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Jun 17, 2019 at 15:25 comment added Emil Jeřábek $\let\fk\mathfrak$... and embeddings $f_n\colon A(\fk M_n)\to A(\fk N_{n+1})$, $g_{n+1}\colon B(\fk N_{n+1})\to B(\fk M_{n+1})$ such that $f_0\let\sset\subseteq\sset f_1\sset f_2\sset\dots$ and $g_1\sset g_2\sset g_3\sset\dots$, and we take $\fk M_\infty$ and $\fk N_\infty$ to be the limits of the chains. The embedings will actually have to preserve $A\exists B\forall$ formulas in a suitable way to make the inductive construction to go through. This will be kind of a mess to set it up properly, but I think there should not be any substantial difficulty.
Jun 17, 2019 at 15:22 comment added Emil Jeřábek $\let\fk\mathfrak$I believe a direct argument proving both Q1 and Q2 for arbitrary languages might go as follows. It is easy to see that it suffices to show the following: if $\fk M_0\Rightarrow_{A\exists B\forall}\fk N_0$, there exist $\fk M_\infty\succeq\fk M_0$ and $\fk N_\infty\succeq\fk N_0$ such that $\fk N_\infty$ is an $A$-super-$B$-substructure of $\fk M_\infty$. Then, similar to common proofs of Robinson’s joint consistency theorem, you build elementary chains $\fk M_0\preceq\fk M_1\preceq\fk M_2\preceq\dots$ and $\fk N_0\preceq\fk N_1\preceq\fk N_2\preceq\dots$, ...
Jun 17, 2019 at 15:12 comment added Emil Jeřábek Yes, you are right, this should work.
Jun 17, 2019 at 15:04 vote accept James E Hanson
Jun 17, 2019 at 15:04 comment added James E Hanson Thank you. For an uncountable language can't you run the same argument in every countable reduct containing the symbols in the sentence $\varphi$ and then take an ultraproduct of the resulting pairs of structures to get a witness that $\varphi$ is not preserved in models of $T$?
Jun 17, 2019 at 13:15 history edited Emil Jeřábek CC BY-SA 4.0
stupid typo
Jun 17, 2019 at 12:23 history edited Emil Jeřábek CC BY-SA 4.0
added 151 characters in body
Jun 17, 2019 at 12:14 history answered Emil Jeřábek CC BY-SA 4.0