# Can we have models that do not satisfy unproved existential sentences in the theories they model?

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?

• Complete theories? For any incomplete theory let $\phi$ be an undecidable sentence and $\psi$ be $\phi\land x=x.$ – spaceisdarkgreen Jan 1 at 23:15
• Why the close vote? The question is fairly trivial, but it's certainly not "off-topic". – Alex Kruckman Jan 1 at 23:16
• @spaceisdarkgreen I see we had the exact same idea! :) – Alex Kruckman Jan 1 at 23:17
• Zuhair: Just out of curiosity, why did you put brackets around A and S in [A]ny and [S]ome? – Alex Kruckman Jan 1 at 23:22
• @AlexKruckman, nothing special, just to emphasize on them. Thanks for the answer. – 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. )
• 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. – Eric Wofsey Jan 1 at 23:19