Proof. It is enough to show that $P_0^{\alpha ,2}$ of $\mathcal {T}_\alpha ^2$ is well-ordered by $<_{lex}$ . We show this using induction on $\alpha $ . Clearly $P_0^{0,2}$ of $\mathcal {T}_0^2$ is well ordered by $<_{lex}$ . Suppose that for all $\beta <\alpha $ , $P^{\beta ,2}_0$ of $\mathcal {T}_{\beta }^2$ is well ordered by $<_{lex}$ . To get a contradiction, we assume $P^{\alpha ,2}_0$ of $\mathcal {T}_\alpha ^2$ is not well ordered by $<_{lex}$ . Then there exists a decreasing $(\eta _i)_{i<\omega }$ in $P^{\alpha ,2}_0$ of $\mathcal {T}_\alpha ^2$ . By finite support, for each $i<\omega $ , there exists $\beta _i<\alpha $ such that $\eta _i(\beta _i)=1$ and $\eta _i(\beta )=0$ for all $\beta _i<\beta <\alpha $ . Note that $\beta _i\ge \beta _{i+1}$ for all $i<\omega $ . Since $\alpha $ is well ordered by $<$ , there exists $i^*<\omega $ such that $\beta _{i^*}=\beta _i$ for all $i>i^*$ . Let $\eta ^{\prime }_i:=\eta _i|_{\beta _{i^*}}$ for each $i>i^*$ . Note that the domain of each $\eta _i$ is $\alpha $ since $\eta _i\in P^{\alpha ,2}_0$ . Thus $\eta ^{\prime }_i\in P^{\beta _{i^{*}}\!,2}_0$ of $\mathcal {T}_{\beta _{i^*}}^2$ for all $i>i^*$ , and hence $(\eta ^{\prime }_i)_{i^*<i<\omega }$ forms a descending sequence in $P^{\beta _{i^{*}}\!,2}_0$ of $\mathcal {T}_{\beta _{i^*}}^2$ , which yields a contradiction with the induction hypothesis.

Notation 3.3. For ordinals $\beta <\alpha $ , let $\Omega _\beta ^{\alpha ,2}$ be the ordinal which is order isomorphic to $(P_\beta ^{\alpha ,2},<_{lex})$ in $\mathcal {T}_\alpha ^2$ , and let $(\eta (\alpha ,\beta ,i))_{i<\Omega ^{\alpha ,2}_\beta }$ be the enumeration of $P_\beta ^{\alpha ,2}$ with respect to $<_{lex}$ .

Fact 3.5. [Reference Simon21, Facts 2.19 and 2.20].

If $(q_i)_{i\in I}$ is a sequence of A-invariant types, then $\bigotimes _{i\in I}q_i(x_i)$ is also an A-invariant type. Moreover, if the invariant types are coheirs, then the type product $\bigotimes _{i\in I}q_i(x_i)$ is also a coheir. Let $(I,<_I)$ and $(J,<_J)$ be linearly ordered sets. Let $I\oplus J$ be a disjoint union of I, J and give a linear order $<$ on $I\oplus J$ by $x<y$ if and only if $x,y\in I$ and $x<_I y$ , $x,y\in J$ and $x<_J y$ , or $x\in I$ and $y\in J$ . Then for given two sequences of global A-invariant types $(q_i)_{i\in I}$ and $(q_j)_{j\in J}$ , $\left (\bigotimes _{j\in J}q_j(x_j)\right )\otimes \left (\bigotimes _{i\in I}q_i(x_i)\right )=\bigotimes _{k\in I\oplus J}q_k(x_k)$ . In other words, the type product $\otimes $ on invariant types is associative.

Notation 3.6. Let $\alpha $ be an ordinal and $p((x_\eta )_{\eta \in \mathcal {T}_\alpha ^2})$ a global type. For a subset X of $\mathcal {T}_\alpha ^2$ , let $p|_X:=p|_{(x_\eta )_{\eta \in X}}$ .

Proof. We use induction on $\alpha \in \text {ON}$ . Let $q_0( (x_\eta )_{\eta \in \mathcal {T}_0^2})=q((x_\eta )_{\eta \in \mathcal {T}_0^2})$ . Suppose that $\alpha>0$ and we have constructed a sequence of global coheirs $(q_\beta ((x_\eta )_{\eta \in \mathcal {T}_\beta ^2}))_{\beta < \alpha }$ satisfying (i)–(iv).

First we assume that $\alpha $ is a successor ordinal, namely, $\alpha =\beta +1$ . Choose any $(a^0_\eta )_{\eta \in \mathcal {T}_\beta ^2}$ , $(a^1_\eta )_{\eta \in \mathcal {T}_\beta ^2}$ , and $a^*$ such that

$$ \begin{align*} (a^0_\eta)_{\eta\in\mathcal{T}_\beta^2}&\models q_\beta |_M,\\ (a^1_\eta)_{\eta\in\mathcal{T}_\beta^2}&\models q_\beta |_{M(a^0_\eta)_{\eta\in\mathcal{T}_\beta^2}},\;\text{and}\\ a^*&\models\;p\;|_{M(a^1_\eta)_{\eta\in\mathcal{T}_\beta^2}(a^0_\eta)_{\eta\in\mathcal{T}_\beta^2}}. \end{align*} $$

Then

$$\begin{align*}q_\beta( (x^1_\eta)_{\eta\in\mathcal{T}_\beta^2})\otimes q_\beta((x^0_\eta)_{\beta\in\mathcal{T}_\beta^2})\supseteq \operatorname{tp}((a^1_\eta)_{\eta\in\mathcal{T}_\beta^2}(a^0_\eta)_{\eta\in\mathcal{T}_\beta^2}/M).\end{align*}$$

Choose any $(\hat {a}^1_\eta )_{\eta \in \mathcal {T}_\beta ^2}$ and $(\hat {a}^0_\eta )_{\eta \in \mathcal {T}_\beta ^2}$ outside the monster model $\mathbb {M}$ such that

$$\begin{align*}(\hat{a}^1_\eta)_{\eta\in\mathcal{T}_\beta^2}(\hat{a}^0_\eta)_{\eta\in\mathcal{T}_\beta^2}\models q_\beta( (x^1_\eta)_{\eta\in\mathcal{T}_\beta^2})\otimes q_\beta((x^0_\eta)_{\beta\in\mathcal{T}_\beta^2}).\end{align*}$$

Then by strong left transitivity of . Since

$$\begin{align*}(\hat{a}^1_\eta)_{\eta\in\mathcal{T}_\beta^2}(\hat{a}^0_\eta)_{\eta\in\mathcal{T}_\beta^2}\equiv_M({a}^1_\eta)_{\eta\in\mathcal{T}_\beta^2}({a}^0_\eta)_{\eta\in\mathcal{T}_\beta^2},\end{align*}$$

we have $a^{**}$ such that

$$\begin{align*}(\hat{a}^1_\eta)_{\eta\in\mathcal{T}_\beta^2}(\hat{a}^0_\eta)_{\eta\in\mathcal{T}_\beta^2}a^{**}\equiv_M({a}^1_\eta)_{\eta\in\mathcal{T}_\beta^2}({a}^0_\eta)_{\eta\in\mathcal{T}_\beta^2}a^*.\end{align*}$$

By left extension of , there exists $\hat {a}^*\equiv _{M(\hat {a}^1_\eta )_{\eta \in \mathcal {T}_\beta ^2}(\hat {a}^0_\eta )_{\eta \in \mathcal {T}_\beta ^2}} a^{**}$ such that

Note that $\hat {a}^*\notin \mathbb {M}$ and $(\hat {a}^1_\eta )_{\eta \in \mathcal {T}_\beta ^2}(\hat {a}^0_\eta )_{\eta \in \mathcal {T}_\beta ^2}\hat {a}^*\equiv _M({a}^1_\eta )_{\eta \in \mathcal {T}_\beta ^2}({a}^0_\eta )_{\eta \in \mathcal {T}_\beta ^2}a^*.$ For each $\eta \in \mathcal {T}_\alpha ^2$ , define $a^\dagger _\eta $ by

$$ \begin{align*} a^\dagger_\eta= \begin{cases} \hat{a}^* & \text{if } \eta=\emptyset \\ \hat{a}^0_\nu & \text{if } \eta=(0){^{\frown}}\nu \\ \hat{a}^1_\nu & \text{if } \eta=(1){^{\frown}}\nu, \\ \end{cases} \end{align*} $$

and let

$$\begin{align*}q_\alpha((x_\eta)_{\eta\in\mathcal{T}_\alpha^2})=\operatorname{tp}((a^\dagger_\eta)_{\eta\in\mathcal{T}_\alpha^2} /\mathbb{M}).\end{align*}$$

Then by Fact 3.5 and the induction hypothesis, $q_\alpha $ satisfies (ii) and (iii). Now we show that $q_\alpha $ satisfies (iv). Choose any A and suppose $(a_\eta )_{\eta \in \mathcal {T}_\alpha ^2}\models q_\alpha |_{MA}$ . Then

$$\begin{align*}(a_\eta)_{\eta\in\mathcal{T}_\alpha^2}\equiv_M\! (a^0_\eta)_{\eta\in\mathcal{T}_\beta^2}(a^1_\eta)_{\eta\in\mathcal{T}_\beta^2}a^*.\end{align*}$$

Thus ( $\ast $ ) is clear. ( $\ast \ast $ ) is by the induction hypothesis and Fact 3.5.

Now we suppose that $\alpha $ is a limit ordinal. Then we just take $q_\alpha :=\bigcup _{\beta <\alpha }p_\beta $ with respect to canonical inclusions. Clearly that $q_\alpha $ satisfies (ii)–(iv) by finite support.

Proof. Suppose $\varphi (x,a)$ Kim-divides over M. Then there exist $k\in \omega $ and a global M-invariant type $p(y)$ containing $\operatorname {tp}(a/M)$ such that the set $\{\varphi (x,a_i)\}_{i\in \omega }$ is k-inconsistent for all $(a_i)_{i\in \omega }\models p^{\otimes \omega }|_M$ . To get a contradiction, we assume that there is no global coheir extension $q(y)$ of $\operatorname {tp}(a/M)$ such that $\{\varphi (x,a_i)\}_{i\in \omega }$ is inconsistent for some (any) $(a_i)_{i\in \omega }\models q^{\otimes \omega }|_M$ . Recall that the number of all global coheir extensions of $\operatorname {tp}(a/M)$ is bounded (Remark 2.9). Let $\theta $ be the cardinality of all global coheir extensions of $\operatorname {tp}(a/M)$ and $\kappa $ be a cardinal such that $\kappa>\theta ^+$ .

Choose any global coheir extension $q(y)$ of $\operatorname {tp}(a/M)$ . By Lemma 3.9, we can find $(q_\beta ( (y_\eta )_{\eta \in \mathcal {T}_\beta ^2}))_{\beta \le \kappa }$ such that:

1. (i) $q_0( (y_\eta )_{\eta \in \mathcal {T}_0^2})=q((y_\eta )_{\eta \in \mathcal {T}_0^2})$ ,

2. (ii) $\bigotimes _{i<\Omega ^{\kappa ,2}_\beta }q_\beta ((y_\eta )_{\eta \in C^{\kappa ,2}_{\eta (\kappa ,\beta ,i)}})\subseteq q_\kappa ( (y_\eta )_{\eta \in \mathcal {T}_\kappa ^2})$ for each $\beta <\kappa $ .

3. (iii) If $(a_\eta )_{\eta \in \mathcal {T}_\kappa ^2}\models q_\kappa ( (y_\eta )_{\eta \in \mathcal {T}_\kappa ^2})|_M$ , then

   1. (*) $a_\eta \models p|_{Ma_{\rhd \eta }}$ for each $\eta \in \mathcal {T}_\kappa ^2$ ,

   2. (**) for each $\beta <\kappa $ , there exists a global coheir $q^{\beta }$ such that $a_\eta \models q^\beta |_{Ma_{\perp _{lex}\eta }}$ for all $\eta \in P^{\kappa ,2}_\beta $ .

Let $(a_\eta )_{\eta \in \mathcal {T}_\kappa ^2}$ be a realization of $q_\kappa |_M$ . Since $\kappa>\theta ^+$ , there exists a set of successor ordinals $K\subseteq \kappa $ such that $|K|=\omega $ and $q^\beta =q^{\beta '}$ for all $\beta ,\beta '\in K$ . Let $\{\beta _i\}_{i<\omega }$ be an enumeration of K such that $\beta _i<\beta _j$ for all $i<j<\omega $ . For each $\eta \in \mathcal {T}_\omega ^2$ , let $f(\eta )$ be an element in $\mathcal {T}_\kappa ^2$ such that $\text {dom}(f(\eta ))=[\beta _{t(\eta )},\kappa )$ and

$$ \begin{align*} f(\eta)(\beta)= \begin{cases} \eta(i) & \text{if } \beta=\beta_{i+1}-1 \\ 0 & \text{otherwise}. \\ \end{cases} \end{align*} $$

Then we can regard f as an $L_1$ -embedding from $\mathcal {T}_\omega ^2$ into $\mathcal {T}_\kappa ^2$ . Let $q^*=q^\beta $ for some (any) $\beta \in K$ . Then $a_{f(\eta )}\models q^*|_{Ma_{\perp _{lex}f(\eta )}}$ for all $\eta \in \mathcal {T}_\omega ^2$ . For each $\eta \in \mathcal {T}_\omega ^2$ , let $a^*_\eta :=a_{f(\eta )}$ . Then for each antichain X in $\mathcal {T}_\omega ^2$ , $(a^*_\eta )_{\eta \in X}$ is a coheir Morley sequence over M generated by $q^*$ with respect to $<_{lex}$ . Thus the set $\{\varphi (x,a^*_\eta )\}_{\eta \in X}$ is consistent. On the other hand, if X is a path in $\mathcal {T}_\omega ^2$ , then $(a^*_\eta )_{\eta \in X}$ is an invariant Morley sequence over M generated by p with respect to $\lhd $ . Thus the set $\{\varphi (x,a^*_\eta )\}_{\eta \in X}$ is k-inconsistent. By compactness and [Reference Ahn, Kim and Lee6, Lemma 3.20], we can construct an antichain tree, which yields a contradiction with the assumption that T is NATP.

Fact 3.12. Let M be a model. Suppose that $\alpha (x,e)\vdash \psi (x,c)\vee \bigvee _{i<n}\varphi _i(x,a_i)$ and $\varphi _i(x,a_i)$ coheir-divides over M for each $i<n$ . Then there exists $e_0,\ldots ,e_m$ such that $e_i\equiv _M e$ for each $i\le m$ and $\{\alpha (x,e_i)\}_{i\le m}\vdash \psi (x,c)$ .

Proof. Suppose a formula $\varphi (x,a)$ Kim-forks over a model M. Then there exist formulas $\psi _0(x,a_0), \ldots ,\psi _n(x,a_n)$ such that $\varphi (x,a)\vdash \bigvee _{i\le n}\psi _i(x,a_i)$ and $\psi _i(x,a_i)$ Kim-divides over M for each $i\leq n$ . By Lemma 3.10, $\psi _i(x,a_i)$ coheir-divides over M for each $i\le n$ . By taking $\psi (x,c):=\perp $ (i.e., $\forall x(x\neq x)$ ), we can apply Fact 3.12. Thus there exist $a_0,\ldots ,a_m$ such that $a_i\equiv _M a$ for each $i\le m$ and $\{\varphi (x,a_i)\}_{i\le m}\vdash \perp $ . So $\{\varphi (x,a_i)\}_{i\le m}$ is inconsistent and $\varphi (x,a)$ quasi-divides over M.

Notation 3.15. Suppose that satisfies quasi-strong finite character and full existence over d. For each a and b, let be the unique $\subseteq $ -maximal partial type over d such that

if and only if

Note that if satisfies monotonicity over d additionally, then for all $a'\subseteq a$ and $b'\subseteq b$ , by the maximality.

Proof. Let $q(x):=\operatorname {tp}(a/M)$ and $\Lambda (x)$ be

First we show that $\Lambda (x)$ is consistent. Suppose not. Then there exist $\varphi _0(x^{\prime }_0,b^{\prime }_0)$ , $\ldots $ , $\varphi _n(x^{\prime }_n,b^{\prime }_n)$ and $\psi _0(x^{\prime \prime }_0,d^{\prime \prime }_0)$ , …, $\psi _m(x^{\prime \prime }_m,d^{\prime \prime }_m)$ such that $\varphi _i(a^{\prime }_i,y^{\prime }_i)$ Kim-forks over M and $a_i'\subseteq a$ for each $i\le n$ , for each $j\le m$ , and

$$\begin{align*}q(x)\vdash \bigvee_{i\le n}\varphi_i(x^{\prime}_i,b^{\prime}_i) \vee \bigvee_{j\le m}\psi_j(x^{\prime\prime}_j,d^{\prime\prime}_j).\end{align*}$$

Note that $x^{\prime }_0,\ldots ,x^{\prime }_n$ , $x^{\prime \prime }_0,\ldots ,x^{\prime \prime }_m$ are finite subtuples of x, and $d^{\prime \prime }_j\subseteq d_j$ for each $j\le m$ . Since $\bigvee _{i\le n}\varphi _i(a^{\prime }_i,y^{\prime }_i)$ also Kim-forks over M, we may assume $n=0$ . Let $a':=a^{\prime }_0$ , $b':=b^{\prime }_0$ , $x':=x_0'$ , $y':=y_0'$ , and $\varphi (x',y'):=\varphi _0(x^{\prime }_0,y^{\prime }_0)$ . Then we have

$$\begin{align*}q(x)\vdash \varphi(x',b') \vee \bigvee_{j\le m}\psi_j(x^{\prime\prime}_j,d^{\prime\prime}_j).\end{align*}$$

By Corollary 3.13, $\varphi (a',y')$ quasi-divides over M, so there exist $a_0,\ldots ,a_k$ such that $\{\varphi (a_i,y)\}_{i\le k}$ is inconsistent and $a_i\equiv _M a'$ for each $i\le k$ . Choose any $\hat {a}_i$ containing $a_i$ such that $\hat {a}_ia_i\equiv _Maa'$ for each $i\le k$ . Let $\hat {a}:=(\hat {a}_0,\ldots ,\hat {a}_k)$ and $r(\hat {x}_0\ldots \hat {x}_k):=\operatorname {tp}(\hat {a}_0\ldots \hat {a}_k/M)$ . For each $i\le k$ and $j\le m$ , let $\hat {x}^{\prime }_i$ and $\hat {x}^{\prime \prime }_{ij}$ be the subtuples of $\hat {x}_i$ whose indices in $\hat {x}_i$ correspond to the indices of $x'$ and $x^{\prime \prime }_j$ in x, respectively. Then for each $i\le k$ ,

$$\begin{align*}r|_{\hat{x}_i}\vdash \varphi(\hat{x}^{\prime}_i,b)\vee \bigvee_{j\le m}\psi_j(\hat{x}^{\prime\prime}_{ij},d^{\prime\prime}_j).\end{align*}$$

Thus

$$\begin{align*}r(\hat{x}_0,\ldots,\hat{x}_k)\vdash\bigwedge_{i\le k}\Big[ \varphi(\hat{x}^{\prime}_i,b)\vee \bigvee_{j\le m}\psi_j(\hat{x}^{\prime\prime}_{ij},d^{\prime\prime}_j)\Big].\end{align*}$$

But

$$\begin{align*}r(\hat{x}_0,\ldots,\hat{x}_k)\vdash \neg\exists y\Big(\bigwedge_{i\le k}\varphi(\hat{x}^{\prime}_i,y)\Big),\end{align*}$$

and hence

$$\begin{align*}r(\hat{x}_0,\ldots,\hat{x}_k)\vdash \bigvee_{i\le k, j\le m}\!\!\!\!\psi_j(\hat{x}^{\prime\prime}_{ij},d^{\prime\prime}_j).\end{align*}$$

By full existence of , there exists $a^*=(a^*_0,\ldots ,a^*_k)$ such that $a^*\equiv _M\hat {a}$ and . Since $a^*\models r$ , there exist $i\le k$ and $j\le m$ such that $\models \psi _j(a^*_{ij},d^{\prime \prime }_j)$ , where $a^*_{ij}$ is the subtuple of $a^*_i$ whose indices in $a^*_i$ correspond to the indices of $\hat {x}^{\prime \prime }_{ij}$ in $\hat {x}_i$ (equivalently, $x^{\prime \prime }_{j}$ in x). Thus . Since $a^*_i\equiv _M \hat {a}_i\equiv _M a$ and $d_j\equiv _M d_j$ , we have , which yields a contradiction with by monotonicity of . Thus $\Lambda (x)$ is consistent.

Choose any global type $p(x)$ which is a completion of $\Lambda (x)$ . Then $p(x)$ satisfies all the conditions we want.

Proof. Apply Corollary 3.16 on .

Fact 4.3 [Reference Chernikov and Kaplan8, Reference Kaplan and Ramsey12, Reference Kim13].

Let b be a tuple of parameters, C a set, and M a model.

1. (i) If T is simple, then every Morley sequence in $\operatorname {tp}(b/C)$ is a witness of dividing of b over C

2. (ii) If T is NSOP $_1$ , then every invariant Morley sequence over M starting with b is a witness of Kim-dividing of b over M.

3. (iii) If T is NTP $_2$ , then every strictly invariant Morley sequence over M starting with b is a witness of dividing of b over M.

Proof. Clearly (i) implies (ii). Suppose (ii). Then by compactness, the partial type

$$ \begin{align*} \begin{aligned} \Lambda(x):=\operatorname{tp}_x(a/Db) {} & \cup \{ \neg\varphi(x,c)\;|\;c\in\mathbb{M},\;\varphi\in\mathcal{L}(D),\;\varphi(a,y)\text{ Kim-forks over }D\} \\ & \cup \{ \neg\psi(x,c)\;|\;c\in\mathbb{M},\;\psi\in\mathcal{L}(D),\;\psi(x,c)\text{ is not realized in }D\} \end{aligned} \end{align*} $$

is consistent. Any global type which is a completion of $\Lambda (x)$ is a strict coheir containing $\operatorname {tp}(a/Db)$ .

Proof. Let p be a global coheir of $\operatorname {tp}(a/M)$ such that $I\models p^{\otimes \omega }|_M$ . Note that the number of all coheirs of $\operatorname {tp}(a/M)$ , say $\kappa $ , is bounded. Choose any $\kappa '>\kappa $ . Let $I'=(a^{\prime }_i)_{i<\kappa '}$ be a coheir Morley sequence generated by p over M. Then $I'$ is also a witness of Kim-dividing of a over M. First we claim that for each $i<\kappa '$ . Choose any $i<\kappa '$ . By Lemma 4.6, it is enough to show that for any b, there exists $b'\equiv _{Ma^{\prime }_{<i}}b$ such that and . Let $\Theta =\kappa '\setminus i$ . Then we can find $J=(a^*_j)_{j\in \Theta }$ such that $a^*_j\models p|_{Ma^{\prime }_{<i}ba^*_{<j}}$ for all $j\in \Theta $ . Since $J\equiv _{Ma^{\prime }_{<i}}(a^{\prime }_j)_{j\geq i}$ , we can find $b'\equiv _{Ma^{\prime }_{<i}}b$ such that $a^{\prime }_j\models p|_{Ma^{\prime }_{<j}b'}$ for all $i\leq j<\kappa '$ . Then $b'$ satisfies . Note that $(a^{\prime }_j)_{j\in \Theta }$ is a witness of Kim-dividing of $a^{\prime }_i$ over M. Thus by indiscernibility of $(a^{\prime }_j)_{j\in \Theta }$ over $a^{\prime }_{<i}b'$ . Note that Kim-dividing and Kim-forking over a model are equivalent since we assume that every type over a model has a global invariant extension that is a witness. Thus .

Since $\kappa '$ is sufficiently large, we may assume that there exists a global strong coheir q of $\operatorname {tp}(a/M)$ such that $\operatorname {tp}(a_i/Ma_{<i})\subseteq q$ for all $i\in \kappa '$ . Thus I is a strict coheir Morley sequence in $\operatorname {tp}(a/M)$ .

Proof. Suppose (i) is true and choose any b. By full existence, there exists a global type $p(x)\supseteq \operatorname {tp}(b/M)$ such that for all $b'\models p|_{MA}$ . Since is stronger than , $p(x)$ is a coheir over M. Let $\varphi (x,b)$ Kim-divide over M. Then by Theorem 3.10, there exists a global coheir $q(x)$ over M and $k<\omega $ such that $\{\varphi (x,b_i)\}_{i<\omega }$ is k-inconsistent for every coheir Morley sequence $(b_i)_{i<\omega }$ generated by $q(x)$ over M.

It is enough to show that $\{\varphi (x,b_i)\}_{i<\omega }$ is inconsistent for any coheir Morley sequence $(b_i)_{i<\omega }$ generated by $p(x)$ over M. To get a contradiction, suppose not. Then $\{\varphi (x,b_i)\}_{i<\omega }$ is consistent for some (any) coheir Morley sequence $(b_i)_{i<\omega }$ generated by $p(x)$ over M.

Claim. For each $n<\omega $ and a small set A, there exists $(b_\eta )_{\eta \in 2^{<n}}$ such that:

1. (*) $b_\eta \models q|_{Mb_{\rhd \eta }}$ for each $\eta \in 2^{<n}$ ;

2. (**) $b_\eta \models p|_{MAb_{\perp _{lex}\eta }}$ for each $\eta \in 2^{<n}$ ;

3. (***) .

Proof of Claim.

We use induction on $n<\omega $ . If $n=0$ , then there is nothing to prove. Suppose $n=1$ and let A be an arbitrary small set. Choose any $b'\models p|_{MA}$ and let $b_{\emptyset }:=b'$ . Then $(b_\eta )_{\eta \in 2^{<n}}$ satisfies ( $\ast $ ), ( $\ast \ast $ ), and ( $\ast \!\ast \!\ast $ ).

Now let n be an arbitrary natural number larger than $0$ and suppose that for any small set, there exists $(b_i)_{\eta \in 2^{<n}}$ satisfying ( $\ast $ ), ( $\ast \ast $ ), and ( $\ast \!\ast \!\ast $ ) over the set. Choose any A. We find $(b_i)_{\eta \in 2^{<{n+1}}}$ satisfying ( $\ast $ ), ( $\ast \ast $ ), and ( $\ast \!\ast \!\ast $ ) over A.

By the induction hypothesis, there exists $(b^0_\eta )_{\eta \in 2^{<n}}$ satisfying ( $\ast $ ), ( $\ast \ast $ ), and ( $\ast \!\ast \!\ast $ ) over A. Let $B^0:=(b^0_\eta )_{\eta \in 2^{<n}}$ . By applying the induction hypothesis again, we can find $(b^1_i)_{\eta \in 2^{<n}}$ satisfying ( $\ast $ ), ( $\ast \ast $ ), and ( $\ast \!\ast \!\ast $ ) over $AB^0$ . Let $B^1:=(b^1_\eta )_{\eta \in 2^{<n}}$ . Then by strong right transitivity.

Let $m=|2^{<{n+1}}|-1=2^{n+1}-2$ and choose any enumeration $\{\eta _i\}_{i<m}$ of $2^{<{n+1}}\setminus \{\emptyset \}$ with $\eta _0(i)=0$ for all $i<n+1$ . Let $B=(b_{\eta _i})_{i<m}$ be an enumeration of $B^0B^1$ such that $b_{\eta _0}=b^0_{\eta ^*}$ , where $\eta ^*\in 2^{n-1}$ and $\eta ^*(i)=0$ for all $i<n-1$ . Then and $b_{\eta _0}\models q|_M$ .

Subclaim. We can continue constructing a sequence

$$\begin{align*}B_0,B_1,\ldots,B_\alpha,B_{\alpha+1},\ldots\end{align*}$$

such that $B_\alpha =(b_{\alpha ,\eta _0},\ldots ,b_{\alpha ,\eta _{m-1}})$ for each ordinal $\alpha $ , $b_{0,\eta _i}=b_{\eta _i}$ for each $i<m$ so that $B_0=B$ , and

1. (†) $B_\alpha \equiv _MB_0$ and for each ordinal $\alpha $ ,

2. (‡) $b_{\alpha ,\eta _0}\models q|_{MB_{<\alpha }}$ for each ordinal $\alpha $ .

Proof of Subclaim.

Let $b_{0,\eta _i}:=b_{\eta _i}$ for each $i<m$ and $B_0:=(b_{0,\eta _0},\ldots ,b_{0,\eta _{m-1}})$ . For each ordinal $\alpha $ , suppose that we have constructed a sequence $(B_\beta )_{\beta <\alpha }$ satisfying ( $\dagger $ ) and ( $\ddagger $ ). Choose any $b'\models q|_{MB_{<\alpha }}$ . Since $b'\equiv _M b_{0,f\eta _0}$ , there exists $B'$ containing $b'$ such that $B'\equiv _MB_0$ . Since , there exists $B"\equiv _{Mb'}B'$ such that by left extension of . Let $B_\alpha =B"$ and $b_{\alpha ,\eta _0}=b'$ . Then the sequence $(B_\beta )_{\beta <\alpha +1}$ satisfies ( $\dagger $ ) and ( $\ddagger $ ). This completes proof of the subclaim.

Proof. Suppose that a formula $\varphi (x,y)$ has k-DCTP $_2$ with $(b_\eta )_{\eta \in 2{^{<\omega }}}$ . Define a map $f:\omega {^{<\omega }}\to 2{^{<\omega }}$ by

$$\begin{align*}f(\eta)= \begin{cases} \emptyset & \text{ if } \eta=\emptyset\\ f(\nu){^{\frown}}(0)^i{^{\frown}} (1) & \text{ if } \eta=\nu{^{\frown}}(i), \end{cases} \end{align*}$$

and let $c_\eta :=b_{f(\eta )}$ for each $\eta \in \omega {^{<\omega }}$ . By using the modeling property, we have a strongly indiscernible $(d_\eta )_{\eta \in \omega ^{<\omega }}$ , which is strongly locally based on $(c_\eta )_{\eta \in \omega ^{<\omega }}$ . Define a map $g:2{^{<\omega }}\to \omega ^{<\omega }$ by

$$\begin{align*}g(\eta)= \begin{cases} \emptyset & \text{ if } \eta=\emptyset\\ g(\nu){^{\frown}}(0) & \text{ if } \eta=\nu{^{\frown}}(1)\\ g(\nu){^{\frown}}(1) & \text{ if } \eta=\nu{^{\frown}}(0), \end{cases} \end{align*}$$

and let $a_\eta :=d_{g(\eta )}$ for each $\eta \in 2{^{<\omega }}$ . Then $(a_\eta )_{\eta \in 2{^{<\omega }}}$ is strongly indiscernible and satisfies (i) and (ii).

Proof. (iii) and (iv) are clear. (i) can be proved by using the argument in [Reference Ahn and Kim4, Proposition 4.6].

To prove (ii), suppose $\varphi (x,y)$ has k-DCTP $_2$ with $(a_\eta )_{\eta \in 2{^{<\omega }}}$ . By Remark 5.6, we may assume that $(a_\eta )_{\eta \in 2{^{<\omega }}}$ is strongly indiscernible. For each $n<\omega $ and $i<2$ , let $b_{n,i}=a_{{\langle 0 \rangle }^n{^{\frown }}\langle 1-i\rangle }$ . Then $(b_{n,i})_{n<\omega , i<2}$ satisfies the conditions in [Reference Kaplan and Ramsey12, Lemma 2.3]. Thus T has SOP $_1$ .

Fact 5.9 [Reference Mutchnik17, Theorem 4.9].

Suppose T is N- $\omega $ -DCTP $_2$ . If $\varphi (x,a)$ coheir-divides over a model M, then for any canonical Morley sequence $(a_i)_{i<\omega }$ over M with $a_0=a$ , the set $\{\varphi (x,a_i)\}_{i<\omega }$ is inconsistent. As a consequence, coheir-dividing and coheir-forking are equivalent over models.

Proof. Suppose $\varphi (x,b)$ Kim-forks over M, a model. Since the theory is N- $\omega $ -DCTP $_2$ , it is NATP. By Theorem 3.10, $\varphi (x,b)$ coheir-forks over M. By Fact 5.9, $\varphi (x,b)$ coheir-divides over M. Thus it Kim-divides over M.

Question 5.11. Is there a natural example of N- $\omega $ -DCTP $_2$ theories having TP $_1$ and TP $_2$ ?

Question 5.12. For theories, is having NATP equivalent to having N- $\omega $ -DCTP $_2$ ?