Existential common-substructure test for quantifier elimination

For a first-order theory $T$, suppose every existential quantifier-free formula over a common substructure has a witness in either of two $T$-models exactly when it has one in the other. Then $T$ has quantifier elimination. One may test a single quantified variable, since elimination of these formulas inductively eliminates all quantifiers. A language with a constant avoids empty-generated-substructure and quantifier-free-sentence conventions.

If $T$ has algebraically prime models and every inclusion between models of $T$ is a simple closure, then $T$ has quantifier elimination. For a common base $A$, embed its algebraically prime extension into both models. Simple closure transfers a witness into that extension; the second embedding transfers it to the other model.