Stationary types are almost orthogonal if and only if the realizers are independent Model theory Wiki (https://modeltheory.fandom.com/wiki/Orthogonal_types_(stability_theory)) states that stationary types are almost orthogonal if and only if the realizers are independent.  It does not provide a proof, which it says is "easy."  I attempted to do this myself just by the axioms of the independence relation, but my proof remains incomplete.  I have Pillay's Introduction to Stability Theory and Tent and Zigler's A Course in Model Theory, but I don't think they contain a proof of this fact.
How can one prove this?  Alternatively: is there a source from which to learn the proof?
 A: Let $p(x)$ and $q(y)$ be types over $C$. Assume that $p(x)$ is stationary and that for all $a$ realizing $p(x)$ and $b$ realizing $q(y)$, we have $a \downarrow_C b$ (here I'm using $\downarrow$ in place of the independence anchor). We would like to show that $p(x)$ and $q(y)$ are almost orthogonal, which means that there is a unique complete type extending $p(x)\cup q(y)$.
So let $ab$ and $a'b'$ be two realizations of $p(x)\cup q(y)$. Since $\text{tp}(b/C) = \text{tp}(b'/C) = q(y)$, there is some automorphism $\sigma\in \mathrm{Aut}(\mathcal{U}/C)$ with $\sigma(b') = b$. Let $a'' = \sigma(a')$. By our hypothesis, $a\downarrow_C b$ and $a'\downarrow_C b'$, so by invariance of $\downarrow_C$, we have $a''\downarrow_C b$. Further, $\text{tp}(a/C) = \text{tp}(a'/C) = \text{tp}(a''/C) = p(x)$, and since $p(x)$ is stationary, $\text{tp}(a/Cb) = \text{tp}(a''/Cb)$. Hence $\text{tp}(ab/C) = \text{tp}(a''b/C)$. We have $\sigma^{-1}(a''b) = a'b'$, so  $\text{tp}(ab/C) = \text{tp}(a'b'/C)$. We've shown that any two realizations of $p(x)\cup q(y)$ have the same type over $C$, so $p(x)\cup q(y)$ has a unique completion.
Note that we only needed to assume that $p(x)$ was stationary, $q(y)$ can be an arbitrary type. And the proof works just as well (using symmetry of $\downarrow$) if only $q(y)$ is assumed to be stationary.

Regarding a reference: You're right that Pillay doesn't prove this in Introduction to Stability Theory, but he does state it twice (Exercise 8.45 and Note 9.30), leaving the proof to the reader. The statement also appears as Exercise 10.2.2 in Tent & Ziegler and as Exercise VI.1.27 in Baldwin's Fundamentals of Stability Theory. It seems the consensus among authors is that the proof is a good exercise!
