A sentence is called existential if it's of the form $\exists x_1 \cdots \exists x_n \varphi(x_1, \cdots, x_n)$, where $\varphi$ is quantifier-free formula.
I'm trying to prove a lemma left as an exercise in my lecture notes that says
Let $C$ be an axiomatizable class. Then the following conditions are equivalent: (i) $C$ is $\exists$-axiomatizable; (ii) If $A \in C$ and $A \leq B$ then $B \in C$.
One direction is simple - if $C$ is $\exists$-axiomatizable then it follows easily by the fact that if $A \models \varphi(a_1, \cdots , a_n)$ for an existential formula $\varphi(v_1, \cdots , v_n)$, then $B \models \varphi(a_1, \cdots , a_n)$.
I'm really not sure how to go about proving the other direction.
I've proved the equivalent version for universal axiomatisation - i.e. that $C$ is $\forall$-axiomatizable iff $B \in C$ and $A \leq B$ then $A \leq C.$ To do that, I showed that Th$(C) \cup $Th$_∃(A)$ is finitely satisfiable (where $A\models$ Th$_\forall$(C) - i.e. $A$ is a model of the universal sentences in the theory of $C$) where Th$_∃(A)$ are the existential sentences of the theory of $A$. It followed from that, and some other results I have, that Th$(C)\cup$Diag$(A)$ was satisfiable, from which it followed there is a model $B$ of Th$(C)$ such that $A \leq B$, which meant by the assumption that $A\in C$ and hence (since $A\models$ Th$_\forall$(C)) that $C$ was universally axiomatisable.
I'm really unsure how I'd go about taking a similar approach for the existential case - I haven't proved any similar results like the one that took me from Th$(C)\cup$Diag$(A)$ being satisfiable to there being a $B$ like I have here, and even if I did I'm not sure how I'd apply a similar kind of result - since in this case I have to start with a smaller model $A$ instead of a bigger one.
Any advice or suggestions you could offer would be much appreciated.
from Hot Weekly Questions - Mathematics Stack Exchange
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