Proof. Part (i): If $Q=M^{\mathcal {T}}_b$ then we can just use standard calculations using core maps (done in the codes given by the theory t, however) to find a tail sequence of extenders used along b, and hence, find b itself, from $(\mathcal {T},Q)$ . So suppose $Q\triangleleft M^{\mathcal {T}}_b$ , so $\rho _\omega ^Q=\delta $ and Q is fully sound.

Suppose first $Q\in\mathfrak{C}_0(M^\mathcal{T}_b)$ . Let $f:\theta\to\delta$ be cofinal in $\delta$ , with , $f$ the least such which is definable over $Q$ (without parameters). Let $\alpha\in b$ be such that $(\alpha, b]^\mathcal{T}$ does not drop, $\delta\in\mathrm{rg}(i^\mathcal{T}_{\alpha b})$ and $\theta<\kappa=\mathrm{cr}(i^\mathcal{T}_{\alpha b})$ , so $Q,f\in\mathrm{rg}(i^\mathcal{T}_{\alpha b})$ and $\mathrm{rg}(f)\subseteq\mathrm{rg}(i^\mathcal{T}_{\alpha b})$ . For $\gamma<\theta$ , let $\beta_\gamma$ be the least $\beta\in[\alpha,\mathrm{lh}(\mathcal{T}))$ such that $\alpha\leq^\mathcal{T}\beta$ and $f(\gamma)<\nu(E^\mathcal{T}_\beta)$ . Then $\beta_\gamma\in b$ . (Suppose not. Let $\xi+1\in b$ be least such that $\xi\geq\beta_\gamma$ . Let $\varepsilon=\mathrm{pred}^\mathcal{T}(\xi+1)$ . So $\alpha\leq^\mathcal{T}\varepsilon<\beta_\gamma\leq\xi$ , so by the minimality of $\beta_\gamma$ ,

$$\begin{align*}\mathrm{cr}(E^{\mathcal{T}}_\xi) =\mathrm{cr}(i^{\mathcal{T}}_{\varepsilon b})<\nu(E^{\mathcal{T}}_\varepsilon)\leq f(\gamma) <\nu(E^{\mathcal{T}}_{\beta_\gamma})\leq\nu(E^{\mathcal{T}}_\xi).\end{align*}$$

But then $f(\gamma )\notin \mathrm {rg}(i^{\mathcal {T}}_{{\varepsilon } b})$ , so $f(\gamma )\notin \mathrm {rg}(i^{\mathcal {T}}_{\alpha b})$ , a contradiction.)

So b is appropriately computable from $(\mathcal {T},t)$ and the parameter $(\alpha, \bar{\delta})$ , where $i^\mathcal{T}_{\alpha b}(\bar{\delta})=\delta$ . But if we define another branch $b'$ from $(\mathcal {T},t)$ , in the same manner, but from some other parameter $(\alpha ',\bar {\delta }')$ , with $b'\neq b$ , then by the Zipper Lemma [Reference Steel16, Theorem 6.10] and variants thereof, $Q(\mathcal {T},b')\neq Q(\mathcal {T},b)$ , and this fact is first-order over $(Q,t)$ , because we can compute the corresponding theory $t'$ of $Q(\mathcal {T},b')$ by consulting the theories of the models along $b'$ . So by demanding that the selected parameter results in a Q-structure whose theory agrees with t, we can actually compute the correct b from $(\mathcal {T},t)$ without the extra parameter.

Suppose now $Q\notin\mathfrak{C}_0(M^\mathcal{T}_b)$ . Then $M^\mathcal{T}_b$ is active type 3 and $\nu(F(M^\mathcal{T}_b))\leq\mathrm{OR}^Q$ ,and since $\rho_\omega^Q=\delta$ is an $M^\mathcal{T}_b$ -cardinal, it follows that $\nu(F(M^\mathcal{T}_b))=\delta$ . The foregoing discussion does not quite work, since there is not literally any $\alpha\in b$ with $\delta\in\mathrm{rg}(i^\mathcal{T}_{\alpha b})$ . However, everything works as above after replacing the use of the maps $i^\mathcal{T}_{\alpha b}$ with the maps

$$\begin{align*}j^\mathcal{T}_{\alpha b}:\mathrm{Ult}_0(M^\mathcal{T}_\alpha,F(M^\mathcal{T}_\alpha))\to\mathrm{Ult}_0(M^\mathcal{T}_b,F(M^\mathcal{T}_b))\end{align*}$$

induced by $i^\mathcal{T}_{\alpha b}$ (where these are defined).

Let us again first suppose $Q\in\mathfrak{C}_0(M^\mathcal{T}_b)$ , and handle the contrary case just as we did above (without further mention). Let $A\subseteq\delta$ be definable over Q without parameters, such that no $\kappa <\delta $ is ${<\delta }$ -A-reflecting. Let C be the set of all limit cardinals $\lambda <\delta $ of Q such that for all $\kappa <\lambda $ , $\kappa $ is not ${<\lambda }$ -A-reflecting. Then C is club in $\delta $ because Q does not singularize $\delta $ . Let $\alpha \in b$ be such that $[\alpha ,b)^{\mathcal {T}}$ is non-dropping and $\delta \in \mathrm {rg}(i^{\mathcal {T}}_{\alpha b})$ . Let $i^{\mathcal {T}}_{\alpha b}(\bar {C})=C$ . For $\gamma \in C$ , let $\beta _\gamma $ be the least $\beta \in [\alpha ,\mathrm {lh}(\mathcal {T}))$ such that $\gamma <\mathrm {lh}(E^{\mathcal {T}}_\beta )$ . Then $\beta _\gamma \in b$ . (Suppose not, and let $\xi \geq \beta _\gamma $ be least with $\xi +1\in b$ . Let $\varepsilon =\mathrm {pred}^{\mathcal {T}}(\xi +1)$ , so $\alpha \leq ^{\mathcal {T}}\varepsilon <\beta _\gamma \leq \xi $ . So

$$\begin{align*}\kappa=\mathrm{cr}(E^{\mathcal{T}}_\xi)<\mathrm{lh}(E^{\mathcal{T}}_{\varepsilon})\leq\gamma<\mathrm{lh}(E^{\mathcal{T}}_{\beta_\gamma} )\leq\mathrm{lh}(E^{\mathcal{T}}_\xi) \end{align*}$$

and $\gamma \leq \nu (E^{\mathcal {T}}_\xi )$ since $\gamma $ is a Q-cardinal. But since $i^{\mathcal {T}}_{\alpha b}(\bar {A})=A$ , we have

$$\begin{align*}i_{E^{\mathcal{T}}_\xi}(A\cap\kappa)\cap\gamma=A\cap\gamma, \end{align*}$$

so by the ISC, restrictions of $E^{\mathcal {T}}_\xi $ witness the fact that $\kappa $ is ${<\gamma }$ -A-strong in Q, so $\gamma \notin C$ , contradiction.) So b is computable from $(\mathcal {T},t)$ and the parameter $(\alpha ,\bar {\delta })$ , and like before, we actually therefore get a computation from $(\mathcal {T},t)$ without the extra parameter.

Part (ii): It seems we can’t quite uniformly tell which of the above three cases holds, nor whether $Q\in\mathfrak{C}_0(M^\mathcal{T}_b)$ or not. But the calculations used in the case that $Q\triangleleft M^{\mathcal {T}}_b$ still work when $Q=M^{\mathcal {T}}_b$ and $\delta $ is not -Woodin, but $\rho _{k+10}^Q=\delta $ . So our $\Sigma _1$ formula seeks either some $k<\omega $ such that Q is not k-sound, and applies the procedure for when $Q=M^{\mathcal {T}}_b$ , or some $k<\omega $ such that Q is $(k+10)$ -sound and $\rho _{k+10}^Q=\delta $ , but $\delta $ is not -Woodin, and then uses the procedure for when $Q\triangleleft M^{\mathcal {T}}_b$ (with complexity say ). We have enough information in some $\mathcal {S}_n(X)$ to verify all the relevant computations, including that Q is the correct direct limit of certain substructures appearing along the branch b. The uniqueness facts established above, along with further slight variants thereof, ensure that this process identifies the correct branch $b$ . (The “slight variants” deal with details ignored above. For example, if $Q\in\mathfrak{C}_0(M^\mathcal{T}_b)$ then again by a Zipper Lemma argument, our $\Sigma_1$ formula will not yield a branch $b'\neq b$ with $Q\triangleleft M^{\mathcal{T}}_{b'}$ but $Q\notin\mathfrak{C}_0(M^\mathcal{T}_{b'})$ , although the discussion above did not literally rule this scenario out.) This yields the desired uniform computation for (ii).

Proof. By [Reference Schlutzenberg10, Theorem 3.11(b)], it suffices to see that is definable over $\left \lfloor M\right \rfloor $ . But because M is proper class, and trees $\mathcal {T}$ on in M are guided by Q-structures of the form $\mathcal {J}_\alpha (M(\mathcal {T}))$ , we get $M\models $ “ is $(\omega ,\omega _1+1)$ -iterable”, so is outright definable over $\left \lfloor M\right \rfloor $ , and hence so is $\mathbb {E}^M$ .

Proof. By [Reference Schindler and Steel6, Theorem 0.2],Footnote ⁷ we may fix $\bar {R}\triangleleft M|\omega _1^M$ such that $\rho _\omega ^{\bar {R}}=\omega $ and $M\models $ “ is above- $\mathrm {OR}^{\bar {R}} (0,\omega _1)$ -iterable”, as witnessed by the restriction of the correct strategy . That is, , where $\zeta =\mathrm {OR}^{\bar {R}}$ . Given R such that , $\Sigma _R^M$ denotes the restriction of this strategy to trees on R.

We say that $(R,S)\in M$ is a conflicting pair iff:

• – R and S are tame $\omega $ -premice,

• – and $\bar {R}\triangleleft S$ and $R|\omega _1^{R}=S|\omega _1^{S}$ but $R\neq S$ , and

• – $M\models$ “ $S$ is above- $\omega_1^S$ $(\omega,\omega_1)$ -iterable”.

If part 2 of the theorem fails for every $\alpha <\omega _1^M$ , then note that for every such $\alpha $ there is a conflicting pair $(R,S)$ with $\alpha <\omega _1^R=\omega _1^S$ . However, for the present we just assume that we have some conflicting pair and work with this, without assuming that part 2 fails for every $\alpha $ .

So fix a conflicting pair $(R_0,S_0)$ . Let $\Gamma _0$ be an above- $\omega_1^{S_0}$ $(\omega,\omega_1^M)$ -strategy for $S_0$ in M. Working in M, we attempt to compare $R_0,S_0$ , via $\Sigma ^M_{R_0},\Gamma _0$ , folding in extra extenders to ensure that for every limit stage $\lambda $ of the comparison, letting

• – $\delta _\lambda =\delta ((\mathcal {T},\mathcal {U})\!\upharpoonright \!\lambda ),$ and

• – $N_\lambda =M((\mathcal {T},\mathcal {U})\!\upharpoonright \!\lambda )$ ,

we have that

1. (*1) $M|\delta _\lambda $ is generic for the meas-lim extender algebra of $N_\lambda $ at $\delta _\lambda $ , and

2. (*2) if $N_\lambda $ is not a Q-structure for $\delta _\lambda $ then $(\mathcal {T},\mathcal {U})\!\upharpoonright \!\lambda \subseteq M|\delta _\lambda $ and $(\mathcal {T},\mathcal {U})\!\upharpoonright \!\lambda $ is definable over $M|\delta _\lambda $ from parameters (and therefore so is $N_\lambda $ ).Footnote ⁸

Note, however, that there need not actually be Woodin cardinals in $R_0,S_0$ , and the trees might drop in model at points. To deal with this correctly, the folding in of extenders for genericity iteration (and other purposes) is done much as in [Reference Schlutzenberg and Steel15], and also in [Reference Schlutzenberg9, Definition 5.4]. We clarify below exactly how this is executed, along with ensuring the definability condition (*2).

We will define the comparison $(\mathcal {T},\mathcal {U})$ in certain blocks, during some of which we fold in short linear iterations. In order to ensure the definability condition (*2) above, initially we must linearly iterate to the point in M which constructs $(R_0,S_0)$ , and following certain limit stages $(\mathcal {T},\mathcal {U})\!\upharpoonright \!(\eta +1)$ ( $\eta $ a limit ordinal) of the comparison, we will fold in a linear iteration out to a segment of M which constructs $(\mathcal {T},\mathcal {U})\!\upharpoonright \!(\eta +1)$ . Overall, we will define a strictly increasing, continuous sequence $\left <\eta _\alpha \right>_{\alpha <\omega _1^M}$ of ordinals $\eta _\alpha $ such that either $\eta _\alpha =0$ or $\eta _\alpha $ is a limit, and simultaneously define $(\mathcal {T},\mathcal {U})\!\upharpoonright \!(\eta _\alpha +1)$ .

Also, we will define $(\mathcal {T},\mathcal {U})\!\upharpoonright \!(\eta +1)$ by induction on $\eta <\omega _1^M$ (refining the recursive construction of blocks just mentioned). Given $(\mathcal {T},\mathcal {U})\!\upharpoonright \!(\eta +1)$ , if this constitutes a successful comparison (i.e., $M^{\mathcal {T}}_\eta \trianglelefteq M^{\mathcal {U}}_\eta $ or vice versa), we stop at stage $\eta $ (and will then derive a contradiction via Claim 2 below). Now suppose otherwise and let $F^{\mathcal {T}}_\eta ,F^{\mathcal {U}}_\eta $ be the extenders witnessing least disagreement between $M^{\mathcal {T}}_\eta ,M^{\mathcal {U}}_\eta $ (as explained below, we might not use these extenders in $\mathcal {T},\mathcal {U}$ , however). We have $F^{\mathcal {T}}_\eta \neq \emptyset $ or $F^{\mathcal {U}}_\eta \neq \emptyset $ . Let $\ell_\eta =\mathrm {lh}(F^{\mathcal {T}}_\eta )$ or $\ell _\eta =\mathrm {lh}(F^{\mathcal {U}}_\eta )$ , whichever is defined, and $K_\eta =M^{\mathcal {T}}_\eta ||\ell _\eta =M^{\mathcal {U}}_\eta ||\ell _\eta $ . If $\eta $ is a limit, let $Q^{\mathcal {T}}_\eta $ be the Q-structure Q for $\delta _\eta $ with $Q\trianglelefteq M^{\mathcal {T}}_\eta $ (so if $\mathcal {T}\!\upharpoonright \!\eta $ is not eventually only padding, then $Q^{\mathcal {T}}_\eta =Q(\mathcal {T}\!\upharpoonright \!\eta ,[0,\eta )_{\mathcal {T}})$ ), and likewise $Q^{\mathcal {U}}_\eta $ the Q-structure Q for $\delta _\eta $ with $Q\trianglelefteq M^{\mathcal {U}}_\eta $ .Footnote ⁹ These exist as $R_0,S_0$ project to $\omega $ and are sound. Also let $Q^{\mathcal {T}}_0=R_0$ and $Q^{\mathcal {U}}_0=S_0$ , and let $N_0=R_0|\omega _1^{R_0}=S_0|\omega _1^{S_0}$ .

We set $\eta _0=0$ . Suppose we have defined $(\mathcal {T},\mathcal {U})\!\upharpoonright \!(\eta _\alpha +1)$ , so $\eta _\alpha <\omega _1^M$ and $\eta _\alpha =0$ or is a limit. We next define $\eta _{\alpha +1}$ and $(\mathcal {T},\mathcal {U})\!\upharpoonright \!\eta _{\alpha +1}$ , and hence $(\mathcal {T},\mathcal {U})\!\upharpoonright \!(\eta _{\alpha +1}+1)$ . In the definition we literally assume that we reach no $\eta <\eta _{\alpha +1}$ such that $(\mathcal {T},\mathcal {U})\!\upharpoonright \!(\eta +1)$ is a successful comparison; if we do reach such an $\eta $ then we stop the construction there. There are three cases to consider.

If $F^{\mathcal {T}}_{\eta _\alpha }$ is defined then $\mathrm {lh}(F^{\mathcal {T}}_{\eta _\alpha })\leq \mathrm {OR}(Q^{\mathcal {T}}_{\eta _\alpha })$ , and likewise for $F^{\mathcal {U}}_{\eta _\alpha }$ . Note that by tameness (or otherwise if $\alpha =0$ ), $\delta _{\eta _\alpha }$ is a strong cutpoint of $Q^{\mathcal {T}}_{\eta _\alpha }$ and of $Q^{\mathcal {U}}_{\eta _\alpha }$ . Now $\mathcal {T}\!\upharpoonright \![\eta _\alpha ,\infty )$ will be based on $Q^{\mathcal {T}}_{\eta _\alpha }$ and above $\delta _{\eta _\alpha }$ , and likewise $\mathcal {U}\!\upharpoonright \![\eta _\alpha ,\infty )$ .

We want to insert a short linear iteration past the point where M constructs $Q^{\mathcal {T}}_{\eta _\alpha },Q^{\mathcal {U}}_{\eta _\alpha }$ , and hence (by (*2) and Lemma 2.1), constructs the branches $[0,\eta _\alpha )_{\mathcal {T}}$ and $[0,\eta _\alpha )_{\mathcal {U}}$ , if $\alpha>0$ . Let $\eta _{\alpha +1}$ be the least limit ordinal $\eta <\omega _1^M$ such that $Q^{\mathcal {T}}_{\eta _\alpha },Q^{\mathcal {U}}_{\eta _\alpha }\in M|(\eta +\omega )$ (clearly if $\alpha>0$ then $\eta _\alpha \leq \delta _{\eta _\alpha }<\eta _{\alpha +1}$ ).

Now $(\mathcal {T},\mathcal {U})\!\upharpoonright \![\eta _\alpha ,\eta _{\alpha +1})$ is given as follows: Let $\eta \in [\eta _\alpha ,\eta _{\alpha +1})$ and suppose we have defined $(\mathcal {T},\mathcal {U})\!\upharpoonright \!(\eta +1)$ . Recall that $K_\eta $ was defined above. If $K_\eta $ has a ( $K_\eta $ -total) measurable $\mu>\delta _{\eta _\alpha }$ then letting $\mu $ be least such, $E^{\mathcal {T}}_\eta =E^{\mathcal {U}}_\eta $ is the unique normal measure on $\mu $ in $\mathbb {E}^{K_\eta }$ . Otherwise, $E^{\mathcal {T}}_\eta =F^{\mathcal {T}}_\eta $ and $E^{\mathcal {U}}_\eta =F^{\mathcal {U}}_\eta $ .

Note that if $\alpha =0$ then $\delta _{\eta _\alpha }=\omega _1^{N_{\eta _{\alpha +1}}}$ , and if $\alpha>0$ then $N_{\eta _{\alpha +1}}\models $ “ $\delta _{\eta _\alpha }$ is Woodin”, and in either case, $N_{\eta _{\alpha +1}}\models $ “there are no measurables or Woodins $>\delta _{\eta _\alpha }$ ”. So $Q^{\mathcal {T}}_{\eta _{\alpha +1}}=N_{\eta _{\alpha +1}}=Q^{\mathcal {U}}_{\eta _{\alpha +1}}$ , and (by tameness) $N_{\eta _{\alpha +1}}$ has no extenders inducing meas-lim extender algebra axioms with index in $[\delta _{\eta _\alpha },\delta _{\eta _{\alpha +1}}]$ .

By tameness, it follows that $\delta _{\eta _\alpha }$ is a cutpoint (maybe not strong cutpoint) of either $M^{\mathcal {T}}_{\eta _\alpha }$ or $M^{\mathcal {U}}_{\eta _\alpha }$ .Footnote ¹⁰ Here $(\mathcal {T},\mathcal {U})\!\upharpoonright \![\eta _\alpha ,\infty )$ will be above $\delta _{\eta _\alpha }$ .

In this case we want to insert a short linear iteration past the point in M which constructs $(\mathcal {T},\mathcal {U})\!\upharpoonright \!\eta _\alpha $ (we will have $\alpha =\eta _\alpha =\delta _{\eta _\alpha }$ and already have

$$\begin{align*}(\mathcal{T},\mathcal{U})\!\upharpoonright\!\eta\in M|\eta_\alpha \end{align*}$$

for every $\eta <\eta _\alpha $ , but it is not clear that $(\mathcal {T},\mathcal {U})\!\upharpoonright \!\eta _\alpha $ is actually definable over $M|\eta _\alpha $ , as it is not clear that the branch choices of $\mathcal {U}$ are appropriately definable).

So let $\eta <\omega _1^M$ be the least limit ordinal such that $(\mathcal {T},\mathcal {U})\!\upharpoonright \!\eta _\alpha \in M|(\eta +\omega )$ (we have $(\mathcal {T},\mathcal {U})\!\upharpoonright \!\eta _\alpha \in \mathrm {HC}^M$ by assumption). Note then that

$$\begin{align*}[0,\eta_\alpha)^{\mathcal{T}},[0,\eta_\alpha)^{\mathcal{U}}\in M|(\eta+\omega) \end{align*}$$

by tameness. We set $\eta _{\alpha +1}=\max (\eta ,\eta _\alpha +\omega )$ .

Now $(\mathcal {T},\mathcal {U})\!\upharpoonright \![\eta _\alpha ,\eta _{\alpha +1})$ is constructed as in the previous case, and note that again, $N_{\eta _{\alpha +1}}$ has no measurables $>\delta _{\eta _\alpha }$ . (Maybe $\delta _{\eta _\alpha }$ itself is measurable. In order to ensure that we get a useful comparison, it is important here that we do not iterate at $\delta _{\eta _\alpha }$ itself during the interval $[\eta _\alpha ,\eta _{\alpha +1})$ .)

We set $\eta _{\alpha +1}=\eta _\alpha +\omega $ . Let $\eta \in [\eta _{\alpha },\eta _{\alpha +1})$ and suppose we have defined $(\mathcal {T},\mathcal {U})\!\upharpoonright \!(\eta +1)$ . If $\eta =\eta _\alpha $ (and in fact in general),

(1) $$ \begin{align} Q^{\mathcal{T}}_{\eta_\alpha}=Q^{\mathcal{U}}_{\eta_\alpha}\triangleleft K_\eta.\end{align} $$

If there is any $E\in \mathbb {E}^{K_\eta }$ such that $\nu _E$ is a $K_\eta $ -inaccessible limit of $K_\eta $ -measurables, and E induces an extender algebra axiom which is false of $\mathbb {E}^M$ , then set $E^{\mathcal {T}}_\eta =E^{\mathcal {U}}_\eta =$ the least such E. Otherwise set $E^{\mathcal {T}}_\eta =F^{\mathcal {T}}_\eta $ and $E^{\mathcal {U}}_\eta =F^{\mathcal {U}}_\eta $ . Set $\ell'_\eta=\mathrm{lh}(E^{\mathcal{T}}_\eta)$ or $\ell'_\eta=\mathrm{lh}(E^{\mathcal{U}}_\eta)$ , whichever is defined. We will have

$$\begin{align*}\mathrm{OR}(Q^{\mathcal{T}}_{\eta_\alpha})=\mathrm{OR}(Q^{\mathcal{U}}_{\eta_\alpha})<\ell'_\eta, \end{align*}$$

by a simple induction, Claim 1 below, line (1), tameness and since $Q^{\mathcal {T}}_{\eta _\alpha }$ projects to $\delta _{\eta _\alpha }$ .Footnote ¹¹

This completes all cases. Of course, limit stages ${<\omega _1^M}$ are taken care of by our strategies. This completes the definition of the comparison.

Proof. Suppose not and let $\alpha $ be least such. So $M^{\mathcal {T}}_\alpha =M^{\mathcal {U}}_\alpha $ , since $R_0,S_0$ are both sound and project to $\omega $ . So letting $C=\mathfrak {C}_\omega (M^{\mathcal {T}}_\alpha )=\mathfrak {C}_\omega (M^{\mathcal {U}}_\alpha )$ , there is $\beta +1<\mathrm {lh}(\mathcal {T},\mathcal {U})$ such that $\beta <^{\mathcal {T}}\alpha $ and letting $\varepsilon =\mathrm {succ}^{\mathcal {T}}(\beta ,\alpha ]$ , $(\varepsilon ,\alpha ]_{\mathcal {T}}\cap \mathscr {D}^{\mathcal {T}}=\emptyset $ and $C=M^{*\mathcal {T}}_\varepsilon \trianglelefteq M^{\mathcal {T}}_\beta $ , and $E^{\mathcal {T}}_\beta \in \mathbb {E}_+^C$ , and likewise there is $\gamma +1<\mathrm {lh}(\mathcal {T},\mathcal {U})$ such that $\gamma <^{\mathcal {U}}\alpha $ and letting $\eta =\mathrm {succ}^{\mathcal {U}}(\gamma ,\alpha ]$ , we have $(\eta ,\alpha ]_{\mathcal {U}}\cap \mathscr {D}^{\mathcal {U}}=\emptyset $ and $C=M^{*\mathcal {U}}_\eta \trianglelefteq M^{\mathcal {U}}_\gamma $ and $E^{\mathcal {U}}_\gamma \in \mathbb {E}_+^C$ .

Since $R_0\neq S_0$ but $R_0|\omega _1^{R_0}=S_0|\omega _1^{S_0}$ , we have $C\neq R_0$ and $C\neq S_0$ , so in fact, $C\triangleleft M^{\mathcal {T}}_\beta $ and $C\triangleleft M^{\mathcal {U}}_\gamma $ .

Now since $E^{\mathcal {T}}_\beta \in \mathbb {E}_+^C$ and $E^{\mathcal {U}}_\gamma \in \mathbb {E}_+^C$ , but $E^{\mathcal {T}}_\beta $ is the least disagreement between C and $M^{\mathcal {T}}_\alpha $ , and $E^{\mathcal {U}}_\gamma $ is the least disagreement between C and $M^{\mathcal {U}}_\alpha =M^{\mathcal {T}}_\alpha $ , we must have $\beta =\gamma $ and $E^{\mathcal {T}}_\beta =E^{\mathcal {U}}_\beta $ . Therefore $E=E^{\mathcal {T}}_\beta =E^{\mathcal {U}}_\beta $ was chosen either for genericity iteration purposes, or for short linear iteration purposes. We have $\beta <\alpha $ , so either $F^{\mathcal {T}}_\beta $ or $F^{\mathcal {U}}_\beta $ is defined; suppose $F=F^{\mathcal {T}}_\beta $ is defined. Since this is least disagreement between $M^{\mathcal {T}}_\beta ,M^{\mathcal {U}}_\beta $ , but $C\triangleleft M^{\mathcal {T}}_\beta $ and $C\triangleleft M^{\mathcal {U}}_\beta $ , we have $\mathrm {OR}^C<\mathrm {lh}(F)$ . We also have $\mathrm {lh}(E)\leq \mathrm {OR}^C$ , and note that by how we chose E, $\nu _E$ is a cardinal of $K_\beta =M^{\mathcal {T}}_\beta ||\mathrm {lh}(F)$ . But $\mathrm {cr}(E^{\mathcal {T}}_{\varepsilon -1})<\nu _E$ , and therefore $E^{\mathcal {T}}_{\varepsilon -1}$ is total over $K_\beta $ , so

$$\begin{align*}M^{*\mathcal{T}}_\varepsilon=C\triangleleft M^{\mathcal{T}}_\beta|\mathrm{lh}(F)\trianglelefteq M^{*\mathcal{T}}_\varepsilon,\end{align*}$$

a contradiction.

Let $N=N_{\omega _1^M}=M(\mathcal {T},\mathcal {U})$ , so $\mathrm {OR}^N=\omega _1^M$ .

Proof. Suppose otherwise. By tameness, we get $\mathcal {T}$ - and $\mathcal {U}$ -cofinal branches $b,c$ . That is, for each $\delta <\omega _1^M$ such that $\delta $ is Woodin in N, $\delta $ is a cutpoint of $(\mathcal {T},\mathcal {U})$ , meaning that there is no extender used in $(\mathcal {T},\mathcal {U})$ which overlaps $\delta $ . But then letting W be the set of all such $\delta $ , $b=\bigcup _{\delta \in W}[0,\delta ]^{\mathcal {T}}$ is a $\mathcal {T}$ -cofinal branch, and likewise for  $\mathcal {U}$ .

Now working in M, we argue much as in the usual proof of termination of comparison/genericity iteration, with one extra observation. For simplicity, let us assume that there is no $\alpha $ such that $\mathrm {OR}^M=\alpha +\omega $ ; in the contrary case, one needs some minor refinements of the discussion to follow. We get some $\lambda <\mathrm {OR}^M$ and some sufficiently elementary $\pi :\bar {M}\to M|\lambda $ with the relevant objects in $\mathrm {rg}(\pi )$ and $\bar {M}$ countable.Footnote ¹² Let $\kappa =\mathrm {cr}(\pi )$ . Then $\kappa =\eta _\kappa =\delta _{\eta _\kappa }$ . Let

$$\begin{align*}\beta+1=\mathrm{succ}^{\mathcal{T}}(\kappa,\omega_1^M)\text{ and }\gamma+1=\mathrm{succ}^{\mathcal{U}}(\kappa,\omega_1^M).\end{align*}$$

Then $E^{\mathcal {T}}_\beta ,E^{\mathcal {U}}_\gamma $ are compatible through $\min (\nu (E^{\mathcal {T}}_\beta ),\nu (E^{\mathcal {U}}_\gamma ))$ . The usual arguments for termination of comparison/genericity iteration show that $\beta =\gamma =\kappa $ and $E=E^{\mathcal {T}}_\kappa =E^{\mathcal {U}}_\kappa $ was chosen for short linear iteration purposes, and $\mathrm {cr}(E)=\kappa $ . Since $N\models $ “There is a proper class of Woodins”, $N_\kappa =N|\kappa $ satisfies the same. But since $\kappa =\eta _\kappa =\delta _{\eta _\kappa }$ and by the rules of choosing $E^{\mathcal {T}}_\kappa $ , we therefore have $\mathrm {cr}(E)>\kappa $ , contradiction.

Using Claim 3 we may fix $\eta ^*<\omega _1^M$ with $\eta ^*$ above all Woodins of N.

Proof. If both such b and c exist in M then we can reach a contradiction much as in the proof of Claim 3.

Now suppose that we have such a branch $b\in M$ , but not c. Let $Q=Q(\mathcal {T},b)$ . If $Q=M^{\mathcal {T}}_b$ then working in M, we can take a hull, and letting $\kappa $ be the resulting critical point, with $\eta ^*<\kappa $ , note that $Q^{\mathcal {T}}_\kappa =M^{\mathcal {T}}_\kappa $ , contradicting Claim 4. So $Q^{\mathcal {T}}_b\triangleleft M^{\mathcal {T}}_b$ . We claim that

$$\begin{align*}\mathrm{branch}(\mathcal{U},Q^{\mathcal{T}}_b)\text{ yields a }\mathcal{U}\text{-cofinal branch }c\in M,\end{align*}$$

a contradiction. For assuming not, again working in M we can take a hull, and letting $\kappa $ be the resulting critical point, note that

$$\begin{align*}\mathrm{branch}(\mathcal{U}\!\upharpoonright\!\kappa,Q^{\mathcal{T}}_\kappa)\text{ does not yield a }\mathcal{U}\!\upharpoonright\!\kappa\text{-cofinal branch,}\end{align*}$$

contradicting Claim 4 and Lemma 2.1.

If instead we have $c\in M$ but no such $b\in M$ , it is symmetric.

We will now give a more thorough analysis of stages of the comparison, and how they relate to the Woodins of the final common model N and segments of M which project to $\omega $ . Let $\left <\beta _\gamma \right>_{\gamma <\Omega }$ enumerate the Woodin cardinals of N in increasing order, and let $\beta _\Omega =\omega _1^M$ . Let

$$\begin{align*}\alpha_\gamma=\sup_{\gamma'<\gamma}\beta_{\gamma'},\end{align*}$$

so $\alpha _\gamma <\beta _\gamma $ and either $\alpha _\gamma =0$ or $\alpha _\gamma $ is Woodin or a limit of Woodins in N, and $\alpha _\Omega $ is the supremum of all Woodins of N. We will show below that for each $\gamma $ , we have [if $\gamma>0$ then $\alpha _\gamma =\eta _{\alpha _\gamma }=\delta _{\eta _{\alpha _\gamma }}$ ], and either:

• – $\gamma =\alpha _\gamma =0$ (and recall that Case 1 attains at stage $0$ of the comparison) and let $\chi _0=\eta _1$ , that is, $\chi _0$ is the least $\chi $ such that $(R_0,S_0)\in M|(\chi +\omega )$ , or

• – $\gamma $ is a successor (so $\alpha _\gamma =\beta _{\gamma -1}$ is Woodin in N), and Case 1 attains at stage $\alpha =\alpha _\gamma $ of the comparison, and let $\chi _\gamma =\eta _{\alpha _\gamma +1}$ , that is, $\chi _\gamma $ is the least $\chi $ such that $Q^{\mathcal {T}}_{\eta _{\alpha _\gamma }},Q^{\mathcal {U}}_{\eta _{\alpha _\gamma }}\in M|(\chi +\omega )$ , or

• – $\gamma $ is a limit (so $\alpha _\gamma $ is a limit of Woodins of N), and Case 2 attains at stage $\alpha =\alpha _\gamma $ of the comparison, and let $\chi _\gamma =\eta _{\alpha _\gamma +1}$ , that is, $\chi _\gamma =\max (\chi ,\eta _{\alpha _\gamma }+\omega )$ , where $\chi $ is least such that $(\mathcal {T},\mathcal {U})\!\upharpoonright \!\eta _{\alpha _\gamma }\in M|(\chi +\omega )$ .

Proof. By induction on $\gamma $ , with a sub-induction on $\zeta $ . Also note that if $N_\zeta $ is a Q-structure for $\delta _\zeta $ then $[0,\zeta )_{\mathcal {T}}$ and $[0,\zeta )_{\mathcal {U}}$ are easily definable from $(\mathcal {T},\mathcal {U})\!\upharpoonright \!\zeta $ .

Note then that parts 1 and 2 follow easily by induction from parts 4 and 5 (we have $0=\alpha _0=\eta _{\alpha _0}$ by definition, and $\delta _0=\omega _1^{R_0}$ ). Consider part 3. If $\gamma =0$ this is just because $R_0,S_0$ are sound and project to $\omega $ . Suppose $\gamma>0$ . Then $M|\alpha _\gamma $ has largest cardinal $\omega $ by induction. Clearly $\rho _\omega (M|\chi _\gamma )\leq \alpha _\gamma $ , so suppose that $\rho _\omega (M|\chi _\gamma )=\alpha _\gamma $ . Then $\alpha _\gamma =\omega _1^{\mathcal {J}(M|\chi _\gamma )}$ and by Lemma 2.1 we have

$$\begin{align*}(\mathcal{T}\!\upharpoonright\!\alpha_\gamma,b),(\mathcal{U}\!\upharpoonright\!\alpha_\gamma,c)\in\mathcal{J}(M|\chi_\gamma), \end{align*}$$

where $b=[0,\alpha _\gamma )^{\mathcal {T}}$ and $c=[0,\alpha _\gamma )^{\mathcal {U}}$ . But then working inside $\mathcal {J}(M|\chi _\gamma )$ , we can use parts of the proofs of Claims 3 and 5 to reach a contradiction.

Now it suffices to verify parts 6–13 for each limit $\zeta \in [\alpha _\gamma ,\beta _\gamma ]$ , since then parts 4 and 5 follow from parts 6 and 13.

If $\zeta =\alpha _\gamma $ then the required facts already hold by induction if $\gamma $ is a successor (as then $\beta _{\gamma -1}=\alpha _\gamma $ ), and trivially if $\gamma $ is a limit. (In the limit case, $N_\zeta \models $ “There is a proper class of Woodins”, so $N_\zeta $ is a Q-structure for itself.)

So suppose $\zeta>\alpha _\gamma $ and that $N_\zeta $ is not a Q-structure for itself; that is, $N_\zeta \models \mathrm {ZFC}$ and $\mathcal {J}(N_\zeta )\models $ “ $\delta _\zeta =\mathrm {OR}(N_\zeta )$ is Woodin”. So $N_\zeta \models $ “There is a proper class of measurables”, so note $\zeta =\eta _\varphi $ for some $\varphi>0$ , and since we integrated genericity iteration into $(\mathcal {T},\mathcal {U})$ , part 6 (genericity of $M|\delta _\zeta $ ) holds, so $\delta _\zeta $ is regular in $\mathcal {J}(N_\zeta )[M|\delta _\zeta ]$ , hence regular in $\mathcal {J}(M|\delta _\zeta )$ , so $M|\delta _\zeta \models \mathrm {ZFC}^-$ .

Let us verify that $(\mathcal {T},\mathcal {U})\!\upharpoonright \!\zeta \subseteq M|\delta _\zeta $ and $(\mathcal {T},\mathcal {U})\!\upharpoonright \!\zeta $ is definable over $M|\delta _\zeta $ from the parameter $x=(\mathcal {T},\mathcal {U})\!\upharpoonright \!(\alpha _\gamma +1)$ . We have $\chi _\gamma <\delta _\zeta $ because $N_\zeta $ is not a Q-structure for itself. So $x\in M|\delta _\zeta $ . But then working in $M|\delta _\zeta $ , which satisfies $\mathrm {ZFC}^-$ , we can define $(\mathcal {T},\mathcal {U})\!\upharpoonright \!\zeta $ , because the extender selection algorithm can be executed in $M|\delta _\zeta $ , in particular since we only need to make $M|\delta _\zeta $ generic in that interval, and at non-trivial limit stages $\zeta '\in (\alpha _\gamma ,\zeta )$ (when $N_{\zeta '}$ is not a Q-structure for itself) we use the inductively established fact that $Q^{\mathcal {T}}_{\zeta '}=Q^{\mathcal {U}}_{\zeta '}$ is computed by P-construction from some proper segment of M, and in fact some proper segment of $M|\delta _\zeta $ (as $Q^{\mathcal {T}}_{\zeta '}\triangleleft N$ in this case).

Now since $M|\delta _\zeta \models \mathrm {ZFC}^-$ and $\delta _\zeta =\delta _{\eta _\varphi }$ , it follows that $\varphi =\delta _{\eta _\varphi }=\delta _\zeta =\zeta $ . It also follows that $\omega $ is the largest cardinal of $M|\zeta =M|\delta _\zeta $ , as otherwise working in $M|\zeta $ , which then satisfies “ $\omega _1$ exists”, we can establish a contradiction to termination of comparison/genericity iteration much as before.

So $(\mathcal {T},\mathcal {U})\!\upharpoonright \!\zeta \subseteq M|\zeta $ and both $(\mathcal {T},\mathcal {U})\!\upharpoonright \!\zeta $ and $N_\zeta $ are definable from x over $M|\zeta $ , and we have extender algebra genericity as stated earlier. So we have established parts 6–10 for $\zeta $ and are now in a position to form the P-construction of segments of M over $N_\zeta $ .

Let $\xi $ be least such that $\xi \geq \zeta $ and $\rho _\omega ^{M|\xi }=\omega $ . Given $\eta \in [\zeta ,\xi ]$ let $P_\eta $ be the P-construction of $M|\eta $ over $N_\zeta $ , if it exists.

Now if $P_\xi $ exists then it must be a Q-structure for $\zeta $ . For otherwise, we have that $\zeta $ is Woodin in $\mathcal {J}(P_\xi )$ , and $M|\zeta $ is generic for the meas-lim extender algebra of $\mathcal {J}(P_\xi )$ at $\zeta $ , so $\zeta $ is regular in $\mathcal {J}(P_\xi )[M|\zeta ]$ , but then $\zeta $ is regular in $\mathcal {J}(M|\xi )$ , contradicting that $\rho _\omega ^{M|\xi }=\omega $ .

Now suppose there is $\eta <\xi $ such that $P_\eta $ exists and either projects $<\zeta $ or is a Q-structure for $N_\zeta $ . If $P_\eta $ projects $<\zeta $ then note that $P_\eta =M^{\mathcal {T}}_\zeta $ . But then working in $M|\xi $ , noting that $\zeta =\omega _1^{M|\xi }$ , we can reach a contradiction as in the proof of Claim 5. So $\rho _\omega ^{P_\eta }=\zeta $ and $P_\eta $ is a Q-structure for $\zeta $ . But here we also reach a contradiction as in the proof of Claim 5.

It follows that $P_\xi $ exists and $P_\xi =Q^{\mathcal {T}}_\zeta $ , giving part 11 for $\zeta $ .

Finally, if $\zeta <\beta _\gamma $ then $Q^{\mathcal {T}}_\zeta =Q^{\mathcal {U}}_\zeta \triangleleft N$ , since $\zeta $ is not Woodin in N by assumption; and if $\zeta =\beta _\gamma <\omega _1^M$ then $Q^{\mathcal {T}}_\zeta \neq Q^{\mathcal {U}}_\zeta $ and hence, $Q^{\mathcal {T}}_\zeta \ntrianglelefteq N$ , since if $Q^{\mathcal {T}}_\zeta =Q^{\mathcal {U}}_\zeta $ then Case 3 would attain at stage $\zeta $ (recall $\alpha _\gamma <\zeta $ ) and then we would have $Q^{\mathcal {T}}_\zeta \triangleleft N$ , contradicting the fact that $\beta _\gamma $ is Woodin in N. This establishes parts 12 and 13 for $\zeta $ , completing the induction.

Write $\mathcal {T}_0=\mathcal {T}$ and $\mathcal {U}_0=\mathcal {U}$ . We now enter a proof by contradiction, by assuming that part 2 of the theorem fails for every $\alpha <\omega _1^M$ . Then we can fix a conflicting pair $(R_1,S_1)$ with $\chi _\Omega <\zeta =_{\mathrm {def}}\omega _1^{R_1}=\omega _1^{S_1}$ . So $R_1\triangleleft M$ and $R_1|\zeta =M|\zeta =S_1|\zeta $ . We have

$$\begin{align*}M|\zeta\models\mathrm{ZFC}^-\wedge \text{"}V=\mathrm{HC}\text{"},\end{align*}$$

so by Claim 7, $N_\zeta $ is not a Q-structure for itself, so the conclusions of Claim 6 hold for $\zeta $ .

Repeat the foregoing comparison with $(R_1,S_1)$ replacing $(R_0,S_0)$ , producing trees $\mathcal {T}_1$ on $R_1$ and $\mathcal {U}_1$ on $S_1$ . Continue in this manner, producing a sequence

$$\begin{align*}\left<R_n,\mathcal{T}_n,S_n,\mathcal{U}_n\right>_{n<\omega}. \end{align*}$$

(It is not relevant whether the sequence is in M.)

Now $\mathcal {T}_1$ is a tree on $R_1$ , above $\zeta =\omega _1^{R_1}$ . By Claim 6, $Q^{\mathcal {T}_0}_\zeta $ is the output of the P-construction of $R_1$ above $N_\zeta $ . So we can translate $\mathcal {T}_1$ into a tree $\mathcal {T}_1'$ on $Q^{\mathcal {T}_0}_\zeta $ ; note this tree is above $\zeta $ . So

$$\begin{align*}\mathcal{X}_1=\mathcal{T}_0\!\upharpoonright\!(\zeta+1)\ \widehat{\ }\ \mathcal{T}_1' \end{align*}$$

is a correct normal tree on $R_0$ , $\zeta $ is a strong cutpoint of $\mathcal {X}_1$ , and $\mathcal {X}_1$ drops in model at $\zeta +1$ , as $\mathcal {X}_1\!\upharpoonright \![\zeta ,\infty )$ is based on $Q^{\mathcal {T}}_\zeta $ and $Q^{\mathcal {T}}_\zeta \triangleleft M^{\mathcal {T}}_\zeta $ by Claim 6.

Continue recursively, defining $\left <\mathcal {X}_n\right>_{n<\omega }$ , by setting $\zeta _n=\omega _1^{R_{n+1}}$ , and translating $\mathcal {T}_{n+1}$ (on $R_{n+1}$ ) into a tree $\mathcal {T}_{n+1}'$ on $Q(\mathcal {X}_n,[0,\zeta _n)^{\mathcal {X}_n})\triangleleft M^{\mathcal {X}_n}_{\zeta _n}$ (there is a natural finite sequence of intermediate translations between $\mathcal {T}_{n+1}$ and $\mathcal {T}_{n+1}'$ ), and setting

$$\begin{align*}\mathcal{X}_{n+1}=\mathcal{X}_n\!\upharpoonright\!(\zeta_n+1)\ \widehat{\ }\ \mathcal{T}_{n+1}'.\end{align*}$$

Let $\mathcal {X}=\liminf _{n<\omega }\mathcal {X}_n$ . Then $\mathcal {X}$ is a correct normal tree on $R_0$ . But it has a unique cofinal branch, which drops in model infinitely often, a contradiction. This completes the proof of the theorem.

Proof. By [Reference Schlutzenberg11, Theorem 12.1], it suffices to see that , and of course we already have . So suppose . We will reach a contradiction via a slight variant of the construction for Theorem 4.2, so we just give a sketch. Recall that by [Reference Schindler and Steel6], we can fix $\xi <\omega _1^M$ such that (see also Definition 4.1 and Theorem 4.2). Also, M is the inductive condensation stack of M above (see [Reference Schlutzenberg10, Theorem 3.11 and Definition 3.12]). So we can fix a name for and a name $\dot {\Sigma }\in W$ for , and may assume that in W, ${\mathbb {P}}$ forces “the universe is that of a tame premouse N such that is tame-iterability-good, $\dot {\Sigma }$ is an above- $\check {\xi }$ - $(\omega ,\omega _1)$ -strategy for , $\dot {\Sigma }$ is consistent with , N is the inductive condensation stack of N above , N satisfies various first order facts established in this article and elsewhere for tame mice, and ”.

Recall $\mathrm {HC}^W=\mathrm {HC}^M$ . Work in W. Fix a strategy $\Psi $ witnessing that ${\mathbb {P}}$ is strategically $\sigma $ -closed. Pick some $(p_0,q_0)\in {\mathbb {P}}\times {\mathbb {P}}$ and some conflicting pair $(R_0,S_0)$ such that “” and “” (where conflicting pair is defined like before, but with $R_0|\xi =S_0|\xi $ and $\xi <\omega _1^{R_0}=\omega _1^{S_0}$ ). Let $p^{\prime }_0=\Psi (\left <p_0\right>)$ . Let $G\times H$ be $(W,{\mathbb {P}}\times {\mathbb {P}})$ -generic with $(p_0',q_0)\in G\times H$ . Note that “ $\check {{\mathbb {P}}}$ is strategically $\sigma $ -closed, as witnessed by $\check {\Psi }$ ”.

Work in $W[G,H]$ . It follows that $\mathrm {HC}^{W[G,H]}=\mathrm {HC}^{W[G]}=\mathrm {HC}^{W[H]}=\mathrm {HC}^M$ , and therefore $\dot {\Sigma }_G,\dot {\Sigma }_H$ are above- $\xi $ - $(\omega ,\omega _1)$ -strategies in $W[G,H]$ . Compare $R_0,S_0$ in the manner of the previous proof, producing trees $(\mathcal {T},\mathcal {U})$ , via $\dot {\Sigma }_G,\dot {\Sigma }_H$ , except that we fold in -genericity instead of -genericity. As before, the comparison lasts $\omega _1^M$ stages and $M(\mathcal {T},\mathcal {U})$ has boundedly many Woodins. Therefore there is some $\xi _1<\omega _1^M$ after which $\mathcal {T},\mathcal {U}$ agree about all Q-structures, and these Q-structures are given by P-construction using proper segments of . Let $\dot {\mathcal {T}}_0,\dot {\mathcal {U}}_0\in W$ be ${\mathbb {P}}\times {\mathbb {P}}$ -names for $\mathcal {T},\mathcal {U}$ . For ${\mathbb {P}}$ -names $\tau $ , let $\tau _{\mathrm {lt}}$ and $\tau _{\mathrm {rt}}$ be the ${\mathbb {P}}\times {\mathbb {P}}$ -names for the interpretation of $\tau $ with respect to the left and right projections of the generic filter, respectively.

Work in W. Let $(p_0",q_0")\leq (p_0',q_0)$ be such that “ $\dot {\mathcal {T}}_0$ is a tree via $\dot {\Sigma }_{\mathrm {lt}}$ of length $\omega _1$ , and for every $\delta \in (\check {\xi }_1,\omega _1)$ , if ” then $\delta =\delta (\dot {\mathcal {T}}_0\!\upharpoonright \!\delta )$ and $Q=Q(\dot {\mathcal {T}}_0\!\upharpoonright \!\delta ,[0,\delta )^{\dot {\mathcal {T}}_0})$ is given by P-construction of $Q'$ above $M(\dot {\mathcal {T}}_0\!\upharpoonright \!\delta )$ , where is the least $\omega $ -premouse such that $\delta \leq \mathrm {OR}^{Q'}$ , and moreover, $Q\triangleleft M^{\dot {\mathcal {T}}_0}_\delta $ ”. Pick some $(p_1^-,q_1)\in {\mathbb {P}}\times {\mathbb {P}}$ and some $(R_1,S_1)$ such that $p_1^-,q_1\leq p_0"$ , and $(R_1,S_1)$ is a conflicting pair with $R_1|\xi _1=S_1|\xi _1$ and $\xi _1<\omega _1^{R_1}=\omega _1^{S_1}$ and “” and “”. So “With $\delta =\omega _1^{\check {R_1}}$ , and as above, then $Q'=\check {R_1}$ ”, and likewise $(q_1,q_0")$ and $S_1$ . Let us now extend $p_1^-$ , so as to instantiate some of these (countable) objects in W. Let $\delta =\omega _1^{R_1}=\omega _1^{S_1}$ . Let $(p_1,q_0"')\leq (p_1^-, q_0")$ and $\bar {\mathcal {T}}_0\in \mathrm {HC}^W$ be such that $\bar {\mathcal {T}}_0$ is an above- $\omega _1^{R_0}$ tree on $R_0$ of length $\delta +1$ with $\delta =\delta (\bar {\mathcal {T}}_0\!\upharpoonright \!\delta )$ , and “ $\check {\bar {\mathcal {T}}}_0\trianglelefteq \dot {\mathcal {T}}_0$ ”. Note then that $Q=Q(\bar {\mathcal {T}}_0\!\upharpoonright \!\delta ,[0,\delta )^{\bar {\mathcal {T}}_0})$ is given by the P-construction of $R_1$ above $M(\bar {\mathcal {T}}_0\!\upharpoonright \!\delta )$ , and moreover, $Q\triangleleft M^{\bar {\mathcal {T}}_0}_\delta $ , and “ $\bar {\mathcal {T}}_0$ is via $\dot {\Sigma }$ ”. Let $p_1'=\Psi (p_0,p_0',p_1)$ .

Carry on in this way, much as before, but also producing the sequence $\left <p_n,p_n'\right>_{n<\omega }$ via $\Psi $ . We can then find $p_\omega \in {\mathbb {P}}$ with $p_\omega \leq p_n$ for all $n<\omega $ . Much as in the proof of Theorem 4.2, we can extend $\bar {\mathcal {T}}_0$ to a tree $\bar {\mathcal {T}}_\omega \in \mathrm {HC}^W$ such that “ $\check {\bar {\mathcal {T}}}_\omega $ is via $\dot {\Sigma }$ , but the only cofinal branch of $\check {\bar {\mathcal {T}}}_\omega $ has infinitely many drops”, a contradiction.

Proof. Part 1: Let . Work in $\left \lfloor M\right \rfloor $ .

Let $\alpha <\omega _1$ be such that and $\rho _\omega ^{M|\alpha }=\omega $ (such an $\alpha $ exists because if then for each $\alpha \in [\beta ,\omega _1)$ ). So and $M|\alpha $ project to $\omega $ and are inter-definable from parameters. Fix a real x coding the pair , x definable over $M|\alpha $ . Let be the translations of to x-premice. Then . So by the relativization to x of [Reference Schlutzenberg10, Theorem 3.11, Definition 3.12], is defined and has universe V. (If $\left \lfloor M\right \rfloor $ has a largest cardinal then $\left \lfloor M_x\right \rfloor \models \mathrm {ZFC}^-$ by assumption, which implies that $M_x$ is tractable in the sense of [Reference Schlutzenberg10, Definition 3.10] (relativized to x), as it satisfies clause (vi) of that definition; note that property is coarse, just dependent on $\left \lfloor M_x\right \rfloor =\left \lfloor M\right \rfloor $ , not $\mathbb {E}^{M_x}$ .) In fact ,