ be as follows:

   1. 1. for all $\alpha \in C$ , $f_\alpha :a\cup \{a\}\twoheadrightarrow M_\alpha $ is given by the soundness;

   2. 2. $g:\omega \twoheadrightarrow a\cup \{a\}$ is an arbitrary enumeration;

   3. 3. for all $\alpha \in C$ , $e_\alpha :=\{(m,n) : f_\alpha (g(m))\in f_\alpha (g(n))\}\subseteq \omega ^2$ .

5. 5^° The sequence $(e_\alpha : \alpha \in C)$ is injective and it belongs to $L(\mathbb {R})$ . This contradicts $L(\mathbb {R})\models \mathsf {AD}$ .

Proof. This follows from Lemma 3.5 and the fact that there exists $\delta <\omega _1$ such that $M/\delta =\mathsf {Lp}(M{\parallel }\delta )$ .

Proof. See the paragraph immediately below Definition 6.8 of [Reference Steel and Woodin18].

Proof.

1. 1^° By Proposition 2.14, we have that $\mathcal {Q}(\mathcal {T},b)$ has an $(\omega _1+1)$ -iteration strategy in $L(\mathbb {R})$ . In particular, $\mathcal Q(\mathcal T,b)/\delta(\mathcal T)$ is $\mathsf {Lp}$ -good and consequently,

   $$ \begin{align*}\mathcal{Q}(\mathcal{T},b)/\delta(\mathcal{T})\unlhd\mathsf{Lp}(\mathcal{M}(\mathcal{T})).\end{align*} $$

2. 2^° Let us now verify that $\Sigma _M(\mathcal {T})=b$ . We assume towards contradiction that the branch $c:=\Sigma _M(\mathcal {T})$ is distinct from b.

3. 3^° The Q-structure $\mathcal {Q}(\mathcal {T},c)$ does not exist, for otherwise it would be $(\omega _1+1)$ -iterable, while there can be only one branch with an iterable Q-structure. Hence, we get that c does not drop and $\mathcal {M}^{\mathcal {T}}_c\models $ “ $\delta (\mathcal {T})$ is Woodin.”

4. 4^° Since c does not drop, we have that $\mathcal {M}^{\mathcal {T}}_c$ is super-suitable.

5. 5^° Putting together ${3}^\circ $ and ${4}^\circ $ , we get that $\mathcal {M}^{\mathcal {T}}_c$ is suitable and that $\delta (\mathcal {T})$ is its unique Woodin. This means that

   $$ \begin{align*}\mathcal{M}^{\mathcal{T}}_c/\delta(\mathcal{T})=\mathsf{Lp} (\mathcal{M}^{\mathcal{T}}_c{\parallel}\delta(\mathcal{T}))=\mathsf{Lp} (\mathcal{M}(\mathcal{T})).\end{align*} $$

6. 6^° Adding to this the conclusion of ${1}^\circ $ , we have that

   $$ \begin{align*}\mathcal Q(\mathcal T,b)/\delta(\mathcal T)\unlhd\operatorname{Lp}(\mathcal M(\mathcal T))=\mathcal M_c^{\mathcal T}/\delta(\mathcal T),\end{align*} $$
   whence $\mathcal Q(\mathcal T,b)\unlhd\mathcal M_c^{\mathcal T}$ .

7. 7^° Since $\delta (\mathcal {T})$ is Woodin in $\mathcal {M}_c^{\mathcal {T}}$ , it follows that $\mathcal {Q}(\mathcal {T},b)=\mathcal {M}_c^{\mathcal {T}}$ .

8. 8^° Now, this would mean that $\mathcal {Q}(\mathcal {T},c)$ does exist (and is equal to $\mathcal {Q}(\mathcal {T},b)$ ), which is in contradiction with ${3}^\circ $ .

Proof.

1. 1^° By induction on $\mathsf {lh}(\mathcal {T})$ , we show that for all $\mathcal {T}\in \mathsf {dom}(\Sigma _M^{\mathrm {st}})$ , if

   $$ \begin{align*}\forall\eta<\mathsf{lh}(\mathcal{T})\text{ limit},\ \mathcal{T}{\restriction}\eta\in\mathsf{dom}(\Sigma_M),\end{align*} $$
   then $\mathcal {T}\in \mathsf {dom}(\Sigma _M)$ and $\Sigma _M(\mathcal {T})=\Sigma _M^{\mathrm {st}}(\mathcal {T})$ .

2. 2^° Since $\mathcal {T}$ is according to $\Sigma _M$ , we immediately have that $\mathcal {T}\in \mathsf {dom}(\Sigma _M)$ .

3. 3^° Let us denote by b the branch $\Sigma _M^{\mathrm {st}}(\mathcal {T})$ . By the definition of $\Sigma _M^{\mathrm {st}}$ , we have that $\mathcal {Q}(\mathcal {T})$ exists and satisfies $\mathcal {Q}(\mathcal {T})\unlhd \mathcal {M}^{\mathcal {T}}_b$ . Thus, $\mathcal {Q}(\mathcal {T},b)=\mathcal {Q}(\mathcal {T})$ exists and is $(\omega _1+1)$ -iterable.

4. 4^° By Proposition 4.7, we must have that $\Sigma _M(\mathcal {T})=b$ , as required.

Proof.

1. 1^° Let $\mathcal {T}$ be a countable normal tree on M according to $\Sigma ^{\mathrm {st}}_M$ . By Proposition 5.3, $\mathcal {T}$ is also according to $\Sigma _M$ .

2. 2^° What needs to be verified depends on the type of the tree $\mathcal {T}$ , as can be seen from Definition 5.2. We have three cases.

3. 3^° Case. $\mathcal {T}$ is of successor length.

   Proof. First, we need to verify that when $\mathcal {T}$ is extended by one extender, we do not get an illfounded model. However, this is obvious since $\mathcal {T}$ is according to $\Sigma _M$ and $\Sigma _M$ is an $(\omega _1,\omega _1+1)$ -iteration strategy for M. The second thing to verify is that if the main branch of $\mathcal {T}$ does not drop, then the last model of $\mathcal {T}$ is suitable. For this, observe that if the main branch of $\mathcal {T}$ does not drop, then the last model of $\mathcal {T}$ is an iterate of a super-suitable premouse M, so it must itself be super-suitable.

4. 4^° Case. $\mathcal {T}$ is short.

   Proof.

   1. 1^′ Being short means that $\mathcal {Q}(\mathcal {T})$ exists, so we only need to verify that there exists a cofinal wellfounded branch b through $\mathcal {T}$ such that $\mathcal {Q}(\mathcal {T})\unlhd \mathcal {M}^{\mathcal {T}}_b$ . We claim that $b:=\Sigma _M(\mathcal {T})$ works.

   2. 2^′ Let us assume towards contradiction that b does not drop. Then $\mathcal {M}^{\mathcal {T}}_b$ is suitable and $\delta (\mathcal {T})$ is its Woodin.

   3. 3^′ By definition, the premouse $\mathcal {Q}(\mathcal {T})$ is $(\omega _1+1)$ -iterable and $\delta (\mathcal {T})$ is its strong cutpoint. This means that $\mathcal {Q}(\mathcal {T})/\delta (\mathcal {T})$ is $\mathsf {Lp}$ -good and consequently,

      $$ \begin{align*}\mathcal{Q}(\mathcal{T})/\delta(\mathcal{T})\unlhd\mathsf{Lp}(\mathcal{M}(\mathcal{T})).\end{align*} $$

   4. 4^′ However, we have by suitability that

      $$ \begin{align*}\mathcal{M}^{\mathcal{T}}_b/\delta(\mathcal{T})=\mathsf{Lp}(\mathcal{M}^{\mathcal{T}}_b{\parallel}\delta(\mathcal{T}))=\mathsf{Lp}(\mathcal{M}(\mathcal{T})),\end{align*} $$
      so we conclude that $\mathcal {Q}(\mathcal {T})\unlhd \mathcal {M}^{\mathcal {T}}_b$ .

   5. 5^′ If it were the case that $o(\mathcal {Q}(\mathcal {T}))< o(\mathcal {M}^{\mathcal {T}}_b)$ , it would hold that $\delta (\mathcal {T})$ is not Woodin in $\mathcal {M}^{\mathcal {T}}_b$ . This being absurd, we conclude that the J-structures underlying premice $\mathcal {Q}(\mathcal {T})$ and $\mathcal {M}^{\mathcal {T}}_b$ are equal.

   6. 6^′ However, this is a contradiction since $\rho (\mathcal {Q}(\mathcal {T}))\leq \delta (\mathcal {T})$ while

      $$ \begin{align*}\rho_\omega(\mathcal{M}^{\mathcal{T}}_b)=\hat o(\mathcal{M}^{\mathcal{T}}_b).\end{align*} $$

   7. 7^′ The conclusion is then that b drops, which means that $\mathcal {Q}(\mathcal {T},b)$ exists.

   8. 8^′ By the uniqueness of the Q-structure above $\mathcal {M}(\mathcal {T})$ (cf. Lemma 2.7), it follows that $\mathcal {Q}(\mathcal {T},b)=\mathcal {Q}(\mathcal {T})$ and consequently, $\mathcal {Q}(\mathcal {T})\unlhd \mathcal {M}^{\mathcal {T}}_b$ , as required.

5. 5^° Case. $\mathcal {T}$ is maximal.

   Proof.

   1. 1^′ In this case, we need to verify that there exists a non-dropping cofinal wellfounded branch b through $\mathcal {T}$ such that $\mathcal {M}^{\mathcal {T}}_b$ is suitable. We claim that $b:=\Sigma _M(\mathcal {T})$ works.

   2. 2^′ If b dropped, then $\mathcal {Q}(\mathcal {T},b)$ would exist and be $(\omega _1+1)$ -iterable. In other words, $\mathcal {Q}(\mathcal {T})$ would exist and be equal to $\mathcal {Q}(\mathcal {T},b)$ , which is not the case.

   3. 3^′ Thus, b does not drop.

   4. 4^′ It follows that $\mathcal {M}^{\mathcal {T}}_b$ is an iterate of a super-suitable premouse M, as witnessed by $\mathcal {T}^\frown b$ , so $\mathcal {M}^{\mathcal {T}}_b$ is super-suitable.

   5. 5^′ In particular, $\mathcal {M}^{\mathcal {T}}_b$ is suitable, as required.

6. 6^° The three cases above cover all possibilities, so the verification is concluded.

Proof. This is immediate from the definition since it subsumes the short tree iterability of M.

Proof.

1. Implication ( $\Rightarrow $ )

2. 1^° The part a follows from Proposition 5.3, while the part b follows from the part a.

3. 2^° We want to verify the part c. In this case, the tree $\mathcal {T}$ is maximal.

4. 3^° If b dropped, then $\mathcal {Q}(\mathcal {T},b)$ would exist and be $(\omega _1+1)$ -iterable, which would mean that $\mathcal {Q}(\mathcal {T})$ exists and is equal to $\mathcal {Q}(\mathcal {T},b)$ . This being contradictory, we conclude that b does not drop.

5. 4^° It follows that $\mathcal {M}^{\mathcal {T}}_b$ is super-suitable and that $\delta (\mathcal {T})$ is its Woodin. In particular,

   $$ \begin{align*}\mathcal{M}^{\mathcal{T}}_b/\delta(\mathcal{T})=\mathsf{Lp}(\mathcal{M}(\mathcal{T}))=N/\delta(\mathcal{T}),\end{align*} $$
   where the second equality follows from Lemma 5.6.

6. 5^° This suffices for the conclusion that $N=\mathcal {M}^{\mathcal {T}}_b$ .

7. Implication ( $\Leftarrow $ )

8. 6^° The assumptions yield that if $\mathcal {T}$ has a last model, then $\mathcal {T}{\restriction }(\mathsf {lh}(\mathcal {T})-1)$ is not maximal.

9. 7^° We also know that there cannot exist $\xi $ such that $\xi +1<\mathsf {lh}(\mathcal {T})$ and $\mathcal {T}{\restriction }\xi $ is maximal. The reason for this is that $\delta (\mathcal {T})=\delta (\mathcal {M}^{\mathcal {T}}_\xi )$ is a strong cutpoint of $\mathcal {M}^{\mathcal {T}}_\xi $ , so any cofinal branch through $\mathcal {T}$ must contain $\xi $ and must drop at the successor of $\xi $ .

10. 8^° These two facts together with Proposition 5.3 imply that $\mathcal {T}$ is according to $\Sigma ^{\mathrm {st}}_M$ . The rest is now a routine verification.

Proof. This follows from Lemma 5.7.

Proof. This follows from [Reference Larson6, Theorem 10.3].

Proof. See [Reference Larson6, Corollary 10.24].

Proof.

1. 1^° Let T be the tree of a $(\Sigma ^2_1)^{L(\mathbb {R})}$ -scale on a universal $(\Sigma ^2_1)^{L(\mathbb {R})}$ -set and let

   $$ \begin{align*}\mu:=\max\{\delta(M),\delta(N)\}^{+L[T,M,N]}.\end{align*} $$

   We want to show that $\mu <\omega _1$ , for once we do this, the conclusion follows from [Reference Steel and Woodin18, Theorem 6.14].

2. 2^° Let x be a real coding M and N. It suffices to show that $\omega _1^{L[T,x]}<\omega _1$ .

3. 3^° The theorem at the bottom of the page 77 of [Reference Steel13] implies that

   $$ \begin{align*}L[T]\cap V_{\boldsymbol\delta^2_1} = \mathsf{HOD} \cap V_{\boldsymbol\delta^2_1}.\end{align*} $$

   We denote this set by X. Since $\boldsymbol \delta ^2_1$ is an inaccessible cardinal in $\mathsf {HOD}$ , it holds that $X \models \mathsf {ZFC}$ .

4. 4^° By Lemmas 5.10 and 5.11, $\mathsf {HOD}_x = \mathsf {HOD} [x]$ , and this class is a ${<}\boldsymbol \delta ^2_1$ -generic extension of $\mathsf {HOD}$ . Consequently, x is generic over X and

   $$ \begin{align*}\mathsf{HOD}_x \cap V_{\boldsymbol\delta^2_1} = \mathsf{HOD} [x] \cap V_{\boldsymbol\delta^2_1} = \left( \mathsf{HOD} \cap V_{\boldsymbol\delta^2_1} \right) [x] = X[x].\end{align*} $$

5. 5^° By ${3}^\circ $ , $L[T]\cap V_{\boldsymbol \delta ^2_1} = X$ and $\boldsymbol \delta ^2_1$ is a beth-fixed point in $L[T]$ , so x is ${<} \boldsymbol \delta ^2_1$ -generic over $L[T]$ and

   $$ \begin{align*}L[T][x] \cap V_{\boldsymbol\delta^2_1} = X[x].\end{align*} $$

6. 6^° We can now conclude that

   $$ \begin{align*}\omega_1^{L[T, x]} = \omega_1^{L[T][x]} = \omega_1^{X[x]} = \omega_1^{\mathsf{HOD}_x} < (\text{"the first measurable"})^{\mathsf{HOD}_x} = \omega_1,\end{align*} $$
   as required.

Proof. Such a tree $\mathcal {T}$ exists by the fact that N is a normal iterate of M. The uniqueness follows from the fact that any such tree must be the first coordinate of the coiteration of $(M,\Sigma _M)$ and $(N,\Sigma _N)$ . We will further elaborate on this last statement. A way to restate it is to say that $(\mathcal {T}, \mathcal {U})$ is the coiteration of $(M,\Sigma _M)$ and $(N,\Sigma _N)$ whenever $\mathcal {T}$ is a normal tree on $(M,\Sigma _M)$ whose last model is N and $\mathcal {U}$ is the trivial tree on $(N,\Sigma _N)$ whose length is $\mathsf {lh}(\mathcal {T})$ .Footnote ¹⁷ This is verified by induction.

Let $\alpha < \mathsf {lh}(\mathcal {T})$ be such that $(\mathcal {T} {\restriction } \alpha , \mathcal {U} {\restriction } \alpha )$ is the coiteration up to stage $\alpha $ . We want to show that $(\mathcal {T} {\restriction } (\alpha + 1), \mathcal {U} {\restriction } (\alpha + 1))$ is the coiteration up to stage $\alpha + 1$ . If $\alpha $ is $0$ , this is trivially true; if $\alpha $ is a limit ordinal, this follows from the fact that $\mathcal {T}$ and $\mathcal {U}$ are built using the same pair of strategies as in the case of the coiteration. So, let us assume that $\alpha $ is a successor ordinal. The least disagreement between $\mathcal {M}^{\mathcal {T}}_{\alpha - 1}$ and $\mathcal {M}^{\mathcal {U}}_{\alpha - 1}$ is exactly the least disagreement between $\mathcal {M}^{\mathcal {T}}_{\alpha - 1}$ and N, which is just another way of saying “the least disagreement between $\mathcal {M}^{\mathcal {T}}_{\alpha - 1}$ and $\mathcal {M}^{\mathcal {T}}_{\mathsf {lh}(\mathcal {T}) - 1}$ .” By [Reference Steel16, Lemma 3.5], this least disagreement occurs at the index $\mathsf {lh}(E^{\mathcal {T}}_{\alpha - 1})$ , and this index is a cardinal in $\mathcal {M}^{\mathcal {T}}_{\mathsf {lh}(\mathcal {T}) - 1}$ ; in other words, the least disagreement is exactly $(E^{\mathcal {T}}_{\alpha - 1}, \emptyset )$ .Footnote ¹⁸ This means that to get from $(\mathcal {T} {\restriction } \alpha , \mathcal {U} {\restriction } \alpha )$ to $(\mathcal {T} {\restriction } (\alpha + 1), \mathcal {U} {\restriction } (\alpha + 1))$ , one indeed uses the least disagreement between $\mathcal {M}^{\mathcal {T}}_{\alpha - 1}$ and $\mathcal {M}^{\mathcal {U}}_{\alpha - 1}$ , so $(\mathcal {T} {\restriction } (\alpha + 1), \mathcal {U} {\restriction } (\alpha + 1))$ remains an initial segment of the coiteration.

The argument just given establishes that $(\mathcal {T}, \mathcal {U})$ is an initial segment of the coiteration of $(M,\Sigma _M)$ and $(N,\Sigma _N)$ . However, the final models of $\mathcal {T}$ and $\mathcal {U}$ are equal (viz., N), so the coiteration terminates exactly at this stage. This shows that $(\mathcal {T}, \mathcal {U})$ is exactly the coiteration of $(M,\Sigma _M)$ and $(N,\Sigma _N)$ .

Proof. The commutativity follows from the normalization (cf. [Reference Schlutzenberg11, Theorem 1.1]), while wellfoundedness follows from the fact that $(\mathcal {I}(M),\leq _{\mathcal {I}})$ is countably directed.

Proof. Let $\sigma \in \eta ^\omega $ be given. Fix a set of reals $A \in L(\mathbb {R})$ of Wadge rank $\eta $ . For every $n < \omega $ , there is a real $x_n$ which codes a continuous reduction of the set A to a set of reals of Wadge rank $\sigma _n$ . Let y be a real that codes the sequence $(x_n : n < \omega )$ . Now, the pair $(A, y)$ belongs to $L(\mathbb {R})$ and codes the sequence $\sigma $ in the following way: from y, one decodes the sequence of reals $(x_n : n < \omega )$ , then the sequence of continuous functions $(f_n : n < \omega )$ from $\mathbb {R}$ to $\mathbb {R}$ , and finally, the sequence

$$ \begin{align*}\sigma=\bigl(\mathrm{rank}_W(f_n^{-1}[A]):n<\omega\bigr).\end{align*} $$

Therefore, $\sigma \in L(\mathbb {R})$ .

Proof.

1. 1^° By [Reference Neeman8, Corollary 4.14], there exists in $L(\mathbb {R})$ a unique supercompactness measure $\mu $ on $[\eta ]^\omega $ . We will be done if we show that $\mu = \mu _\eta $ .

2. 2^° Since $\mu $ is an $L(\mathbb {R})$ -ultrafilter, it suffices to show that every $S \in \mu $ contains a club (in V).

3. 3^° For $x \in \mathbb {R}$ , we denote by $\mathcal {M}_x$ the x-premouse $\mathcal {M}_\omega (x) {\parallel } \tau $ , where $\tau := \delta (\mathcal {M}_\omega (x))^{+\mathcal {M}_\omega (x)}$ , and by $\Sigma _{\mathcal {M}_x}$ the $(\omega _1, \omega _1 + 1)$ -iteration strategy for $\mathcal {M}_x$ derived from a fixed $(\omega _1, \omega _1 + 1)$ -iteration strategy for $\mathcal {M}_\omega (x)$ . An x-super-suitable premouse is a $\Sigma _{\mathcal {M}_x}$ -iterate of $\mathcal {M}_x$ . For an x-super-suitable premouse M, we denote by $\Sigma _M$ the strategy for M induced by $\Sigma _{\mathcal {M}_x}$ .

4. 4^° By repeating the definitions given above while replacing $\mathcal {M}$ and $\Sigma _{\mathcal {M}}$ by $\mathcal {M}_x$ and $\Sigma _{\mathcal {M}_x}$ (resp.), super-suitable by x-super-suitable, etc., we obtain the following objects:

   • • $(\mathcal {I}_x(M), \leq _{\mathcal {I}_x})$ ,

   • • $\pi ^x_{M, N}: M \to N$ ,

   • • ,

   • • $\mathcal {M}_\infty ^x$ ,

   • • $\pi ^x_{M, \infty }$ ,

   where x is a real, M is x-super-suitable, and N is a normal iterate of M. Furthermore, the facts of this section remain true when conditioned on a real x (and when $\mathsf {HOD}$ is replaced by $\mathsf {HOD}_x$ ).

5. 5^° For $x \in \mathbb {R}$ , a sequence of the form $(M_n, \mathcal {T}_n: n < \omega )$ is an x-super-correct iteration sequence iff $M_0$ is x-super-suitable and for all $n < \omega $ , $\mathcal {T}_n$ is a countable normal tree on $M_n$ according to $\Sigma_{M_n}$ with the last model $M_{n + 1}$ ; we say that this x-super-correct iteration sequence is on $M_0$ . For an x-super-correct iteration sequence $\sigma $ , we denote by $a(\sigma )$ the countable set

   $$ \begin{align*}\bigcup_{n < \omega} \mathsf{ran} (\pi^x_{M_n, \infty}) \cap \eta.\end{align*} $$

6. 6^° For $x \in \mathbb {R}$ and for an x-super-suitable M, let $C^x_M$ consist of all $a(\sigma )$ where $\sigma $ is an x-super-correct iteration sequence on M. By [Reference Neeman8, Remark 4.11], there exist $x \in \mathbb {R}$ and an x-super-suitable M such that $C^x_M \subseteq S$ .

7. 7^° We will be done if we show that $C^x_M$ contains a club. By [Reference Kasum and Sargsyan4, Proposition 3.14], it suffices to show that Player II has a winning quasi-strategy in $\Game ^{\mathrm {club}}(\eta , C^x_M)$ .

8. 8^° Let $\Game ^*$ be a length $\omega $ two player game of the form

   $$ \begin{align*} \begin{array}{c|ccccc} \text{I} & \alpha_0 & & \alpha_1 & & \dots\\ \hline \text{II} & & \beta_0, \mathcal{T}_0, M_1 & & \beta_1, \mathcal{T}_1, M_2 & \dots \end{array} \end{align*} $$
   where $M_0 := M$ by convention and for all $n < \omega $ ,

   • • $\alpha _n, \beta _n < \eta $ ;

   • • $\mathcal {T}_n$ is a normal tree on $M_n$ according to $\Sigma _{M_n}$ with the last model $M_{n + 1}$ .

   Player II wins iff

   $$ \begin{align*}a \left( (M_n, \mathcal{T}_n: n < \omega) \right) = \{\alpha_n, \beta_n: n < \omega\}.\end{align*} $$

9. 9^° Observe that for every play $(\alpha _n, (\beta _n, \mathcal {T}_n, M_{n + 1}): n < \omega )$ of $\Game ^*$ which is won by Player II, $(\alpha _n, \beta _n: n < \omega )$ is a play of $\Game ^{\mathrm {club}}(\eta , C^x_M)$ which is won by Player II. So, it suffices to show that Player II has a winning quasi-strategy in the game $\Game ^*$ .

10. 10^° This readily follows: a winning quasi-strategy for Player II in $\Game ^*$ consists in

    • • ensuring that for all $n < \omega $ , $\alpha _n \in \mathsf {ran}(\pi^x_{M_{n+1},\infty})$ ;

    • • fixing an enumeration $\{\gamma ^n_i: i < \omega \}$ of $\mathsf {ran}(\pi ^x_{M_n, \infty }) \cap \eta $ for every $n < \omega $ ;

    • • using a bookkeeping device when choosing $\beta _n$ ’s so as to ensure that

      $$ \begin{align*}\{\beta_n : n < \omega\} = \{\gamma^n_i: n, i < \omega\}.\end{align*} $$

    Player II indeed wins in this way: in the end, we have

    $$ \begin{align*}\{\alpha_n, \beta_n : n < \omega\} = \{\gamma^n_i: n, i < \omega\} = a \left( (M_n, \mathcal{T}_n : n < \omega) \right).\end{align*} $$

Proof.

1. 1^° Recall that M is an initial segment of an iterate of $\mathcal {M}_\omega ^\sharp $ . This means that no extender on the M-sequence, or on the sequence of any of its iterates, overlaps a Woodin cardinal.

2. 2^° Assume towards a contradiction that the conclusion fails and let $\alpha < \mathsf {lh}(\mathcal {T}) - 1$ be the least such that $\delta (\mathcal {M}^{\mathcal {T}}_\alpha )$ is defined and $\mathsf {lh}(E^{\mathcal {T}}_\alpha ) \geq \delta (\mathcal {M}^{\mathcal {T}}_\alpha )$ .

3. 3^° By [Reference Steel16, Lemma 3.5], we get that $\mathcal {M}^{\mathcal {T}}_\alpha $ and the last model of $\mathcal {T}$ (viz., $\mathcal {M}_\infty $ ) agree below $\mathsf {lh}(E^{\mathcal {T}}_\alpha )$ and $\mathsf {lh}(E^{\mathcal {T}}_\alpha )$ is a cardinal in $\mathcal {M}_\infty $ .

4. 4^° Since $E^{\mathcal {T}}_\alpha $ does not overlap $\delta (\mathcal {M}^{\mathcal {T}}_\alpha )$ , we have that $\mathsf {crit}(E^{\mathcal {T}}_\alpha ) \geq \delta (\mathcal {M}^{\mathcal {T}}_\alpha )$ .

5. 5^° The last two points yield that in $\mathcal {M}_\infty $ , there is a measurable Woodin cardinal or there is a measurable cardinal above a Woodin cardinal—either case is absurd.

Proof. Assume towards contradiction that $\mathsf {lh}(E^{\mathcal {T}}_{\xi }) \geq \Theta $ . By [Reference Steel16, Lemma 3.5], $\mathsf {lh}(E^{\mathcal {T}}_\xi )$ is a successor cardinal in the last model of $\mathcal {T}$ , that is, in $\mathcal {M}_\infty $ . This is a contradiction: $\mathcal {M}_\infty $ has no successor cardinals which are $\geq \Theta $ .

Proof. By Lemma 7.2, $\delta (\mathcal {T} {\restriction } \lambda ) \leq \Theta $ , so in fact, $\delta (\mathcal {T} {\restriction } \lambda ) = \Theta $ . Any extender after stage $\lambda $ would have the index $\geq \Theta $ , which is excluded as a possibility by Lemma 7.2. Therefore, $\mathsf {lh}(\mathcal {T}) = \lambda + 1$ .

Proof.

1. 1^° We only need to show that $\mathcal {Q}(\mathcal {T} {\restriction } \lambda , b)$ exists and is an initial segment of $\mathsf {HOD} {\parallel } \Theta $ . The uniqueness follows from Proposition 2.5.

2. 2^° Let $\delta :=\delta (\mathcal {T} {\restriction } \lambda )=\bigcup _{i<\lambda }\mathsf {lh}(E^{\mathcal {T}}_i)$ . So, we are assuming that $\delta <\Theta $ .

3. 3^° If $[0, \lambda )_{\mathcal {T}}$ drops, then $\mathcal {Q}(\mathcal {T} {\restriction } \lambda , [0, \lambda )_{\mathcal {T}})$ exists.

4. 4^° Claim. Even if $[0, \lambda )_{\mathcal {T}}$ does not drop, $\mathcal {Q}(\mathcal {T} {\restriction } \lambda , [0, \lambda )_{\mathcal {T}})$ exists.

   Proof. It suffices to prove that $\delta $ is strictly less than the least Woodin cardinal of $\mathcal {M}^{\mathcal {T}}_\lambda $ , viz. $\delta (\mathcal {M}^{\mathcal {T}}_\lambda )$ . Indeed, then $\delta $ is not a Woodin cardinal in $\mathcal {M}^{\mathcal {T}}_\lambda $ , so $\mathcal {Q}(\mathcal {T} {\restriction } \lambda , [0, \lambda )_{\mathcal {T}})$ exists.

   Notice that

   $$ \begin{align*}\delta = \bigcup_{\alpha \in [0, \lambda)_{\mathcal{T}}} \mathsf{lh}(E^{\mathcal{T}}_\alpha).\end{align*} $$

   Now, by Lemma 7.1, we get that for all $\alpha \in [0, \lambda )_{\mathcal {T}}$ ,

   $$ \begin{align*}\mathsf{lh}(E^{\mathcal{T}}_\alpha) < \delta(\mathcal{M}^{\mathcal{T}}_\alpha) \leq i^{\mathcal{T}}_{\alpha, \lambda}(\delta(\mathcal{M}^{\mathcal{T}}_\alpha)) = \delta(\mathcal{M}^{\mathcal{T}}_\lambda),\end{align*} $$

   so we conclude that $\delta \leq \delta (\mathcal {M}^{\mathcal {T}}_\lambda )$ .

   Assume towards contradiction that $\delta = \delta (\mathcal {M}^{\mathcal {T}}_\lambda )$ . Then

   $$ \begin{align*}\delta(\mathcal{M}^{\mathcal{T}}_\lambda) < \Theta = \delta(\mathcal{M}_\infty),\end{align*} $$

   which means that $\mathcal {M}^{\mathcal {T}}_\lambda $ is not the last model of $\mathcal {T}$ . Hence, there is $E^{\mathcal {T}}_\lambda $ and by Lemma 7.1,

   $$ \begin{align*}\mathsf{lh}(E^{\mathcal{T}}_\lambda) < \delta (\mathcal{M}^{\mathcal{T}}_\lambda) = \delta = \delta(\mathcal{T} {\restriction} \lambda).\end{align*} $$

   This is, of course, a contradiction.

   Therefore, $\delta < \delta (\mathcal {M}^{\mathcal {T}}_\lambda )$ , as required.

5. 5^° Thus, in any case, $\mathcal {Q}(\mathcal {T} {\restriction } \lambda ,[0, \lambda )_{\mathcal {T}})$ exists. Let us denote it by Q. Our goal is to show that $Q\unlhd \mathsf {HOD}{\parallel }\Theta $ .

6. 6^° Since $\mathcal {T}$ is normal, $\mathsf {lh}(E^{\mathcal {T}}_\lambda ) \geq \delta $ .

7. 7^° Since for all $i<\lambda $ , $\mathsf {lh}(E^{\mathcal {T}}_i)$ is a cardinal of $\mathcal {M}^{\mathcal {T}}_\lambda $ , we know that $\delta $ is a limit cardinal of $\mathcal {M}^{\mathcal {T}}_\lambda $ . This means that there are no extenders in $\mathcal {M}^{\mathcal {T}}_\lambda $ indexed at $\delta $ and consequently, it must be the case that $\mathsf {lh}(E^{\mathcal {T}}_\lambda )$ is strictly greater than $\delta $ .

8. 8^° Since $\mathsf {lh}(E^{\mathcal {T}}_\lambda )$ is a cardinal in $\mathsf {HOD}$ ,Footnote ²¹ we conclude that

   $$ \begin{align*}\delta^{+\mathsf{HOD}} \leq \mathsf{lh}(E^{\mathcal{T}}_\lambda).\end{align*} $$

9. 9^° Since $\mathcal {M}^{\mathcal {T}}_\lambda {\parallel } \mathsf {lh}(E^{\mathcal {T}}_\lambda ) = \mathsf {HOD} {\parallel } \mathsf {lh}(E^{\mathcal {T}}_\lambda )$ ,Footnote ²² it follows that

   $$ \begin{align*}\mathsf{HOD} {\parallel} \delta^{+\mathsf{HOD}} \unlhd \mathcal{M}^{\mathcal{T}}_\lambda.\end{align*} $$

10. 10^° Since $\delta $ is not WoodinFootnote ²³ in $\mathsf {HOD}$ , there exists the smallest $\zeta $ such that

    $$ \begin{align*}\mathsf{HOD}{\parallel}(\zeta+1)\models"\delta\text{ is not Woodin}\text{."}\end{align*} $$

11. 11^° Observe that $\zeta <\delta ^{+\mathsf {HOD}}$ , so $9^\circ $ implies that

    $$ \begin{align*}\mathsf{HOD}{\parallel}(\zeta+1) = \mathcal{M}^{\mathcal{T}}_\lambda {\parallel} (\zeta + 1).\end{align*} $$

    Therefore, $Q=\mathsf {HOD}{\parallel }\zeta $ , as required.

Proof. Part a is immediate once we have Lemmas 7.1 and 7.2. On the other hand, part b is verified as follows: Corollary 7.3 yields that $\delta (\mathcal {T} {\restriction } \lambda ) < \Theta $ ; the rest follows from Lemma 7.4.

Proof.

1. 1^° Let us work in a fixed generic extension of V and let $\mathcal {T}$ and $\mathcal {U}$ be trees on M whose last models are both $\mathcal {M}_\infty $ , whose main branches do not drop and whose main-branch embeddings are equal to $\pi _{M, \infty }$ . We want to show that $\mathcal {T} = \mathcal {U}$ .

2. 2^° We will prove by induction on $\alpha \leq \min \{\mathsf {lh}(\mathcal {T}), \mathsf {lh}(\mathcal {U})\}$ that $\mathcal {T} {\restriction } \alpha = \mathcal {U} {\restriction } \alpha $ . Once we do this, we will know that either $\mathcal {T} \unlhd \mathcal {U}$ or $\mathcal {U} \unlhd \mathcal {T}$ . Since $\mathcal {T}$ and $\mathcal {U}$ end with the same model, it follows that $\mathcal {T} = \mathcal {U}$ .

3. 3^° Let us proceed with the induction. We are assuming that $\alpha \leq \min \{\mathsf {lh}(\mathcal {T}), \mathsf {lh}(\mathcal {U})\}$ and that for all $\beta < \alpha $ , $\mathcal {T} {\restriction } \beta = \mathcal {U} {\restriction } \beta $ , and we need to show that $\mathcal {T} {\restriction } \alpha = \mathcal {U} {\restriction } \alpha $ .

4. 4^° If $\alpha $ is $0$ or $1$ or a limit ordinal, there is nothing to be proved. Hence, the remaining cases are “ $\alpha $ is the successor of a limit ordinal” and “ $\alpha $ is the successor of a successor ordinal.” In either case, let us denote $\beta := \alpha - 1$ .

5. 5^° Case 1. $\alpha $ is the successor of a limit ordinal.

   Proof. In this case, we have to show that $[0, \beta )_{\mathcal {T}} = [0, \beta )_{\mathcal {U}}$ . If $\delta (\mathcal {T} {\restriction } \beta ) = \delta (\mathcal {U} {\restriction } \beta ) < \Theta $ , then by Lemma 7.4, both $[0, \beta )_{\mathcal {T}}$ and $[0, \beta )_{\mathcal {U}}$ are the unique wellfounded cofinal branch b through the tree $\mathcal {T} {\restriction } \beta = \mathcal {U} {\restriction } \beta $ which has a Q-structure which is an initial segment of $\mathsf {HOD} {\parallel } \Theta $ ; in particular, $[0, \beta )_{\mathcal {T}} = [0, \beta )_{\mathcal {U}}$ .

   It remains to verify the conclusion in the case that the ordinal $\delta (\mathcal {T} {\restriction } \beta ) = \delta (\mathcal {U} {\restriction } \beta ) = \Theta $ . By Corollary 7.3, $\beta $ is the last index in $\mathcal {T}$ and $\mathcal {U}$ , that is,

   $$ \begin{align*}\mathsf{lh}(\mathcal{T}) = \mathsf{lh}(\mathcal{U}) = \beta + 1 = \alpha.\end{align*} $$

   This means that the branches $[0, \beta )_{\mathcal {T}}$ and $[0, \beta )_{\mathcal {U}}$ do not drop and $i^{\mathcal {T}}_{0, \beta } = \pi _{M, \infty }$ and $i^{\mathcal {U}}_{0, \beta } = \pi _{M, \infty }$ ; in other words, branches $[0, \beta )_{\mathcal {T}}$ and $[0, \beta )_{\mathcal {U}}$ through the tree $\mathcal {T} {\restriction } \beta = \mathcal {U} {\restriction } \beta $ give rise to the same embedding. [Reference Steel and Woodin18, Lemma 2.6] now implies that these two branches are equal.

6. 6^° Case 2. $\alpha $ is the successor of a successor ordinal.

   Proof. In this case, both $\mathcal {T} {\restriction } \alpha $ and $\mathcal {U} {\restriction } \alpha $ are obtained from the tree $\mathcal {T} {\restriction } \beta = \mathcal {U} {\restriction } \beta $ by applying the same extender (Proposition 7.5). Therefore, $\mathcal {T} {\restriction } \alpha = \mathcal {U} {\restriction } \alpha $ .

7. 7^° The two cases conclude the induction.

Proof.

1. 1^° Taking into account Lemma 7.6, we only need to verify existence.

2. 2^° Let $\Psi $ be the restriction of $\Sigma _M$ to the stacks of normal trees in $H_{\omega _1}$ which have no unnecessary drops. This $(\omega _1, \omega _1)^*$ -iteration strategy for M is coded by a $<\lambda$ -universally Baire set by Theorem 1.8, where $\lambda$ is the least limit of Woodin cardinals.

3. 3^° Let g be V-generic for $\mathsf {Col}(\omega , \mathbb {R})$ and let us work in $V[g]$ . We have that $\Psi _g \cap H_{\omega _1}^V = \Psi $ .Footnote ²⁴

4. 4^° Claim. There exists a stack $(\mathcal {T}_i)_{i < \omega }$ on $(\mathcal {M}, \Psi _g)$ whose last model is $\mathcal {M}_\infty $ , whose cofinal branch does not drop and gives rise to the embedding $\pi _{M, \infty }$ .

   Proof. In V, $\mathbb {R}$ surjects onto the directed system for $\mathcal {M}_\infty $ . Hence, this directed system is countable in $V[g]$ . It follows that there is a cofinal sequence of indices $(M_i : i < \omega )$ in the directed system, satisfying $M_0 = M$ . In V, each $M_{i + 1}$ is the last model of a normal tree $\mathcal {T}_i$ on $M_i$ according to $\Sigma _{M_i}$ ; now, $\Sigma _{M_i}$ is a tail strategy of $\Sigma _M$ , so $\mathcal {T}_i$ is according to $\Psi $ . We have thus obtained a stack $(\mathcal {T}_i: i < \omega )$ on M according to $\Psi _g$ with all desired properties.

5. 5^° Claim. There exists an $(\omega _1, \omega _1 + 1)^*$ -iteration strategy $\mathrm {X}$ for M such that $\Psi _g \subseteq \mathrm {X}$ .

   Proof. Let s be a stack on $(M, \Psi _g)$ of total length $\omega _1$ . We want to define $\mathrm {X}(s)$ . Let h be $V[g]$ -generic for $\mathsf {Col}(\omega , \omega _1)$ . The stack s is on $(M, \Psi _{g*h})$ , so $b := \Psi _{g*h}(s)$ is a cofinal wellfounded branch through s. Since $\mathsf {Col}(\omega , \omega _1)$ is homogeneous in $V[g]$ , we have that $b \in V[g]$ . We may define $\mathrm {X}(s) := b$ .

6. 6^° By [Reference Schlutzenberg11, Theorem 1.1] applied to $(M, \mathrm {X})$ , there is a normal tree $\mathcal {T}$ on $(M, \Psi _g)$ whose last model is $\mathcal {M}_\infty $ , whose main branch does not drop, and whose main-branch embedding is equal to that of the stack $(\mathcal {T}_i)_{i < \omega }$ , that is, to $\pi _{\mathcal {M}, \infty }$ .

7. 7^° By Lemma 7.6, $\mathcal {T}$ is the unique normal tree on M whose last model is $\mathcal {M}_\infty $ , whose main branch does not drop and whose main-branch embedding is equal to $\pi _{M, \infty }$ .

8. 8^° Homogeneity of $\mathsf {Col}(\omega , \mathbb {R})$ implies that $\mathcal {T} \in V$ , which concludes the proof.