$(1)\implies (2)$: Let $\{\bar b_i : i \in I\}$ containing $\bar b$ be order indiscernible over $A$. Suppose $\displaystyle \bigcup_{i\in I} \operatorname{tp}(\bar c / A,\bar b_i)$ is not satisfiable. :By compactness, some finite subset $\{\phi_1(x, \bar b_{i_1}), \dots, \phi_k(x, \bar b_{i_k})\}$ is not satisfiable. :Let $\phi$ be $\phi_1 \wedge \cdots \wedge \phi_k$. :Note that $\phi(x, \bar b)$ is in $\operatorname{tp}(\bar c / A,\bar b)$. :$\{\phi(x, \bar b_i ) : i\in I\}$ is not satisfiable since it implies $\{\phi_1(x, \bar b_{i_1}), \dots, \phi_k(x, \bar b_{i_k}))$. :Again by compactness, some finite subset $\{\phi(x, \bar b_{j_1}), \dots, \phi(x, \bar b_{j_h})\}$ is inconsistent. :Since all of the $b_i$ are order-indiscernibles, this means that any cardinality $h$ subset of $\{\phi(x, \bar b_i ) : i\in I\}$ is not satisfiable. :This contradicts the assumption that $\operatorname{tp}(\bar c / A,\bar b)$ does not divide over $A$. Thus $\displaystyle \bigcup_{i\in I} \operatorname{tp}(\bar c / A,\bar b_i)$ is satisfiable, and since $\mathfrak C$ is a monster model, it is satisfiable in $\mathfrak C$. By Infinite Ramsey's Theorem and indiscernibility of $\{\bar b_i : i \in I\}$ 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 $\displaystyle \bigcup_{i\in I} \operatorname{tp}(\bar c / A, \bar b_i)$, and :for any $i_1 < \cdots < i_k$ and $j_1 < \cdots < j_k$ in $I$, $\mathfrak C \models \phi(b_{i_1}, \dots, b_{i_k}, \bar c_\Delta) \leftrightarrow  \phi(b_{j_1}, \dots, b_{j_k}, \bar c_\Delta)$ for all $\phi \in \Delta$. Since $b$ is one of the $b_i$, we have that each $c_\Delta$ realizes $\operatorname{tp}(\bar c / A, \bar b)$. By compactness, since there is a $c_\Delta$ with $\mathfrak C \models \phi(b_{i_1}, \dots, b_{i_k}, \bar c_\Delta) \leftrightarrow  \phi(b_{j_1}, \dots, b_{j_k}, \bar c_\Delta)$ for all  $\phi \in \Delta$ for each finite $\Delta$, there is a $\bar c'$ with type $\operatorname{tp}(\bar c / A, \bar b)$ and which satisfies $\mathfrak C \models \phi(b_{i_1}, \dots, b_{i_k}, \bar c') \leftrightarrow \phi(b_{j_1}, \dots, b_{j_k}, \bar c')$ for all $\phi$. But this holding for every $\phi$ means that $\{b_i : i\in I\}$ is order-indiscernible over $A,\bar c'$. $(2)\implies (1)$: We prove this case by contradiction. Suppose $\operatorname{tp}(\bar c / A,\bar{b})$ divides over $A$. By definition, this means that $\operatorname{tp}(\bar c / A,\bar{b})$ implies some $\phi(x, \bar{b})$ with parameters from $A$ which divides over $A$. By definition of dividing, this means that for some $k \in \N$ there is a sequence $(\bar{b}_i)_{i\in \N}$  such that $\operatorname{tp}(\bar{b}_i /  A) = \operatorname{tp}(\bar{b} / A)$ for each $i\in\N$, and for any distinct $k$-many terms $\bar b_{i_1}, \dots, \bar b_{i_k}$ of  the sequence, the set $\{\phi(\bar{x}, \bar{b}_{i_1}), \dots,  \phi(\bar{x}, \bar{b}_{i_k})\}$ is not satisfiable in $\mathfrak C$. Since $\operatorname{tp}(\bar{b}_i /  A) = \operatorname{tp}(\bar{b} / A)$ for each $i$, we can assume (using an argument involving Infinite Ramsey's Theorem and partitions based on the formula satisfied by the $b_i$) that $\{\bar b_i : i\in I\}$ is order-indiscernible over $A$. Then, by $(2)$, there is $\bar c'$ with $\operatorname{tp}(\bar c' / A,\bar b) = \operatorname{tp}(\bar c / A,\bar b)$ such that $\{\bar b_i : i\in I\}$ is order-indiscernible over $A,\bar c'$. However, $\operatorname{tp}(\bar c' / A,\bar b) = \operatorname{tp}(\bar c / A,\bar b)$ gives us that $\mathfrak C \models \phi(\bar c', \bar{b})$. So, $\{\bar b_i : i\in I\}$ being order-indiscernible over $A,\bar c'$ gives us that $\mathfrak C \models \phi(\bar c', \bar{b_i})$ for all $i$. This contradicts the non-satisfiability of $\{\phi(\bar{x}, \bar{b}_{i_1}), \dots,  \phi(\bar{x}, \bar{b}_{i_k})\}$. Thus, $\operatorname{tp}(\bar c /  A,\bar{b})$ does not divide over $A$. $(2)\implies (3)$: Since $c$ and $c'$ have the same type over $A,b$, the function $\bar c' \mapsto \bar c$ is partial elementary. Since $\mathfrak C$ is homogeneous, there is an $A, b$-automorphism $f$ which extends $\bar c' \mapsto \bar c$. Let $\{f(\bar b_i) : i \in I\}$ be the image of $\{\bar b_i : i \in I\}$. $\{f(\bar b_i) : i \in I\}$ is order-indiscernible over $A,\bar c$ since if $i_1 < \cdots < i_k$ and $j_1 < \cdots < j_k$, then: {{begin-eqn}} {{eqn | l = \mathfrak C \models \phi(f(\bar b_{i_1}), \dots, f(\bar b_{i_k}), \bar c)       | o = \leadstoandfrom       | r = \mathfrak C \models \phi(\bar b_{i_1}, \dots, \bar b_{i_k}, \bar c')       | c = as $f$ is an automorphism (fixing any parameters in $\phi$ from $A, b$) }} {{eqn | o = \leadstoandfrom       | r = \mathfrak C \models \phi(\bar b_{j_1}, \dots, \bar b_{j_k}, \bar c')       | c = as the original sequence is order-indiscernible over $A, \bar c'$ }} {{eqn | o = \leadstoandfrom       | r = \mathfrak C \models \phi(f(\bar b_{j_1}), \dots, f(\bar b_{j_k}), \bar c)       | c = as $f$ is an automorphism }}  {{end-eqn}} $(3)\implies (2)$: Let $\bar c' = f^{-1}(\bar c)$. $\{\bar b_i : i \in I\}$ is order-indiscernible over $A, \bar c'$ since if $i_1 < \cdots < i_k$ and $j_1 < \cdots < j_k$, then: {{begin-eqn}} {{eqn | l = \mathfrak C \models \phi(\bar b_{i_1}, \dots, \bar b_{i_k}, \bar c')       | o = \leadstoandfrom       | r = \mathfrak C \models \phi(f(\bar b_{i_1}), \dots, f(\bar b_{i_k}), \bar c)       | c = as $f$ is an automorphism (fixing any parameters in $\phi$ from $A, b$) }} {{eqn | o = \leadstoandfrom       | r = \mathfrak C \models \phi(f(\bar b_{j_1}), \dots, f(\bar b_{j_k}), \bar c)       | c = as the image sequence is order-indiscernible over $A, \bar c$. }} {{eqn | o = \leadstoandfrom       | r = \mathfrak C \models \phi(\bar b_{j_1}, \dots, \bar b_{j_k}, \bar c')       | c = as $f$ is an automorphism }}  {{end-eqn}} {{qed}} Category:Model Theory i7s6qjbnrrzd6csyj45oh1fvllrn4wh