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Mar 6, 2014 at 17:11 comment added Alex Kruckman Exactly. I just rephrased it because I wanted to emphasize that the nice property of superstable theories (nonforking always descends to a finite base) is true for stable theories if you only care about a single (or finitely many) formulas.
Mar 6, 2014 at 11:33 comment added TimZ Thanks, in my notation take the $\phi$-type $tp^\phi(a_0/Aa_1)$. It is definable over some finite set $A_0$ hence non-forking over $A_0$. Now any non-forking extension of this $\phi$-type does the job, we choose a Morley sequence $/A_0$ in one of this nf-extension.
Mar 6, 2014 at 11:27 vote accept TimZ
Mar 3, 2014 at 3:41 history answered Alex Kruckman CC BY-SA 3.0