Why are there no chains of forking extensions in the random graph? I am trying to show that the theory of the random graph does not have SU-rank bigger than 1. By the definition of SU-rank from the book of Tent and Ziegler, which I am using, this amounts to showing that there is not sequence of types $p \subset q \subset r$ such that $q$ is a forking extension of $p$ and $r$ is a forking extension of $q$. As far as I can see in Tent and Ziegler, and also in another definition of the SU-rank I checked out, there is not requirement that the types in question must be $1$-types. Now I do not understand why the following is not a counterexample (I know that the SU-rank of the random graph should be 1):
Let $a_0$ and $a_1$ be vertices in (the monster model of) the random graph. We can choose them such that $a_0Ea_1$, but that does not really matter I think. Now let $p := \text{tp}(a_0 a_1 / \emptyset)$, $q:= \text{tp}(a_0 a_1 / a_0)$, $r:= \text{tp}(a_0 a_1 / a_0 a_1)$.
Then $q$ contains the formula $x_0 = a_0$, which it seems to me divides over the empty set: $a_1$ has infinitely many neighbours $(a_0^j \,|\, j \in \omega)$, and since the theory of the random graph has quantifier elimination, we have $\text{tp}(a_0a_1) = \text{tp}(a_0^ja_1)$ for all $j$. So the sequence $(a_0^j a_1 \,|\, j \in \omega)$ witnesses the dividing. (Obviously the set $\left\{ x_0 = a_0^j \,|\, j \in \omega \right\}$ is 2-inconsistent).
In the same vein, $r$ contains the formula $x_1 = a_1$, which it seems divides over $a_0$: $a_0$ has infinitely many neighbours $(a_1^j \,|\, j \in \omega)$, and again by quantifier elimination we have $\text{tp}(a_0 a_1^j / a_0) = \text{tp}(a_0 a_1 / a_0)$ - I do not see that the type $\text{tp}(a_0 a_1 / a_0)$ can express anything about $a_1$ that $\text{tp}(a_0 a_1)$ cannot express as well. So the sequence $(a_0 a_1^j \,|\, j \in \omega)$ witnesses the dividing.
This would establish that $q$ is a forking extension of $p$ and that $r$ is a forking extension of $q$ and that the SU-rank must be $\geq 2$. Where is the flaw in my argument?
 A: The SU-rank is defined first for complete types, and then for formulas, by $$\mathrm{SU}(\varphi(\overline{x})) = \sup\{\mathrm{SU}(p)\mid p\text{ a complete type in the variables }\overline{x},\text{with }\varphi(\overline{x})\in p\}.$$
For any tuple of variables $\overline{x}$, there is a tautological formula $\top(\overline{x})$. Most people write this formula as $\bigwedge_{i=1}^n x_i = x_i$ (I prefer to include $\top$ as an atomic logical connective, but that's neither here nor there). Now there are two reasonable ways to define the SU-rank of a theory $T$. You could set $\mathrm{SU}(T) = \mathrm{SU}(\top(x))$, or you could set $\mathrm{SU}(T) = \sup_{n\in \omega} \mathrm{SU}(\top(x_1,\dots,x_n))$. In my experience, most people mean the former when they talk about the rank of a theory, though I've also seen the latter. 
Here's one reason to prefer the former definition: if $T_1$ has $\mathrm{SU}(\top(x)) = 1$ and $T_2$ has $\mathrm{SU}(\top(x)) = 2$, then the former definition will tell them apart, while according to the latter definition, $\mathrm{SU}(T_1) = \mathrm{SU}(T_2) = \omega$. 
You've shown that in the theory of the random graph, $\mathrm{SU}(p) = 2$, so $\mathrm{SU}(\top(x_1,x_2)) \geq 2$. In fact it's true that $\mathrm{SU}(\top(x_1,x_2)) = 2$. But in the single variable context, $\mathrm{SU}(\top(x)) = 1$. This is what people mean when they say the random graph has SU-rank $1$.
