For any first order theory $T$, let's say that $M$ is an ideal existential model of $T$ if and only if, for every formula $\varphi(x)$ in the language of $T$ in which only the symbol "$x$" appears free, we have: $$M \models \exists x (\varphi(x)) \text{ if and only if }T\vdash \exists x (\varphi(x)). $$ In English, there exists $x$ fulfilling $\varphi(x)$ in the domain of the model $M$ of $T$ if and only if $T$ proves the existence of $x$ for which $\varphi(x)$ holds.

Question: Do ideal existential models exist? And if so, then can they exist for [A]ny first order theory? If only [S]ome, then which theories can have such models?

  • $\begingroup$ Complete theories? For any incomplete theory let $\phi$ be an undecidable sentence and $\psi$ be $\phi\land x=x.$ $\endgroup$ – spaceisdarkgreen Jan 1 at 23:15
  • $\begingroup$ Why the close vote? The question is fairly trivial, but it's certainly not "off-topic". $\endgroup$ – Alex Kruckman Jan 1 at 23:16
  • $\begingroup$ @spaceisdarkgreen I see we had the exact same idea! :) $\endgroup$ – Alex Kruckman Jan 1 at 23:17
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    $\begingroup$ Zuhair: Just out of curiosity, why did you put brackets around A and S in [A]ny and [S]ome? $\endgroup$ – Alex Kruckman Jan 1 at 23:22
  • $\begingroup$ @AlexKruckman, nothing special, just to emphasize on them. Thanks for the answer. $\endgroup$ – Zuhair Jan 2 at 17:14

A theory $T$ has an ideal existential model if and only if $T$ is complete (and if $T$ is complete, then every model is ideal existential!).

Indeed, if $T$ is complete, then for any sentence $\psi$ and any $M\models T$, we have $M\models \psi$ if and only if $T\vdash \psi$.

Conversely, suppose $M$ is an ideal existential model of $T$. Note that for any sentence $\psi$, we can let $x$ be a variable which doesn't occur in $\psi$, and let $\widehat{\psi}(x)$ be the formula $(\psi\land (x = x))$. Then $\exists x\, \widehat{\psi}(x)$ is logically equivalent to $\psi$.

Now if $M\models \psi$, then $M\models \exists x\, \widehat{\psi}(x)$, so $T\vdash \exists x\, \widehat{\psi}(x)$, so $T\vdash \psi$. Since either $M\models \psi$ or $M\models \lnot \psi$, we have $T\vdash \psi$ or $T\vdash \lnot \psi$. Hence $T$ is complete.

(In this answer, I have ignored empty models - if your version of first-order logic allows empty models, then some details above have to be adjusted slightly, since $\psi$ and $\exists x\, \widehat{\psi}(x)$ are only logically equivalent in non-empty models. See Eric Wofsey's comment below. )

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    $\begingroup$ And of course if you do allow empty models, empty models are always ideal existential models. So $T$ has an ideal existential model iff either it is complete or has an empty model. $\endgroup$ – Eric Wofsey Jan 1 at 23:19

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