In order to prove (the non trivial part of) Chang-Los-Suszko's theorem [1], I'm struggling with the following lemma :

Lemma. Let $T$ be a $\mathcal L$-theory and $T_{\forall\exists}$ the set of $\forall\exists$-sentences consequences of $T$. Then for every model $\mathfrak M \models T_{\forall\exists}$, there exists $\mathfrak N \models T$ such that $\mathfrak M \subseteq \mathfrak N$ and where the inclusion is essentially closed.

I recall here that an inclusion $\mathfrak M \subseteq \mathfrak N$ is essentially closed if for every quantifier-free formula $\varphi(\bar x, \bar y)$, and for every tuple $\bar m$ of elements of $M$, one has $$ \mathfrak N \models \exists \bar x \varphi(\bar x,\bar m) \quad \text{implies} \quad \mathfrak M \models \exists \bar x \varphi(\bar x,\bar m) .$$

My thought was to get $\mathfrak N$ by showing that the $\mathcal L_M$-theory $$ T \cup \Delta(\mathfrak M) \cup \{ (\exists \bar x\varphi(\bar x,\bar m)) \to \varphi(\bar a_{\varphi(\bar x, \bar m)},\bar m) : \varphi \ \text{quantifier-free}, \bar m \in M \} $$ is consistent, where $\Delta(\mathfrak M)$ is the simple diagram of $\mathfrak M$, and where $\bar a_{\varphi(\bar x, \bar m)}$ is a choice of a tuple of elements of $M$ such that $\mathfrak M \models \varphi(\bar a_{\varphi(\bar x,\bar m)},\bar m)$ if it exists and else is arbitrary. And to do so, of course I would use compactness and only show that the $$ T \cup \{ \psi(\bar m) \wedge \bigwedge_{i=1}^n\left((\exists \bar x\varphi_i(\bar x,\bar m_i)) \to \varphi(\bar a_{\varphi(\bar x, \bar m_i)},\bar m_i) \right)\} $$ are consistent (with $\psi$ and the $\varphi_i$ quantifier-free). Here I'm kind of stuck…

I want to express the latter formula (between the braces) as some $\exists \bar z \forall \bar t \chi(\bar z,\bar t)$ with $\chi$ quantifier-free : then $\mathfrak M$ will satisfy it and so prevent $T$ from showing its negation, yielding a model $\mathfrak N$ of $T$ satisfying $\exists \bar z \forall \bar t \chi(\bar z,\bar t)$. CQFD.

It does not seem far from reach but I just can't compute that $\chi$. I hope someone can give me a hint (or tell me that I'm completely off).

[1] For those who wonder, Chang-Los-Suszko's theorem states that a theory $T$ is inductive (i.e. its class of models is closed under taking increasing union) if and only if $T$ admits a $\forall\exists$-axiomatization.


Hint: Let's slightly modify the construction. Show that $$T \cup \Delta(M) \cup \{\forall x \phi(x, \bar m) : M \models \forall x \phi(x, \bar m), \phi \text { quantifier free}\}$$ is consistent.

  • $\begingroup$ Thanks, I think I got it now. Just to be sure, can you check it ? Should one make conjunctions, one can assume a finite fragment of your theory to be included in $T \cup \{\psi(\bar m') \wedge \forall \bar x \varphi(\bar x,\bar m)$ with $M$ a model of the latter formula. Then $M \models \exists \bar y \, \bar y' \forall \bar x (\psi(\bar y') \wedge \varphi(\bar x, \bar y))$, preventing $\forall \bar y \, \bar y' \exists \bar x (\neg \psi(\bar y') \vee \neg \varphi(\bar x, \bar y))$ to be consequence of $T$, so there is a model of our finite fragment. […] $\endgroup$ – Pece Feb 4 '14 at 18:07
  • $\begingroup$ […] Hence there is a model of your theory and such a model $N$ contains $M$ as a substructure (because model $\Delta(M)$) and if $N \models \exists \bar x \varphi(\bar x, \bar m)$ then $N \not \models \forall \bar x \neg \varphi(\bar x, \bar m)$ and so necessarily $M \not \models \forall \bar x \neg \varphi(\bar x, \bar m)$ either (by the latter part of your theory). Which conclude that $N$ is an eventually closed extension of $M$. $\endgroup$ – Pece Feb 4 '14 at 18:10
  • $\begingroup$ That is exactly how you proceed. $\endgroup$ – Levon Haykazyan Feb 4 '14 at 18:17
  • $\begingroup$ Ok, so my mistake was to force a witness in $M$ for every witness in $N$, while you make sure there is no witness in $N$ when there is no in $M$. Thank you again. $\endgroup$ – Pece Feb 4 '14 at 18:23

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