# Is nfcp equivalent to stable + eliminates $\exists^\infty$?

Let $T$ be a complete first-order theory. Recall that a formula $\phi(\overline{x},\overline{y})$ has the finite cover property (fcp) if for all $n$, there exist $\overline{a}_1,\dots,\overline{a}_n$ such that $$T\models \lnot\exists \overline{x} \bigwedge_{i = 1}^n \phi(\overline{x},\overline{a}_i),$$ but for all $1\leq j \leq n$, $$T\models \exists \overline{x} \bigwedge_{i \neq j} \phi(\overline{x},\overline{a}_i).$$

As usual, a theory has nfcp if no formula has fcp. This condition was introduced by Keisler (in the context of Keisler's order) and studied by Shelah in Classification Theory, where it is shown that nfcp implies stability.

It is also easy to see that if $T$ has nfcp, then $T$ eliminates the quantifer $\exists^\infty$ (there exist infinitely many). That is, for every formula $\phi(\overline{x},\overline{y})$, there is a natural number $n$ such that for all $\overline{a}$, if $|\phi(M,\overline{a})|>n$, then $|\phi(M,\overline{a})|$ is infinite.

Now I have heard that there is a converse to these two facts. In particular, a partial converse (with the additional assumption of elimination of imaginaries) is stated as Theorem 2.8 in this paper, without proof or a reference:

Theorem: If $T=T^{eq}$ is stable and eliminates $\exists^\infty$, then $T$ has nfcp.

Can anyone provide a proof or a reference? The issue doesn't appear to be addressed in Classification Theory.

Assertion (8) exactly says that $T^{eq}$ does not eliminate $\exists^{\infty}$ and the theorem says that for stable $T$, this is equivalent to having the f.c.p.