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












0












$begingroup$


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: $$M models exists x (varphi(x)) text{ if and only if }Tvdash exists x (varphi(x)). $$ 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?







first-order-logic model-theory







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$endgroup$












  • $begingroup$
    Complete theories? For any incomplete theory let $phi$ be an undecidable sentence and $psi$ be $philand x=x.$
    $endgroup$
    – 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






  • 3




    $begingroup$
    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
















0












$begingroup$


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: $$M models exists x (varphi(x)) text{ if and only if }Tvdash exists x (varphi(x)). $$ 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?







first-order-logic model-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Complete theories? For any incomplete theory let $phi$ be an undecidable sentence and $psi$ be $philand x=x.$
    $endgroup$
    – 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






  • 3




    $begingroup$
    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














0












0








0





$begingroup$


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: $$M models exists x (varphi(x)) text{ if and only if }Tvdash exists x (varphi(x)). $$ 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?







first-order-logic model-theory







share|cite|improve this question











$endgroup$




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: $$M models exists x (varphi(x)) text{ if and only if }Tvdash exists x (varphi(x)). $$ 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?







first-order-logic model-theory




first-order-logic model-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 23:19









Alex Kruckman

26.9k22556




26.9k22556










asked Jan 1 at 23:02









ZuhairZuhair

256112




256112












  • $begingroup$
    Complete theories? For any incomplete theory let $phi$ be an undecidable sentence and $psi$ be $philand x=x.$
    $endgroup$
    – 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






  • 3




    $begingroup$
    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


















  • $begingroup$
    Complete theories? For any incomplete theory let $phi$ be an undecidable sentence and $psi$ be $philand x=x.$
    $endgroup$
    – 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






  • 3




    $begingroup$
    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
















$begingroup$
Complete theories? For any incomplete theory let $phi$ be an undecidable sentence and $psi$ be $philand x=x.$
$endgroup$
– spaceisdarkgreen
Jan 1 at 23:15




$begingroup$
Complete theories? For any incomplete theory let $phi$ be an undecidable sentence and $psi$ be $philand x=x.$
$endgroup$
– 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




$begingroup$
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




$begingroup$
@spaceisdarkgreen I see we had the exact same idea! :)
$endgroup$
– Alex Kruckman
Jan 1 at 23:17




3




3




$begingroup$
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




$begingroup$
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




$begingroup$
@AlexKruckman, nothing special, just to emphasize on them. Thanks for the answer.
$endgroup$
– Zuhair
Jan 2 at 17:14










1 Answer
1






active

oldest

votes


















6












$begingroup$

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 $Mmodels T$, we have $Mmodels psi$ if and only if $Tvdash psi$.



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 $(psiland (x = x))$. Then $exists x, widehat{psi}(x)$ is logically equivalent to $psi$.



Now if $Mmodels psi$, then $Mmodels exists x, widehat{psi}(x)$, so $Tvdash exists x, widehat{psi}(x)$, so $Tvdash psi$. Since either $Mmodels psi$ or $Mmodels lnot psi$, we have $Tvdash psi$ or $Tvdash lnot psi$. 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. )






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    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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1 Answer
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6












$begingroup$

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 $Mmodels T$, we have $Mmodels psi$ if and only if $Tvdash psi$.



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 $(psiland (x = x))$. Then $exists x, widehat{psi}(x)$ is logically equivalent to $psi$.



Now if $Mmodels psi$, then $Mmodels exists x, widehat{psi}(x)$, so $Tvdash exists x, widehat{psi}(x)$, so $Tvdash psi$. Since either $Mmodels psi$ or $Mmodels lnot psi$, we have $Tvdash psi$ or $Tvdash lnot psi$. 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. )






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    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
















6












$begingroup$

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 $Mmodels T$, we have $Mmodels psi$ if and only if $Tvdash psi$.



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 $(psiland (x = x))$. Then $exists x, widehat{psi}(x)$ is logically equivalent to $psi$.



Now if $Mmodels psi$, then $Mmodels exists x, widehat{psi}(x)$, so $Tvdash exists x, widehat{psi}(x)$, so $Tvdash psi$. Since either $Mmodels psi$ or $Mmodels lnot psi$, we have $Tvdash psi$ or $Tvdash lnot psi$. 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. )






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    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














6












6








6





$begingroup$

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 $Mmodels T$, we have $Mmodels psi$ if and only if $Tvdash psi$.



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 $(psiland (x = x))$. Then $exists x, widehat{psi}(x)$ is logically equivalent to $psi$.



Now if $Mmodels psi$, then $Mmodels exists x, widehat{psi}(x)$, so $Tvdash exists x, widehat{psi}(x)$, so $Tvdash psi$. Since either $Mmodels psi$ or $Mmodels lnot psi$, we have $Tvdash psi$ or $Tvdash lnot psi$. 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. )






share|cite|improve this answer











$endgroup$



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 $Mmodels T$, we have $Mmodels psi$ if and only if $Tvdash psi$.



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 $(psiland (x = x))$. Then $exists x, widehat{psi}(x)$ is logically equivalent to $psi$.



Now if $Mmodels psi$, then $Mmodels exists x, widehat{psi}(x)$, so $Tvdash exists x, widehat{psi}(x)$, so $Tvdash psi$. Since either $Mmodels psi$ or $Mmodels lnot psi$, we have $Tvdash psi$ or $Tvdash lnot psi$. 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. )







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 0:16

























answered Jan 1 at 23:14









Alex KruckmanAlex Kruckman

26.9k22556




26.9k22556








  • 1




    $begingroup$
    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














  • 1




    $begingroup$
    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








1




1




$begingroup$
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




$begingroup$
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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