Forking in simple theories A complete first-order theory $T$ is said to be simple (supersimple) if each type does not fork over some subset $A$ of its domain where $|A|\leq |T|$ ($|A|<\aleph_0$).
Define an abstract independence relation as follows:  $a\downarrow_A b$ iff $tp(a/Ab)$ does not divide over $A$.
I am trying to prove the following statements:


*

*The theory of random graph is simple (actually supersimple).

*If $G$ is the generic model of random graph  and $A,B,C\subset G$, then $A\downarrow_C B$ iff $A\cap B \subseteq C$. 


In particular (2) implies for $a,b\in G$ and $A\subset G$ we have $a\downarrow_A b$ iff either $a\neq b$ or $a=b\in A$. This result is so strange for me because if $aRb$ (i.e. $a$ and $b$ are adjacent) then they are independent!!
I would appreciate if you give a proof with details.
 A: A motivating "philosophy" of simple theories is: $$\text{simple} = \text{stable}+\text{random}$$
One way of making this philosophy precise is the stable forking conjecture (which remains a major open problem). This conjecture states that in a simple theory, forking is always witnessed by a stable formula. Precisely, if a type $p(x)\in S_x(B)$ forks over a subset $A\subseteq B$, then there is a stable formula $\varphi(x,y)$ such that $\varphi(x,b)\in p(x)$ and $\varphi(x,b)$ forks over $A$.
So the idea here is that a simple theory has some "stable part", controlled by the stable formulas, which completely determines forking, plus some "random part" on top, which doesn't have any impact on forking. 
The main difference in the theory of forking between stable and simple theories is that forking is stationary in stable theories: at least over $\text{acl}^{\text{eq}}$-closed sets, non-forking extensions are unique. That is, if $A$ and $B$ are independent over $C$, then we know exactly how $A$ and $B$ interact. In a simple theory, we lose stationary. Intuitively, this is because we can see arbitrary interactions between $A$ and $B$ in the "random part" without forking. 
In the case of the random graph, the "stable part" is trivial (just the theory of an infinite set in the language $\{=\}$), and the "random part" is the edge relation. The stable forking conjecuture holds here because (as we will see below), the only instances of forking come from positive instances of equality, while the edge relation has no impact on forking. 

Let $T$ be the theory of the random graph. The strategy is to first characterize forking in models of $T$ (question 2), and then conclude that $T$ is supersimple (question 1). I'll assume you know $T$ has quantifier elimination. We'll also use the fact that for any subset $A$ of a model $M\models T$, $\text{acl}(A) = A$. These properties are easy to show directly; more abstractly, they follow from the observation that the class of finite graphs has the disjoint (a.k.a. strong) amalgamation property, and $T$ is the theory of the Fraïssé limit of this class.  


*

*One direction of 2 is easy: If $A\cap B\not\subseteq C$, then $\text{tp}(A/BC)$ contains a formula $x = b$ where $b\in (B\setminus \text{acl}(C))$, using the fact that $\text{acl}(C) = C$. Such a formula always divides (and hence forks) over $C$. Indeed, since $b\notin \text{acl}(C)$, we can find an infinite sequence $(b_i)_{i\in \omega}$ of distinct realizations of $\text{tp}(b/C)$, and the set $\{x = b_i\mid i\in \omega\}$ is $2$-inconsistent. 

*Conversely, suppose $A\cap B \subseteq C$, and we want to show that $p(x) = \text{tp}(A/BC)$ does not divide over $C$. By replacing $B$ with $B\cup C$, we may assume $C\subseteq B$. Let $(B_i)_{i\in \omega}$ be an indiscernible sequence over $C$ with $B_0 = B$. Let $p_i(x)$ the type obtained by replacing the parameters from $B$ in $p(x)$ with the corresponding parameters from $B_i$. It suffices to show that the partial type $\bigcup_{i\in \omega} p_i(x)$ is consistent. The intuitive idea here is that by quantifier elimination, this partial type just describes a graph with domain $A'\cup \bigcup_{i\in \omega} B_i$ (where $A'$ is a set realizing the partial type), and since we're in the theory of the random graph, this is consistent. But there are some things to check here... I can write out a more careful argument, if you're not able to work it out for yourself.

*In fact, if $A\cap B \subseteq C$, then $\text{tp}(A/BC)$ does not fork over $C$ either. Proof: If $\text{tp}(A/BC)$ forks over $C$, then it implies a finite disjunction of formulas (possibly with parameters outside $BC$) which divide over $C$. Let $D$ be the finite set of parameters appearing in this disjunction. Then since $C = \text{acl}(C)$, we can find some realization $A'$ of $\text{tp}(A/BC)$ such that $A'\cap D\subseteq C$. But then $\text{tp}(A'/D)$ does not divide over $C$, contradicting the fact that it must contain one of the finitely many formulas in the disjunction.

*We've characterized forking and dividing in models of $T$ (and we've shown that forking = dividing). Having established this, it's easy to see that $T$ is supersimple. Let $p(x)\in S_x(B)$ be a type over $B$ in a finite tuple of variables $x$. For each variable $x_i$ in $x$, there is at most one element $b\in B$ such that $x_i = b$ is in $p(x)$. Let $C$ be the set of all $b\in B$ such that $x_i = b$ is in $p(x)$, for some $x_i$ in $x$. Note that $C$ is finite, since $x$ is a finite tuple. And $p(x)$ does not fork over $C$, since for any realization $a$ of $p(x)$, letting $A$ be the set of all elements in the tuple $a$, we have $A\cap B = C$.  

Another way to approach your questions is by the Kim-Pillay characterization of forking in simple theories (Theorem 4.2 in Simple Theories). In this approach, we define a notion of independence between subsets of models of $T$ by $$A\downarrow_C B \text{ iff } A\cap B\subseteq C$$
Then we check that it satisfies the six axioms in Definition 4.1 of the Kim-Pillay paper. The conclusion is that $T$ is simple and $\downarrow = \downarrow^f$, i.e. the independence relation we defined is forking independence. So we don't have to do any of the work characterizing dividing from my answer above. Of course, to conclude that $T$ is supersimple, we still need strengthen local character (axiom 2) as in the argument in the last bullet point above. 
Often, the Kim-Pillay theorem is the most efficient way to show that a theory is simple and to understand forking in that theory. In the random graph, it's already easy to characterize dividing directly, so I think both approaches are just as good.
