Fact 3.2. For a condition $p \in PT_f$ the set $\operatorname {\mathrm {split}}_i(p)$ is a front in p, i.e., it is an antichain in $(p, \lhd )$ with

$$\begin{align*}\forall b \in [p]: |b \cap \operatorname{\mathrm{split}}_i(p)| = 1. \end{align*}$$

Call it the ith splitting front of p.

Proof We proceed by induction on i:

• • $i = 0$ : Trivial.

• • $i \to i+1$ : The map $\eta \mapsto \min \{\nu \lhd \eta : {\mathrm {ht^{s}}}_p(\nu ) = i+1\}$ is bijection between $\operatorname {\mathrm {split}}_{i+1}(p)$ and $\bigcup _{\eta \in \operatorname {\mathrm {split}}_i(p)} \operatorname {\mathrm {succ}}_p(\eta )$ . By the inductive hypothesis and the fact that p is ${<}\kappa $ -splitting, the latter set has size $<\kappa $ .

• • $\lambda $ is a limit: Since every $\eta \in \operatorname {\mathrm {split}}_\lambda (p)$ is the limit of a sequence $(\eta _j)_{j < \lambda }$ with $\eta _j \in \operatorname {\mathrm {split}}_j(p)$ , we have $|\operatorname {\mathrm {split}}_\lambda (p)| \leq |\prod _{j < \lambda } \operatorname {\mathrm {split}}_j(p)| < \kappa $ by the inaccessibility of $\kappa $ .

Proof Assume $p \Vdash \dot {g} \in \kappa ^\kappa $ and $\dot {g}(i)$ is decided by an antichain $A_{i+1}$ for every $i < \kappa $ . Construct a fusion sequence $(q_{i})_{i < \kappa }$ below p by setting $q_0 := p$ and finding a $q_{i+1} \leq _i q_i$ with $|A_{i+1} {\restriction } q_{i+1}| < \kappa $ in successor steps. In limit steps $\lambda $ , set $q_\lambda $ to be a fusion limit of $(q_i)_{i < \lambda }$ . The fusion limit $q_\kappa $ of the whole sequence will force $q_\kappa \Vdash \dot {g} \leq \check {h}$ for some $h \in \kappa ^\kappa $ in the ground model.

Proof If $(p_i)_{i < \delta }$ with $\delta < \kappa $ is a decreasing sequence, set $q := \bigcap _{i < \delta } p_i$ . We check that q is a condition; only (S2) is nontrivial, so we assume that all other axioms hold.

Let thus $\eta \in q$ . For some $b \in [q]$ with $\eta \lhd b$ (recall that by (S4) + (S5) such a b exists) consider the sets

$$\begin{align*}C_i := \{j < \kappa : b {\restriction} j \in \operatorname{\mathrm{split}}(p_i)\}. \end{align*}$$

By (S2) and (S6), $C_i$ is a club subset of $\kappa $ . Thus $\bigcap _{i < \delta } C_i$ is a club and yields a $\nu $ with $\eta \lhd \nu $ and $\nu \in \operatorname {\mathrm {split}}(q)$ .

Fact 3.9. The following are equivalent:

1. 1. $q \leq _i p.$

2. 2. $q \leq _{PT_f} p$ and $\forall j \leq i: \operatorname {\mathrm {split}}_j(p) = \operatorname {\mathrm {split}}_j(q).$

3. 3. $q \leq _{PT_f} p$ and $\forall \eta \in p: {\mathrm {ht^{s}}}_p(\eta ) \leq i \Rightarrow \operatorname {\mathrm {succ}}_p(\eta ) \subseteq q.$

4. 4. $q \leq _{PT_f} p $ and $\operatorname {\mathrm {split}}_{i+1}(p) \subseteq q.$

Proof If $\delta < \kappa $ , the intersection q from Lemma 3.6 can be seen to also be a fusion limit.

Otherwise once again set $q = \bigcap _{j < \kappa } p_j$ and follow the proof of Lemma 3.6; along a branch $b \in [q]$ again define the sets

$$\begin{align*}C_j := \{\ell < \kappa : b {\restriction} \ell \in \operatorname{\mathrm{split}}(p_j)\}. \end{align*}$$

By using the fact that $(p_j)_{j < \kappa }$ is a fusion sequence, one can arrive at

$$\begin{align*}\left(\Delta_{j < \kappa} C_j\right) \cap \{\lambda < \kappa: \lambda \textrm{ limit}\} \subseteq \bigcap_{j < \kappa} C_j, \end{align*}$$

which is enough to conclude that $\bigcap _{j < \kappa } C_j$ is also a club by the closure of the club filter under diagonal intersections. It can easily be seen that q is a fusion limit.

Proof For any $r \leq p$ we may extend the stem of r in the following way: take any branch $b \in [r] \subseteq [p]$ ; since we then know $|b \cap \operatorname {\mathrm {split}}_{i}(p)| = 1$ , we see that there is a unique $\nu $ with $\nu \in b \cap r \cap \operatorname {\mathrm {split}}_i(p)$ . Then $r^{[\nu ]}$ is $(p,i)$ -determined.

Proof It remains to show the antichain condition. To this end, let A be a maximal antichain, $p \in PT_f$ and $i < \kappa $ . Enumerate $\operatorname {\mathrm {split}}_{i+1}(p)$ as $(\eta _j)_{j < \delta }$ with $\delta < \kappa $ . We will decompose p into $|\delta |$ many parts, each of which will be thinned out above the $(i+1)$ th splitting front.

Proceed by finding for each $j < \delta $ a condition $s_j \leq p^{[\eta _j]}$ such that $|A {\restriction } s_j| = 1$ . Set

$$\begin{align*}q := \bigcup_{j < \delta} s_j. \end{align*}$$

Then $q \in PT_f$ is a condition with $\operatorname {\mathrm {split}}_{i+1}(p) \subseteq q$ and thus $q \leq _i p$ . To prove $|A {\restriction } q| < \kappa $ , let $r \in A$ be compatible with q. By the previous lemma we may pick a $t_r$ that is $(p,i+1)$ -determined with $t_r \leq r,q$ and hence $t_r \cap \operatorname {\mathrm {split}}_{i+1}(p) = \{\eta _{j_r}\}$ for some $j_r < \delta $ . But since $t_r \leq q$ , we can conclude $t_r \leq s_{j_r}$ and thus $r \parallel s_{j_r}$ . We have thus found a function from $A {\restriction } q$ to $\delta $ , mapping $r \mapsto j_r$ , which is injective (since $|A {\restriction } s_j| = 1$ for all $j < \delta $ ). The desired conclusion $|A {\restriction } q| < \kappa $ follows.

Fact 4.1. $\mathbb {P}$ is ${<}\kappa $ -closed.

Fact 4.3. Forcing notions that are ${<}\kappa ^+$ -closed or have the $\kappa ^+$ -c.c. are $\kappa $ -proper. Furthermore, $\kappa $ -proper forcing notions do not collapse $\kappa ^+$ .

Fact 4.5. Property B* implies $\kappa $ -properness and $\kappa ^\kappa $ -bounding.

Proof We construct q inductively such that $\mathcal {P}_\alpha \ni q {\restriction } \alpha $ is a fusion limit of $\langle p_i {\restriction } \alpha , F_i \cap \alpha : i < \delta \rangle $ for each $\alpha \leq \zeta $ .

Assume $q {\restriction } \alpha $ has been defined for $\alpha < \zeta $ . To define $q(\alpha )$ , distinguish three cases:

• • $\alpha \in \bigcup _{j < \delta } \operatorname {\mathrm {supp}}(p_j) \wedge \alpha \in \bigcup _{j < \delta } F_j$ : Find $j^*(\alpha )$ minimal such that $\alpha \in F_{j^*(\alpha )}$ . Now $q {\restriction } \alpha \Vdash " (p_j(\alpha ))_{j \geq j^*(\alpha )} \textrm { is a fusion sequence"}$ , so let $q(\alpha )$ be a fusion limit of that sequence.

• • $\alpha \in \bigcup _{j < \delta } \operatorname {\mathrm {supp}}(p_j) \wedge \alpha \notin \bigcup _{j < \delta } F_j$ : Note that this case may only occur for $\delta < \kappa $ , thus we may use ${<}\kappa $ -closure of $\dot {\mathcal {Q}}_\alpha $ to construct $q(\alpha )$ from $(p_j(\alpha ))_{j < \delta }$ .

• • $\alpha \notin \bigcup _{j < \delta } \operatorname {\mathrm {supp}}(p_j)$ : Set .

To see that $q {\restriction } \gamma \in \mathcal {P}_\gamma $ for limit $\gamma $ , merely note $\operatorname {\mathrm {supp}}(q {\restriction } \gamma ) \subseteq \bigcup _{i < \delta } \operatorname {\mathrm {supp}}(p_i {\restriction } \gamma )$ .

Proof Enumerate F as an increasing sequence $(\beta _j)_{j < \delta }$ with $\delta < \kappa $ and set $\beta _\delta := \zeta $ . For an $r \leq p$ we will inductively construct a decreasing sequence $(s_j)_{j \leq \delta }$ below r and a $\subseteq $ -increasing sequence $(g_j)_{j \leq \delta }$ with $g_j \in \prod _{\beta \in F \cap \beta _j} \kappa ^{<\kappa }$ such that $s_j \textrm { is } (p, F \cap \beta _j, i) \textrm {-determined}$ following $g_j$ .

• • $j=0$ : Set $s_{0} := r$ .

• • $j \to j+1$ : Since $s_j {\restriction } \beta _j \Vdash s_j(\beta _j) \leq _{\dot {\mathbb {Q}}_{\beta _j}} p(\beta _j)$ , we may use Lemma 3.13 to find $\mathbb {P}_{\beta _j}$ -names $\dot {t}, \dot {\eta }_{\beta _j}, \dot {\nu }_{\beta _j} $ with

  $$\begin{align*}s_j {\restriction} \beta_j \Vdash \dot{t} \in \mathbb{Q}_{\beta_j} \wedge \dot{t} \leq_{\dot{\mathbb{Q}}_{\beta_j}} s_j(\beta_j) \end{align*}$$
  and
  $$ \begin{gather*} s_j {\restriction} \beta_j \Vdash \dot{t} \cap \operatorname{\mathrm{split}}_i(p) = \{\dot{\eta}_{\beta_j}\} \wedge \operatorname{\mathrm{succ}}_{\dot{t}}(\dot{\eta}_{\beta_j}) = \{\dot{\nu}_{\beta_j}\}. \end{gather*} $$
  Find a stronger condition $\tilde {s}_j \leq s_j {\restriction } \beta _j$ that decides the names $\dot {\eta }_{\beta _j}, \dot {\nu }_{\beta _j}$ as $\eta _{\beta _j}, \nu _{\beta _j}$ . Define $s_{j+1} := \tilde {s}_j ^\frown \dot {t} ^\frown (s_j {\restriction } [\beta _j + 1, \zeta ))$ and $g_{j+1} := g_j \cup \{(\beta _j, \nu _{\beta _j})\}$ .

• • $\lambda \leq \delta $ is a limit: By ${<}\kappa $ -closure we can find a lower bound $s_\lambda $ of the sequence $(s_\ell )_{\ell < \lambda }$ . Define $g_\lambda := \bigcup _{\ell < \lambda } g_\ell $ . Clearly, $s_\lambda $ is $(p, F \cap \beta _\lambda , i)$ -determined following $g_\lambda $ .

Now $s_\delta \leq r$ is $(p,F,i)$ -determined following $g_\delta $ . Lastly, if s is $(p,F,i)$ -determined following g, then clearly any $s' \leq s$ is as well.

Fact 4.10. If $p' \leq _{F,i} p$ and $s \leq p'$ , then s is $(p,F,i)$ -determined iff it is $(p', F,i)$ -determined.

Proof Construct $p^{(s)}$ by induction such that for each $\alpha \leq \zeta $ we have $p^{(s)} {\restriction } \alpha \in \mathbb {P}_\alpha $ and $p^{(s)} {\restriction } \alpha \leq _{F \cap \alpha , i} p {\restriction } \alpha $ .

Assume $p^{(s)} {\restriction } \alpha $ has been defined; to define $p^{(s)}(\alpha )$ , there are two cases to distinguish:

• • If $\alpha \notin F$ , set $p^{(s)}(\alpha ) := \begin {cases} s(\alpha ), & \textrm { if } s {\restriction } \alpha \in \dot {G}_\alpha , \\ p(\alpha ), & \textrm { otherwise.} \end {cases}$

• • If $\alpha \in F$ , set $p^{(s)}(\alpha ) := \begin {cases} s(\alpha ) \cup (p(\alpha ) \backslash p(\alpha )^{[g(\alpha )]}), & \textrm { if } s {\restriction } \alpha \in \dot {G}_\alpha , \\ p(\alpha ), & \textrm { otherwise.} \end {cases}$

  Note that we have $s {\restriction } \alpha \Vdash g(\alpha ) \in p(\alpha )$ and

  $$\begin{align*}p^{(s)} {\restriction} \alpha \Vdash p^{(s)}(\alpha) \leq_{i} p(\alpha). \end{align*}$$

To see that $p^{(s)} {\restriction } \gamma \in \mathbb {P}_\gamma $ for $\gamma $ limit, we note that

$$\begin{align*}\operatorname{\mathrm{supp}}(p^{(s)} {\restriction} \gamma) \subseteq \operatorname{\mathrm{supp}}(s {\restriction} \gamma). \end{align*}$$

Furthermore, we clearly have $s \leq p^{(s)}$ .

It remains to check the second requirement. Take some $(p,F,i)$ -determined $s'$ following g with $s' \leq p^{(s)}$ . Assume inductively that $s' {\restriction } \alpha \leq s {\restriction } \alpha $ . Since the case $\alpha \notin F$ is trivial, we may restrict our attention to the case $\alpha \in F$ . Then we have $s' {\restriction } \alpha \Vdash s'(\alpha ) \leq _{\mathbb {Q}_\alpha } p^{(s)}(\alpha ) = s(\alpha ) \cup (p(\alpha ) \backslash p(\alpha )^{[g(\alpha )]})$ . But then we already have $s' {\restriction } \alpha \Vdash s'(\alpha ) \leq _{\mathbb {Q}_\alpha } s(\alpha )$ . In conclusion, $s' \leq s$ , which finishes the proof of the lemma.

Proof Construct $\tilde {s}$ by induction such that for all $\alpha \leq \zeta $ we have $\tilde {s} {\restriction } \alpha \in \mathbb {P}_\alpha $ , $\tilde {s} {\restriction } \alpha \leq q {\restriction } \alpha $ and $\tilde {s} {\restriction } \alpha $ is $(q {\restriction } \alpha , F \cap \alpha , i)$ -determined following $g {\restriction } \alpha $ .

Assume $\tilde {s} {\restriction } \alpha $ has been defined; define $\tilde {s}(\alpha )$ as

$$\begin{align*}\tilde{s}(\alpha) := \begin{cases} q(\alpha)^{[g(\alpha)]}, & \textrm{ if } \alpha \in F, \\ q(\alpha),& \textrm{ otherwise.} \end{cases} \end{align*}$$

If $\alpha \notin F$ , there is nothing to prove. For $\alpha \in F$ , observe that since $\tilde {s} {\restriction } \alpha \leq q {\restriction } \alpha $ is $(q {\restriction } \alpha , F \cap \alpha , i)$ -determined following $g {\restriction } \alpha $ and $q \leq _{F,i} p^{(s)}$ , so by the above remark we can conclude $\tilde {s} {\restriction } \alpha \leq s {\restriction } \alpha $ . But

$$\begin{align*}s {\restriction} \alpha \Vdash \exists \nu: g(\alpha) \in \operatorname{\mathrm{succ}}_p(\nu) \wedge \nu \in \operatorname{\mathrm{split}}_i(p(\alpha)) \end{align*}$$

and $q {\restriction } \alpha \Vdash \operatorname {\mathrm {split}}_i(p(\alpha )) = \operatorname {\mathrm {split}}_i(q(\alpha ))$ , hence $\tilde {s}(\alpha )$ is well-defined. The other two properties follow easily.

If $\gamma $ is a limit, then we have $\operatorname {\mathrm {supp}}(\tilde {s} {\restriction } \gamma ) \subseteq \operatorname {\mathrm {supp}}(q) \cup F$ , hence $\tilde {s} {\restriction } \gamma \in \mathbb {P}_\gamma $ is a condition.

Knowing $\tilde {s}$ to be well-defined, one can easily see that for each $s' \leq q$ that is $(q,F,i)$ -determined following g we have $s' \leq \tilde {s}$ .

Fact 4.14. $(\mathbb {P}_\zeta , \leq _{F,i})$ is ${<}\kappa $ -closed for all $\zeta , F,i$ .

Fact 4.16. If $p \in \mathbb {P}_\zeta $ is $(F,i)$ -bounded and $p' \leq _{F,i} p$ , then $p'$ is as well.

Fact 4.18. If $p \in \mathbb {P}_\zeta $ is $(D,F,i)$ -complete as witnessed by $(s_g)_{g \in C}$ , then $(s_g)_{g \in C}$ is a maximal antichain below p.

Proof Assume that $g \in C'$ and find a $t \leq s^{\prime }_{g}$ with $t \in D$ (note that $s^{\prime }_g \leq p$ ). Then $t \leq p$ is $(p,F,i)$ -determined following g by Fact 4.10 and thus $g \in C$ and $t \leq s_g$ by the second requirement in the definition of completeness. If $D' \subseteq D$ , we may take $t = s^{\prime }_{g}$ and get $s^{\prime }_{g} \leq s_{g}$ .

Proof We proceed by induction on $\alpha \leq \zeta $ .

• • $\alpha = 1$ : Trivial by the inaccessibility of $\kappa $ .

• • $\alpha \to \alpha + 1$ : Let $p \in \mathbb {P}_{\alpha +1}, F \in [\alpha + 1]^{<\kappa }$ and $i< \kappa $ be given. Since $\mathbb {P}_\alpha $ is ${<}\kappa $ -closed, $\kappa $ remains inaccessible in $V^{\mathbb {P}_\alpha }$ . Thus

  $$\begin{align*}\Vdash_{\mathbb{P}_\alpha} \forall \beta \in F\, \exists \mu_\beta < \kappa:\ \operatorname{\mathrm{split}}_i(p(\beta)) \subseteq \mu_\beta^{<\mu_\beta} \end{align*}$$
  and considering $\sup _{\beta \in F} \mu _\beta $ we can find a name $\dot {\mu }$ for an ordinal less than $\kappa $ with
  $$\begin{align*}\Vdash_{\mathbb{P}_\alpha} \forall \beta \in F:\ \operatorname{\mathrm{split}}_i(p(\beta)) \subseteq \dot{\mu}^{<\dot{\mu}}. \end{align*}$$
  Let now $A \subseteq \mathbb {P}_\alpha $ be a maximal antichain deciding $\dot {\mu }$ ; we may find a $\mathbb {P}_\alpha \ni \hat {q} \leq _{F \cap \alpha ,i} p {\restriction } \alpha $ with $|A {\restriction } \hat {q}| < \kappa $ . Thus
  $$\begin{align*}\hat{q} \Vdash \dot{\mu} < \mu_q \end{align*}$$
  for some $\mu _q < \kappa $ and therefore
  $$\begin{align*}\forall \beta \in F:\ \hat{q} {\restriction} \beta \Vdash \operatorname{\mathrm{split}}_i(p(\beta)) \subseteq \mu_q^{<\mu_q}. \end{align*}$$
  Setting $q := \hat {q} ^\frown p(\alpha )$ and noting that since $\hat {q} \leq _{F \cap \alpha , i} p {\restriction } \alpha $ we have $q {\restriction } \beta \Vdash \operatorname {\mathrm {split}}(q(\beta )) = \operatorname {\mathrm {split}}(p(\beta ))$ for all $\beta \in F$ , so it follows that q is $(F,i)$ -bounded.

• • $\gamma \leq \zeta $ is a limit: Let $p \in \mathbb {P}_{\gamma }, F \in [\gamma ]^{<\kappa }$ and $i< \kappa $ be given. Using ${<}\kappa $ -closure of $(\mathbb {P}_\gamma , \leq _{F,i})$ (see Fact 4.14) and the inductive assumption, we can construct a $\leq _{F,i}$ -decreasing sequence $(q_\beta )_{\beta \in F}$ in $\mathbb {P}_\gamma $ with the following properties:

  1. – $\forall \beta \in F\, \forall \beta ' \in F \cap \beta :\ q_\beta \leq _{F,i} q_{\beta '} \leq _{F,i} p.$

  2. – $\forall \beta \in F\, \exists \mu _\beta < \kappa \, \forall \beta ' \in F \cap (\beta +1):\ q_\beta {\restriction } \beta ' \Vdash _{\mathbb {P}_{\beta '}} \operatorname {\mathrm {split}}_i(q_\beta (\beta ')) \subseteq \mu _\beta ^{<\mu _\beta }.$

  Again using ${<}\kappa $ -closure of $(\mathbb {P}_\gamma , \leq _{F,i})$ , set q to a $\leq _{F,i}$ -lower bound of $(q_\beta )_{\beta \in F}$ and $\mu := \sup _{\beta \in F} \mu _\beta $ . Now $q \leq _{F,i} p$ and

  $$\begin{align*}\forall \beta \in F:\ q {\restriction} \beta \Vdash \operatorname{\mathrm{split}}_i(q(\beta)) \subseteq \mu^{<\mu}.\\[-34pt] \end{align*}$$

Proof By assumption p is $(F,i)$ -bounded, hence we can find a $\mu $ such that

$$\begin{align*}\forall \beta \in F:\ p {\restriction} \beta \Vdash \operatorname{\mathrm{split}}_i(p(\beta)) \subseteq \mu^{<\mu}. \end{align*}$$

Our strategy is to consider all possible choices a $(p,F,i)$ -determined condition might make at the ith splitting front of p and then interpolate on the witnesses of such choices. Since we have a uniform bound $\mu $ on the respective splitting fronts, this will require us to only iterate through ${<}\kappa $ many possibilities. Set $\tilde {\mu }_\beta := \sup _{j \leq \mu } f_\beta (j)$ and consider the set

$$\begin{align*}\tilde{C} := \prod_{\beta \in F} \tilde{\mu}_\beta^{\leq \mu}. \end{align*}$$

Whenever s is $(p,F,i)$ -determined following some g, then $g \in \tilde {C}$ . Enumerate $\tilde {C}$ as $(g_{j+1})_{j < \delta }$ with $\delta < \kappa $ . We now construct a $\leq _{F,i}$ -decreasing sequence $(t_j)_{j < \delta }$ :

• • $j = 0$ : Set $t_0 := p$ .

• • $j \to j+1$ : If there exists an $s \leq t_j$ that is $(p,F,i)$ -determined following $g_{j+1}$ , pick an arbitrary such condition from D (this is possible, since D is dense below p) and call it $\tilde {s}_{g_{j+1}}$ . Set $t_{j+1} := t_j^{(\tilde {s}_{g_{j+1}})}$ . If there is no such s, simply set $t_{j+1} := t_j$ . In any case we have $t_{j+1} \leq _{F,i} t_j$ .

• • $\lambda $ is a limit: Set $t_\lambda $ to a $\leq _{F,i}$ -lower bound of $(t_j)_{j < \lambda }$ (see Fact 4.14).

Set q to a $\leq _{F,i}$ -lower bound of $(t_j)_{j < \delta }$ . We know $q \leq _{F,i} p$ . Now let

$$\begin{align*}C := \left\{g \in \tilde{C}:\ \tilde{s}_g \textrm{ exists} \right\}, \end{align*}$$

i.e., C is the set of all $g_{j+1}$ for which a witness was found in the inductive step $j \to j+1$ . We have $|C| < \kappa $ . Finally, for each $g = g_{j+1} \in C$ apply Lemma 4.13 to $p = t_j$ , $s = \tilde {s}_{g_{j+1}}$ and $q = q$ to construct the condition $s_g$ . We have $s_g \in D$ since $s_g \leq \tilde {s}_g \in D$ and D is open.

We verify that q is $(D,F,i)$ -complete, witnessed by $(s_g)_{g \in C}$ . The first condition in the definition of completeness follows by construction. The second follows immediately from Lemma 4.13 by noting that if $s \leq q$ is $(q,F,i)$ -determined following g, then $g = g_{j+1}$ for some $j < \delta $ , and thus a witness was found in the inductive step $j \to j+1$ ; hence $g \in C$ .

Proof We have seen in Lemma 4.6 that the fusion condition is always fulfilled. We will now prove that $\mathbb {P}_{\zeta }$ fulfils the antichain condition: let $A \subseteq \mathbb {P}_\zeta $ be a maximal antichain, $p \in \mathbb {P}_\zeta , F \in [\zeta ]^{<\kappa }$ and $i < \kappa $ . Find a $q \leq _{F,i} p$ that is $(D,F,i)$ -complete, where

$$\begin{align*}D = \{r \in \mathbb{P}_\zeta:\ |A {\restriction} r| = 1\} \end{align*}$$

and let $(s_g)_{g \in C}$ witness this. Since $(s_g)_{g \in C}$ is a maximal antichain below q by Fact 4.18, it is easy to see that

$$\begin{align*}A {\restriction} q \subseteq \{r \in A:\ \exists g \in C:\ A {\restriction} s_g = \{r\}\} \end{align*}$$

and thus $|A {\restriction } q| \leq |C| < \kappa $ .

Proof We prove the statements by induction on $\alpha $ .

• • $\alpha =1$ : We have $H_1 = \mathbb {P}_1$ and $|\mathbb {P}_1| = |PT_{f_0}| = \kappa ^{+}$ .

• • $\alpha \to \alpha + 1$ : Let $p \in \mathbb {P}_{\alpha +1}$ , $F \in [\alpha +1]^{<\kappa }$ and $i < \kappa $ . Using the inductive hypothesis, we may assume $p {\restriction } \alpha \Vdash _\alpha \dot {c}_\alpha (p(\alpha )) = \{\{j\} \times A_j: j < \kappa \}$ with $A_j \subseteq H_\alpha $ for all $j < \kappa $ . Additionally using Property B*, construct a fusion sequence $\langle q_j, F_j: j < \kappa \rangle $ with:

  1. – $\forall j < \kappa : q_j \in H_\alpha $ and $|A_j {\restriction } q_j| < \kappa $ ,

  2. – $q_0 \leq _{F \cap \alpha , i} p {\restriction } \alpha $ ,

  3. – $\forall j < \ell < \kappa : q_\ell \leq _{F_j, i + j} q_j$ and $F \cap \alpha \subseteq F_j$ ,

  where the $F_j$ are constructed using a bookkeeping argument. Let $q_\kappa $ be a canonical fusion limit of this sequence that is a member of $H_\alpha $ . By the first property of the fusion sequence,

  $$\begin{align*}\dot{\tau} = \{\{j\} \times (A_j {\restriction} q_\kappa): j < \kappa\}\end{align*}$$
  is an $\alpha $ -good name and $q_\kappa \Vdash _\alpha \dot {c}_\alpha (p(\alpha )) = \dot {\tau }$ . Let thus $\dot {r}$ be the $\preceq $ -least $\mathbb {P}_\alpha $ -name that satisfies $q_\kappa \Vdash _\alpha \dot {c}_\alpha (\dot {r}) = \dot {\tau }$ ; now we can (pedantically, using Remark 4.26) conclude $H_{\alpha + 1} \ni (q_\kappa \, ^\frown \dot {r}) \leq _{F,i} p$ .Since $|H_\alpha | = \kappa ^+$ and there are only $|(\kappa ^+)^\kappa |=\kappa ^+$ many $\alpha $ -good names for reals, we get $|\tilde {H}_{\alpha +1}| = \kappa ^+$ and therefore also $|H_{\alpha +1}| = \kappa ^+$ by standard arguments.

• • $\gamma $ is a limit: If $\operatorname {\mathrm {cf}}(\gamma ) = \kappa ^+$ , density is trivial and $|H_\gamma | \leq |\bigcup _{\beta < \gamma } H_\beta | \leq \kappa ^+$ .

  Assume $\operatorname {\mathrm {cf}}(\gamma ) = \delta \leq \kappa $ and let furthermore $p \in \mathbb {P}_{\gamma }$ , $F \in [\gamma ]^{<\kappa }$ and $i < \kappa $ be given. For a cofinal sequence $(\beta _j)_{j < \delta }$ construct a fusion sequence $\langle q_j, F_j: j < \delta \rangle $ with:

  1. – $\forall j < \delta : F \cap \beta _j \subseteq F_j \subseteq \beta _j$ ,

  2. – $\forall j < \delta : q_j {\restriction } \beta _j \leq _{F_j, i + j} p {\restriction } \beta _j$ ,

  3. – $\forall j < \ell < \delta : q_\ell \leq _{F_j, i+j} q_j$ ,

  4. – $\forall j < \delta : q_j \in H_\gamma $ , which may be achieved by having $q_j {\restriction } \beta _j \in H_{\beta _j}$ and letting $q_j(\beta )$ be the trivial condition for $\beta \geq \beta _j$ .

  The $F_j$ are again constructed using a bookkeeping argument. Set $q_\delta $ to be a fusion limit contained in $H_\gamma $ ; then we have $H_\gamma \ni q_\delta \leq _{F,i} p$ . Lastly, using Remark 4.26, we get $|H_\gamma | \leq \left | \prod _{j < \delta } H_{\beta _j} \right | \leq \kappa ^+$ .

Proof See [Reference Jech12, Theorem 16.30].

Proof of Theorem 4.24

By Lemma 4.28, each $\mathbb {P}_\alpha $ has a dense subset of size ${\leq }\kappa ^{+}$ and therefore satisfies the $\kappa ^{++}$ -c.c.; our desired conclusion thus follows easily from Lemma 4.29 and by noting that the set $\{\gamma < \kappa ^{++}: \operatorname {\mathrm {cf}}(\gamma ) = \kappa ^{+}\}$ is stationary in $\kappa ^{++}$ .

Proof Suppose $\alpha < \kappa ^{++}$ . Let $\dot {\tau }$ be a $\mathbb {P}_\alpha $ -name and $p \in \mathbb {P}_\alpha $ force $\dot {\tau }$ to be a subset of $\kappa $ . Without loss of generality assume $\dot {\tau } = \{\{j\} \times A_j: j < \kappa \}$ is a nice name with $A_j \subseteq H_\alpha $ for all $j < \kappa $ . Just like in Lemma 4.28, construct a fusion sequence $\langle q_j, F_j: j < \kappa \rangle $ below p with $|A_j {\restriction } q_j| < \kappa $ for all $j < \kappa $ . The fusion limit $q_\kappa $ forces $\dot {\tau }$ to be equal to an $\alpha $ -good name, of which there are only $\kappa ^+$ many. If we additionally assume $\operatorname {\mathrm {cf}}(\alpha )> \kappa $ , then $q_\kappa $ forces $\dot {\tau }$ to be equal to a $\mathbb {P}_\gamma $ -name for some $\gamma < \alpha $ . The first two statements thus follow by a density argument.

The last point follows immediately from the previous two.

Proof Working within $V^{\mathbb {P}}$ , take $\alpha < \kappa ^{++}$ and $f \in \kappa ^\kappa $ . Since $\mathbb {P}$ is $\kappa ^\kappa $ -bounding by Fact 4.5, we find an $h \in \kappa ^\kappa \cap V$ with $f \leq h$ and $\beta> \alpha $ with $f_\beta (i) = |2^{h(i)}|$ for all $i < \kappa $ . In V we may construct bijections $c_\ell : |2^\ell | \to 2^\ell $ for $\ell < \kappa $ .

Working now in $V^{\mathbb {P}_\beta }$ , define the function $\dot {\sigma }(i) = c_{h(i)}(\dot {s}_\beta (i))$ . For $x \in 2^\kappa \cap V^{\mathbb {P}_\alpha }$ the set

$$\begin{align*}D_x := \{p \in \mathbb{Q}_\beta:\ \exists i < \kappa: p \Vdash \dot{\sigma}(i) = x {\restriction} h(i)\} \end{align*}$$

is dense; in fact, it is easy to see that for any $p \in \mathbb {Q}_\beta $ and $\eta \in \operatorname {\mathrm {split}}(p), j = \operatorname {\mathrm {dom}}(\eta )$ we have $p^{[\eta ^\frown c^{-1}_{h(j)}(x {\restriction } h(j)) ]} \in D_x$ . Here $c^{-1}_{h(j)}(x {\restriction } h(j))$ is well-defined, since $2^{<\kappa } \cap V = 2^{<\kappa } \cap V^{\mathbb {P}_\beta }$ . Hence $(\sigma (i))_{i < \kappa }$ provides the required covering for the challenge f and $2^\kappa \cap V^{\mathbb {P}_\alpha } \in \mathcal {SN}$ follows.

Proof Suppose not. Then we can find $\alpha , \dot {\tau }, F,i$ and p with

$$\begin{align*}\forall \delta < \kappa\, \exists \eta_\delta \in 2^\delta:\ \neg(\exists q \leq_{F,i} p:\ q \Vdash \eta \ntriangleleft \dot{\tau}). \end{align*}$$

Set $T:= \{\eta _\delta {\restriction } \ell : \ell \leq \delta < \kappa \}$ . By virtue of the preparation of $\kappa $ ,

$$\begin{align*}V^{\mathbb{P}_\alpha} \models \kappa \textrm{ is weakly compact} \end{align*}$$

and therefore, since T is a ${<}\kappa $ -splitting tree of height $\kappa $ , it has a branch $b^*$ in $V^{\mathbb {P}_\alpha }$ . Since p forces $\dot {\tau } \notin V^{\mathbb {P}_\alpha }$ , there is a $\mathbb {P}_{\alpha , \kappa ^{++}}$ -name $\dot {\ell }$ for an ordinal less than $\kappa $ such that $p \Vdash \dot {\tau } {\restriction } \dot {\ell } \neq b^* {\restriction } \dot {\ell }$ . As $\mathbb {P}_{\alpha , \kappa ^{++}}$ satisfies Property B*, there is a $q \leq _{F,i} p$ and $\ell ^* < \kappa $ with $q \Vdash \dot {\ell } < \ell ^*$ .

Since $b^* {\restriction } \ell ^* \in T$ , there is a $\delta \geq \ell ^*$ such that $b^* {\restriction } \ell ^* = \eta _\delta {\restriction } \ell ^*$ . But this means $q \Vdash \dot {\tau } {\restriction } \ell ^* \neq \check {\eta }_\delta {\restriction } \ell ^*$ and therefore $q \Vdash \eta _\delta \ntriangleleft \dot {\tau }$ , a contradiction.

Fact 5.7.

$$ \begin{align*} X \in \mathcal{SN} &\Leftrightarrow \forall D \,\mathrm{dominating} \,\,\, \exists H \, \mathrm{with\ index} \, D \colon X \subseteq \bigcap_{f \in D} \bigcup_{\alpha< \kappa} [h_f(\alpha)]. \end{align*} $$

Proof We prove the claim with a density argument, let therefore $p \in \mathbb {P}_{\alpha , \kappa ^{++}}$ be arbitrary. Working in $V^{\mathbb {P}_\alpha }$ , we will construct an increasing sequence $(\delta _i)_{i < \kappa }$ of ordinals less than $\kappa $ . On the tree

$$\begin{align*}T := \{g \in \prod_{j < i} 2^{\delta_j}: i < \kappa\} \end{align*}$$

we shall construct a mapping $\mathfrak {q}: T \to \mathbb {P}_{\alpha , \kappa ^{++}}$ and a sequence of increasing sets $(F_i)_{i < \kappa }$ such that whenever $b \in \prod _{j < \kappa } 2^{\delta _j}$ is a branch of T in $V^{\mathbb {P}_\alpha }$ , then

$$\begin{align*}\langle \mathfrak{q}(b {\restriction} i), F_i:\ i < \kappa \rangle \end{align*}$$

is a fusion sequence below p. Each condition $\mathfrak {q}(g)$ will carry some information about an increasingly long initial segment of $\dot {\tau }$ . More specifically, we will ensure that for all $i < \kappa $ and $g \in \prod _{j < i} 2^{\delta _j}$ we have

$$\begin{align*}\mathfrak{q}(g) \Vdash \forall j < i:\ g(j) \ntriangleleft \dot{\tau}. \end{align*}$$

We define $\mathfrak {q}(g)$ for $g \in \prod _{j < i} 2^{\delta _j}$ by induction in i.

• • $i=0$ : Set $\mathfrak {q}(\emptyset ) := p$ and $F_0 := \emptyset $ .

• • $i \to i+1$ : Assume $\mathfrak {q}(g)$ is defined for every $g \in \prod _{j < i} 2^{\delta _j}$ . Using Lemma 5.5 we can define $\delta _{i} := \sup \Big (\{\delta _{\mathfrak {q}(g), F_i, i}: g \in \prod _{j < i} 2^{\delta _j}\} \cup \{\delta _j + 1: j < i\} \Big )$ and for every $g \in \prod _{j < i} 2^{\delta _j}, \eta _{i} \in 2^{\delta _{i}}$ find a condition $\mathfrak {q}(g ^\frown \eta _{i}) \leq _{F_i,i} \mathfrak {q}(g)$ with

  $$\begin{align*}\mathfrak{q}(g ^\frown \eta_{i}) \Vdash \eta_{i} \ntriangleleft \dot{\tau}. \end{align*}$$
  Now since $\mathfrak {q}(g ^\frown \eta _{i}) \leq \mathfrak {q}(g)$ , we also have
  $$\begin{align*}\mathfrak{q}(g ^\frown \eta_{i}) \Vdash \forall j < i:\ g(j) \ntriangleleft \dot{\tau}. \end{align*}$$
  Use a bookkeeping argument to define $F_{i+1}$ .

• • $\lambda < \kappa $ is a limit: By construction, for every $g \in \prod _{j < \lambda } 2^{\delta _j}$ the sequence $(\mathfrak {q}(g {\restriction } j))_{j < \lambda }$ is a fusion sequence. Set $\mathfrak {q}(g)$ to be a fusion limit of said sequence and $F_\lambda := \bigcup _{j < \lambda } F_j$ . Note that we have

  $$\begin{align*}\mathfrak{q}(g) \Vdash \forall j < \lambda:\ g(j) \ntriangleleft \dot{\tau}. \end{align*}$$

This concludes the construction of $\mathfrak {q}$ . Let now $f \in D$ dominate the function $i \mapsto \delta _i$ and set $\eta _i := h_f(i) {\restriction } \delta _i$ . Now $(\mathfrak {q}(g_i))_{i < \kappa }$ with $g_i = (\eta _j)_{j < i}$ is a fusion sequence and has a fusion limit $q_\kappa $ . It follows that

$$\begin{align*}q_\kappa \Vdash \eta_i \ntriangleleft \dot{\tau} \end{align*}$$

for each $i < \kappa $ and therefore $q_\kappa \Vdash \dot {\tau } \notin \bigcap _{f\in D} \bigcup _{i <\kappa } [h_f(i)]$ . Thus the set of conditions that force $\dot {\tau } \notin \bigcap _{f\in D} \bigcup _{i <\kappa } [h_f(i)]$ is dense in $\mathbb {P}_{\alpha , \kappa ^{++}}$ .

Proof Since we already saw one inclusion in Theorem 5.2, it suffices to show $V^{\mathbb {P}} \models \mathcal {SN} \subseteq [2^\kappa ]^{\leq \kappa ^+}$ . Let now $X \in V^{\mathbb {P}}$ be of size $\kappa ^{++}$ and D be a dominating family in $V^{\mathbb {P}}$ which lies in V. We will show that there is no $H \in V^{\mathbb {P}}$ with index D such that

$$\begin{align*}X \subseteq \bigcap_{f \in D} \bigcup_{i < \kappa} [h_f(i)], \end{align*}$$

hence X is not strong measure zero by Fact 5.7. Towards a contradiction, assume such an H exists. Note that since D is in V, the set H can have cardinality at most $\kappa ^+$ ; since $\mathbb {P}$ fulfils the $\kappa ^{++}$ -c.c., we know H must already appear in some intermediate model $V^{\mathbb {P}_\alpha }$ . But $|X| = \kappa ^{++}$ , thus there must be an $x \in X$ with $x \notin V^{\mathbb {P}_\alpha }$ .

Working in $V^{\mathbb {P_\alpha }}$ , let $\dot {x}$ and $\dot {X}$ be $\mathbb {P}_{\alpha , \kappa ^{++}}$ -names for x and X, respectively, so that we have

$$\begin{align*}\Vdash_{\mathbb{P}_{\alpha, \kappa^{++}}} \dot{x} \in \dot{X} \wedge \dot{x} \notin V^{\mathbb{P}_\alpha}. \end{align*}$$

Then by Lemma 5.8 we have

$$\begin{align*}\Vdash_{\mathbb{P}_{\alpha, \kappa^{++}}} \dot{x} \notin \bigcap_{f\in D} \bigcup_{i <\kappa} [h_f(i)], \end{align*}$$

and thus $X \nsubseteq \bigcap _{f \in D} \bigcup _{i < \kappa } [h_f(i)]$ in $V^{\mathbb {P}}$ , a contradiction.

Proof See [Reference Friedman, Khomskii and Kulikov7].

Proof If we set $g(y) := \bigcup \{\eta \in 2^{<\kappa }: y \in [T_\eta ]\}$ , then it is easy to see that $g : [T] \to 2^\kappa $ is a well-defined continuous function and $g^{-1}([\eta ]) = [T_\eta ]$ for all $\eta \in 2^{<\kappa }$ . Since $([\eta ])_{\eta \in 2^{<\kappa }}$ forms a clopen basis of $2^\kappa $ , uniqueness is given.

Fact 6.6. The following statements are $\mathbf {\Pi }_1^1(\kappa )$ and therefore absolute for ${<}\kappa $ -closed forcing extensions:

• • c is a code for a continuous function.

• • “ $[T] = [T']$ ” for trees $T,T'$ .

• • “ $c \preccurlyeq c'$ ” for codes $c,c'$ .

• • “ $g_c$ is a total function” for a code c.

• • “ $\operatorname {\mathrm {ran}} (g_c) \subseteq [T]$ ” for a code c and a tree $T.$

• • “ $g_c$ is uniformly continuous with modulus of continuity $i \mapsto j(i)$ ” for a code $c.$

Proof If necessary, we may strengthen p twice in the following manner:

• • Firstly, since $|Y| < \kappa $ and $p \Vdash \dot {\tau } \notin \check {Y}$ , we may find a name $\dot {\ell }$ for an ordinal less than $\kappa $ such that

  $$\begin{align*}p \Vdash \forall y \in \check{Y}:\ \dot{\tau} {\restriction} \dot{\ell} \neq y {\restriction} \dot{\ell}. \end{align*}$$
  Property B* enables us to find a $p' \leq _{F,i} p$ and $\ell ^* < \kappa $ with
  $$\begin{align*}\forall y \in Y: p' \Vdash \ \dot{\tau} {\restriction} \ell^* \neq y {\restriction} \ell^* \end{align*}$$
  by restricting a maximal antichain deciding $\dot {\ell }$ .

• • Secondly, we can find a $p" \leq _{F,i} p'$ that is $(F,i)$ -bounded (see Definition 4.15).

So without loss of generality assume that p already has both these properties. We construct the sequence $(q_j)_{j < \kappa }$ inductively:

• • $j=0$ : Set $q_0 := p$ .

• • $j \to j+1$ : Since

  $$\begin{align*}D_{j+1} := \{r \leq q_j: r \textrm{ decides } \dot{\tau} {\restriction} (j+1)\} \end{align*}$$
  is open dense below $q_j$ , we may apply Lemma 4.22 to $q_j, F,i$ and $D_{j+1}$ Footnote ⁵ to get $q_{j+1}$ and $(s_g^{j+1})_{g \in C_{j+1}}$ such that $q_{j+1}$ is $(D_{j+1},F,i)$ -complete as witnessed by $(s_g^{j+1})_{g \in C_{j+1}}$ . Note that we have $q_{j+1} \leq _{F,i} q_j \leq _{F,i} p$ .

• • $\delta $ is a limit: Find a $\leq _{F,i}$ -lower bound $\tilde {q}_\delta $ of $(q_\ell )_{\ell < \delta }$ . Just as in the successor step, apply Lemma 4.22 to $\tilde {q_\delta }, F,i$ and

  $$\begin{align*}D_\delta := \{r \leq \tilde{q_\delta}: r \textrm{ decides } \dot{\tau} {\restriction} \delta\} \end{align*}$$
  to get $q_\delta $ and $(s_g^\delta )_{g \in C_\delta }$ .

By Lemma 4.19 we know that $(C_j)_{j < \kappa }$ is a decreasing sequence of non-empty sets of size less than $\kappa $ ; as such, the sequence is eventually constant. Let $j^*$ be the index at which this happens.

Now define

$$\begin{align*}X := \{x \in 2^\kappa: \exists g \in C_{j^*}\, \forall j < \kappa:\ s_{g}^j \Vdash \dot{\tau} {\restriction} j = x {\restriction} j\}. \end{align*}$$

For $g \in C_{j^*}$ the sequence $(s_g^j)_{j < \kappa }$ is decreasing by Lemma 4.19. Hence each $g \in C_{j^*}$ successfully interprets $\dot {\tau }$ as some unique $x \in X$ , i.e.,

$$\begin{align*}\forall g \in C_{j^*}\, \exists !\, x \in X\, \forall j < \kappa:\ s_g^j \Vdash \dot{\tau} {\restriction} j = x {\restriction} j. \end{align*}$$

Since $ \forall y \in Y:\ p \Vdash \dot {\tau } {\restriction } \ell ^* \neq y {\restriction } \ell ^*$ , we know that $X \cap Y = \emptyset $ .

Suppose now that $j \geq j^*$ and $r \leq q_{j}$ . Then r is compatible with $s_{g}^j$ for some $g \in C_{j} = C_{j^*}$ and we can find a $t \leq r, s_{g}^j$ . But then $\exists x \in X:\ t \Vdash \dot {\tau } {\restriction } j = x {\restriction } j$ , so we can conclude

$$\begin{align*}q_j \Vdash \exists x \in \check{X}:\ \dot{\tau} {\restriction} j = x {\restriction} j. \end{align*}$$

Since $|Y| < \kappa $ , we may for each $j < j^*$ pick an arbitrary $x_j \in [\sigma _j] \backslash Y$ , where

$$\begin{align*}q_j \Vdash \dot{\tau} {\restriction} j = \check{\sigma}_j \end{align*}$$

and add those $x_j$ to X, thereby ensuring that

$$\begin{align*}q_j \Vdash \exists x \in \check{X}:\ \dot{\tau} {\restriction} j = x {\restriction} j \end{align*}$$

holds for $j < j^*$ as well.

Proof Enumerate $\operatorname {\mathrm {split}}_i(p(0))$ as $(\eta _k)_{k < \delta }$ with $\delta < \kappa $ . We inductively construct sequences $((t_j^k)_{j < \kappa })_{k < \delta }$ and a sequence of sets $(X_k)_{k < \delta }$ : assuming that $X_m$ has been constructed for $m < k$ , apply Lemma 7.1 to $p^{[\eta _k]}$ and $Y := \bigcup _{m < k} X_m$ to get a sequence of conditions $(t_j^k)_{j < \kappa }$ and a set $X_k$ .

Now let $\ell ^*> \ell $ be an ordinal large enough such that whenever $j_1 \neq j_2$ for $j_1, j_2 < \delta $ and $x_1 \in X_{j_1}, x_2 \in X_{j_2}$ then $x_1 {\restriction } \ell ^* \neq x_2 {\restriction } \ell ^*$ . This is possible, since the $(X_k)_{k < \delta }$ are disjoint and of size less than $\kappa $ . This allows us to define

$$\begin{align*}A_{\eta_k} := \bigcup_{x \in X_k} [x {\restriction} \ell^*]. \end{align*}$$

Now we glue the conditions $t^k_{\ell ^*}$ together in the following way: Set

$$\begin{align*}q(0) := \bigcup_{k < \delta} t^k_{\ell^*}(0) \end{align*}$$

and for $\beta> 0$ define $q(\beta )$ inductively; assuming $q {\restriction } \beta $ has been defined, set

$$\begin{align*}q(\beta) := t^{\dot{k}}_{\ell^*}(\beta), \textrm{ where}\ \dot{k}\ \textrm{is the unique ordinal less than}\ \delta\ \textrm{such that}\ t^{\dot{k}}_{\ell^*} {\restriction} \beta \in \dot{G}_\beta. \end{align*}$$

Note that $(t_{\ell ^*}^k {\restriction } \beta )_{k< \delta }$ is a maximal antichain below $q {\restriction } \beta $ . For limit $\beta $ observe that $\operatorname {\mathrm {supp}}(q {\restriction } \beta ) \subseteq \bigcup _{k < \delta } \operatorname {\mathrm {supp}}(t^k_{\ell ^*} {\restriction } \beta )$ . By Lemma 7.1 we know that $t_{\ell ^*}^k \leq _{F,i} p^{[\eta _k]}$ for each $k < \delta $ , and since $\operatorname {\mathrm {split}}_i(p(0)) = \{\eta _k : k < \delta \}$ , we can inductively conclude $q {\restriction } \beta \leq _{F \cap \beta ,i} p {\restriction } \beta $ for all $\beta \leq \kappa ^{++}$ .

To see the last claim, only note that $q^{[\eta ]} = t_{\ell ^*}^k$ for some $k < \delta $ , therefore by Lemma 7.1 we have $t_{\ell ^*}^k \Vdash \exists x \in \check {X_k}:\ \dot {\tau } {\restriction } \ell ^* = x {\restriction } \ell ^*$ and thus

$$\begin{align*}q^{[\eta]} \Vdash \dot{\tau} \in A_\eta \end{align*}$$

by definition of $A_\eta $ .