Let $T$ be a stable $L$-theory with elimination of imaginaries. We work in the monster model $\mathfrak C$ of $T$. Let $A$ be a small (infinite) set of the monster, $\phi(x,y)$ be a $L(A)$-formula and $a\_i:i\in \omega$ be a Morley sequence over $A$ such that $\mathfrak C\models \phi(a\_0,a\_1)$.


Can we find a Morley sequence $b\_i:i\in \omega$ over a finite set $A\_0$ such that $\mathfrak C\models \phi(b\_0,b\_1)$? 


Note:This is of course true for a superstable theory and if one replaces $\phi$ by a type then this fails in non-superstable theories. 



Replace Morley sequence over some set by one over a finite set, s.t. they both satiesfy a certain formula
Your question comes down to asking whether given a formula $\phi(x,b)$ which does not fork over $A$, there is a finite subset of $A$, $A\_0$, such that $\phi(x,b)$ does not fork over $A\_0$. 


The answer is yes. The reason depends on your preferred definition of forking, of course. Using the definability of types definition, we just have to take $a$ independent of $b$ over $A$ and take $A\_0$ to be the parameters appearing in the $\phi$-definition of $\text{tp}^\phi(a/Ab)$.


