Theorem: Statements Equivalent to Non-Dividing Type

Let $T$ be a complete $\LL$-theory.
Let $\mathfrak C$ be a monster model for $T$.
Let $A$ be a subset of the universe of $\mathfrak C$.
The following are equivalent:
:$(1): \quad$ $\map {\mathrm {tp} } {\bar c / A, \bar b}$ does not divide over $A$.
:$(2): \quad$ For every $\set {\bar b_i : i \in I}$ containing $\bar b$ which is order indiscernible over $A$, there is some $\bar c'$ with $\map {\mathrm {tp} } {\bar c' / A, \bar b} = \map {\mathrm {tp} } {\bar c / A, \bar b}$ such that $\set {\bar b_i : i \in I}$ is order indiscernible over $A,\bar c'$.
:$(3): \quad$ For every $\set {\bar b_i : i \in I}$ containing $\bar b$ which is order indiscernible  over $A$, there is some $A, \bar b$-automorphism $f$ such that $\set {\map f {\bar b_i} : i \in I}$ is order indiscernible over $A, \bar c$.

Proof:
=== $(1) \implies (2)$ ===
Let $\set {\bar b_i : i \in I}$ containing $\bar b$ be order indiscernible over $A$.
{{AimForCont}} $\ds \bigcup_{i \mathop \in I} \map {\mathrm {tp} } {\bar c / A, \bar b_i}$ is not satisfiable.
By the Compactness Theorem, some finite subset $\set {\map {\phi_1} {x, \bar b_{i_1} }, \dots, \map {\phi_k} {x, \bar b_{i_k} } }$ is not satisfiable.
Let $\phi$ be $\phi_1 \wedge \cdots \wedge \phi_k$.
Note that $\map \phi {x, \bar b}$ is in $\map {\mathrm {tp} } {\bar c / A, \bar b}$.
$\set {\map \phi {x, \bar b_i}: i \in I}$ is not satisfiable since it implies $\set {\map {\phi_1} {x, \bar b_{i_1} }, \dots, \map {\phi_k} {x, \bar b_{i_k} } }$.
Again by the Compactness Theorem, some finite subset $\set {\map \phi {x, \bar b_{j_1} }, \dots, \map \phi {x, \bar b_{j_h} } }$ is inconsistent.
Since all of the $b_i$ are order-indiscernibles, this means that any cardinality $h$ subset of $\set {\map \phi {x, \bar b_i} : i \in I}$ is not satisfiable.
This contradicts the assumption that $\map {\mathrm {tp} } {\bar c / A, \bar b}$ does not divide over $A$.
Thus $\ds \bigcup_{i \mathop \in I} \map {\mathrm {tp} } {\bar c / A, \bar b_i}$ is satisfiable.
Because $\mathfrak C$ is a monster model, it is satisfiable in $\mathfrak C$.
By Infinite Ramsey's Theorem and indiscernibility of $\set {\bar b_i : i \in I}$ over $A$, for each finite set $\Delta$ of formulas with parameters from $A$, there is some $\bar c_\Delta$ such that:
:$\bar c_\Delta$ realizes $\ds \bigcup_{i \mathop \in I} \map {\mathrm {tp} } {\bar c / A, \bar b_i}$
and:
:for any $i_1 < \cdots < i_k$ and $j_1 < \cdots < j_k$ in $I$, $\mathfrak C \models \map \phi {b_{i_1}, \dots, b_{i_k}, \bar c_\Delta} \leftrightarrow  \map \phi {b_{j_1}, \dots, b_{j_k}, \bar c_\Delta}$ for all $\phi \in \Delta$.
Because $b$ is one of the $b_i$, we have that each $c_\Delta$ realizes $\map {\mathrm {tp} } {\bar c / A, \bar b}$.
By the Compactness Theorem, since there is a $c_\Delta$ with $\mathfrak C \models \map \phi {b_{i_1}, \dots, b_{i_k}, \bar c_\Delta} \leftrightarrow \map \phi {b_{j_1}, \dots, b_{j_k}, \bar c_\Delta}$ for all $\phi \in \Delta$ for each finite $\Delta$, there is a $\bar c'$ with type $\map {\mathrm {tp} } {\bar c / A, \bar b}$ and which satisfies $\mathfrak C \models \map \phi {b_{i_1}, \dots, b_{i_k}, \bar c'} \leftrightarrow \map \phi {b_{j_1}, \dots, b_{j_k}, \bar c'}$ for all $\phi$.
But this holding for every $\phi$ means that $\set {b_i : i \in I}$ is order-indiscernible over $A, \bar c'$.