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Definition. We say that $\mathcal{M} \models T$ is existentially closed if whenever $\mathcal{N} \models T$, $\mathcal{M}\subseteq \mathcal{N}$, and $\mathcal{N}\models \exists x \phi(x, a)$, where $a \in M$ and $\phi$ is quantifier-free,then $\mathcal{M} \models ∃x φ(x, a)$.

Definition. If $\mathcal{M}$ and $\mathcal{N}$ are $\mathcal{L}$-structures, then we say that $\mathcal{M}$ is an elementary substructure of $\mathcal{N}$ and write $\mathcal{M}\prec \mathcal{N}$ if

$$\mathcal{M} \models \phi(\bar{a}) \Leftrightarrow \mathcal{N} \models \phi(\bar{a})$$

for all $\mathcal{L}$-formulas $\phi(\bar{x})$ and all $\bar{a}\in M$.

Tarski-Vaught test. Suppose that $\mathcal{M}$ is a substructure of $\mathcal{N}$. Then, $\mathcal{M}$ is an elementary substructure if and only if, for any formula $\phi(v,w)$ and $a \in M$, if there is $b \in N$ such that $\mathcal{N} \models \phi(b, a)$, then there is $c \in M$ such that $\mathcal{M} \models \phi(c, a)$.

Question. Does Tarski-Vaught test imply that $\mathcal{M}$ is existentially closed model of $T$ if it is an elementary substructure of any model of $T$ containing $\mathcal{M}$? If yes, why do model theorist use different names/words?

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    $\begingroup$ The definition of "existentially closed" refers only to quantifier-free formulas $\phi$, whereas the Tarski-Vaught test is about arbitrary formulas. $\endgroup$ – Andreas Blass Nov 11 '18 at 1:18

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