Equivalence between tree of parameters and dividing sequences I have been stuck with an equivalence for a few days now and would like some pointers on how to resolve my issues.
Proposition : Let $(\phi_i(x,y_i) : i < \alpha)$ be a sequence of L-formulas and let $(k_i : i < \alpha)$ be a sequence of natural numbers. For any partial type $\pi(x)$ over A the following are equivalent :

*

*There is a sequence $(b_i : i < \alpha)$ such that $\pi(x) \cup \{\phi_i(x,b_i): i < \alpha\}$ is consistent and for each $i < \alpha, \phi_i(x,b_i), k_i$-divides over $Ab_{<i}$.

*There is a tree $(a_s : s \in \omega^{\leq \alpha})$ such that for each $f \in \omega^\alpha$ $$ \pi(x) \cup \{\phi_i(x,a_{f|i+1}) : i < \alpha\}$$ is consistent and for each $i < \alpha$, for each $s \in \omega^i$, $\{\phi_i(x,a_{sj}) : j < \omega\}$ is $k_i$ inconsistent.

The beginning of the suggested proof for $1 \Rightarrow 2 $ says :
Observe that $a_s$ plays no role if the length of s is $0$ or a limit ordinal. We construct $a_s$ for $s \in \omega^i$ by induction in $i \leq \alpha$ with the addition property that $(a_{s|j+1} : j < i) \equiv_A (b_j : j < i)$.
Assume $s \in \omega^i$ and that $a_s$ has already been obtained. Choose c such that $$ (a_{s|j+1} : j < i)c \equiv_A (b_j : j < i)b_i$$. Then $\phi_i(x,c), k_i$-divides over $A' = A\{(a_{s|j+1} : j < i)\}$[...]
Now I have multiple questions :

*

*When defining a tree for instance $(a_s : s \in \omega^{\leq \omega})$ what is exactly the root ? Since $\omega^{\leq \omega}$ represent finite strings of natural numbers it seems that s $s \in \omega$ gives us child of a root node, but what is it exactly ? Is this what is meant by "Observe that $a_s$ plays no role when the length of s is $0$" ?

*Now for the proof itself, I tried to reduce the induction to the case $i = 1$ by choosing $s \in \omega$ such that $s = 1$ for instance. We assume $a_1$ has been defined such that $a_1 \equiv_A b_0$ (is it the correct interpretation of $(a_{s|j+1} : j < i) \equiv_A (b_j : j < i)$ ? Now we choose $c$ such that $$a_1c \equiv_A b_0b_1$$, I think this is possible by homogeneity of the monster model. Now I do not understand why in this case $\phi_1(x,c)$ divides over $Aa_1$.

What I have tried :
Since $\phi_1(x,b_1)$ divides over $Ab_0$ we can find $(b'_k : k < \omega)$ such that $b_1 \equiv_{Ab_0} b'_k$ and $\{\phi_1(x,b'_k) : k < \omega\}$ is $k_1-$ inconsistent.
We want to show ideally that $c \equiv_{Aa_1} b'_k$ for all $k$ since in this case $\phi_1(x,c)$ would divide.
To this end let $\theta(x) \in tp(c,Aa_1)$. This means that $\models \theta(c,\bar aa_1)$ for some $\bar a \in A$. We would like to show that $\models \theta(b'_k,\bar aa_1)$.
Since $a_1c \equiv_A b_0b_1$  we have $\models \theta(b_1, \bar ab_0)$.
Since $b_1 \equiv_{Ab_0} b'_k$ we have $\models \theta(b'_k, \bar ab_0)$.
At this point I don't know how to introduce $a_1$ and eliminate $b_0$. I am tempted to use the fact that $a_1 \equiv_A b_0$ but this is insufficient since $\theta(b'_k, \bar ab_0)$ involves $b'_k$ and any automorphism of the monster model that fixes $A$ will not fix $b'_k$necessarily.
Thank you for any advice or pointer on how to resolve this.
in the previous text $\equiv_A$ means the types are the same over $A$.
 A: Matching the numbering of your two subquestions.
Subquestion 1. The elements of $\omega^{\leq \alpha}$ are sequences of natural numbers of length $\leq \alpha$. The root is the empty sequence, i.e. the sequence of length $0$. If you look carefully at point 2 in the proposition, then you will see that the only $a_s$ that are actually used are of the form $a_{f|i+1}$ or $a_{sj}$. Here "$sj$" stands for "the sequence $s$ with $j$ added to the end". So we only use $a_s$ where the length of $s$ is a successor ordinal. Hence $a_0$ and $a_\ell$ for $\ell$ a limit are not used.
Subquestion 2. You essentially already answered this yourself, but you were trying something too difficult after that. Let's just look at the general case, I do not think working with specific values gives any more insight.  By induction we have $(a_{s|j+1} : j < i) \equiv_A (b_j : j < i)$. So there is an automorphism $f$ of the monster model that fixes $A$ pointwise and such that $a_{s|j+1} = f(b_j)$ for all $j < i$. Then we can define $c = f(b_i)$ and so we get
$$
(a_{s|j+1} : j < i)c \equiv_A (b_j : j < i)b_i.
$$
By assumption $\phi_i(x, b_i)$ $k_i$-divides over $Ab_{<i}$, so $\phi(x, f(b_j))$ $k_i$-divides over $f(Ab_{<i})$. The latter is by construction precisely saying that $\phi(x, c)$ $k_i$-divides over $A\{a_{s|j+1} : j < i\}$.
