A theory $T$ is model-complete if the union of $T$ with an atomic diagram is complete












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$begingroup$


Let $T$ be a theory in first order logic over some language $L$. Let $mathfrak A$ be some structure over $L$ with $mathfrak A models T$ and with $A$ be its universe. Then consider every $a in A$ as a constant and look at the enriched language $L(A) = L cup A$ with the $L(A)$-structure $mathfrak A_A = (mathfrak A, a)_{ain A}$. A formula over $L(A)$ is called basic if it is an atomic formula. The set
$$
operatorname{Diag}(mathfrak A) = { varphi mbox{ is a basic $L(A)$-sentence } mid mathfrak A_A models varphi }
$$

is called the atomic diagram of $mathfrak A$.



A theory $T$ is called model-complete if every substructure relation between two models is actually an elementary embedding.




Then $T$ is model-complete if and only if for any $mathfrak A models T$ the theory $T cup operatorname{Diag}(mathfrak A)$ is complete.




These definitions are from A course in model theory by K.Tent/M.Zeigler.



I do not understand the quoted statement. For let $t_1 = t_2$ and $t_3 ne t_4$ be two atomic sentences for terms $t_1, t_2,t_3,t_4$ in $L(A)$. Then Set $varphi = (t_1 = t_2) land (t_3 ne t_4)$. Now suppose in the terms we have some constants from $A$. Then neither $varphi$ nor $neg varphi$ is in $T cup operatorname{Diag}(mathfrak A)$ as it is not in $operatorname{Diag}(mathfrak A)$ for it is not atomic, nor is it in $T$ as it is a statement over the enrichted language $L(A)$, but not over $L$. Could someone please explain the above statement (and what I oversee here...)?










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$endgroup$

















    0












    $begingroup$


    Let $T$ be a theory in first order logic over some language $L$. Let $mathfrak A$ be some structure over $L$ with $mathfrak A models T$ and with $A$ be its universe. Then consider every $a in A$ as a constant and look at the enriched language $L(A) = L cup A$ with the $L(A)$-structure $mathfrak A_A = (mathfrak A, a)_{ain A}$. A formula over $L(A)$ is called basic if it is an atomic formula. The set
    $$
    operatorname{Diag}(mathfrak A) = { varphi mbox{ is a basic $L(A)$-sentence } mid mathfrak A_A models varphi }
    $$

    is called the atomic diagram of $mathfrak A$.



    A theory $T$ is called model-complete if every substructure relation between two models is actually an elementary embedding.




    Then $T$ is model-complete if and only if for any $mathfrak A models T$ the theory $T cup operatorname{Diag}(mathfrak A)$ is complete.




    These definitions are from A course in model theory by K.Tent/M.Zeigler.



    I do not understand the quoted statement. For let $t_1 = t_2$ and $t_3 ne t_4$ be two atomic sentences for terms $t_1, t_2,t_3,t_4$ in $L(A)$. Then Set $varphi = (t_1 = t_2) land (t_3 ne t_4)$. Now suppose in the terms we have some constants from $A$. Then neither $varphi$ nor $neg varphi$ is in $T cup operatorname{Diag}(mathfrak A)$ as it is not in $operatorname{Diag}(mathfrak A)$ for it is not atomic, nor is it in $T$ as it is a statement over the enrichted language $L(A)$, but not over $L$. Could someone please explain the above statement (and what I oversee here...)?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $T$ be a theory in first order logic over some language $L$. Let $mathfrak A$ be some structure over $L$ with $mathfrak A models T$ and with $A$ be its universe. Then consider every $a in A$ as a constant and look at the enriched language $L(A) = L cup A$ with the $L(A)$-structure $mathfrak A_A = (mathfrak A, a)_{ain A}$. A formula over $L(A)$ is called basic if it is an atomic formula. The set
      $$
      operatorname{Diag}(mathfrak A) = { varphi mbox{ is a basic $L(A)$-sentence } mid mathfrak A_A models varphi }
      $$

      is called the atomic diagram of $mathfrak A$.



      A theory $T$ is called model-complete if every substructure relation between two models is actually an elementary embedding.




      Then $T$ is model-complete if and only if for any $mathfrak A models T$ the theory $T cup operatorname{Diag}(mathfrak A)$ is complete.




      These definitions are from A course in model theory by K.Tent/M.Zeigler.



      I do not understand the quoted statement. For let $t_1 = t_2$ and $t_3 ne t_4$ be two atomic sentences for terms $t_1, t_2,t_3,t_4$ in $L(A)$. Then Set $varphi = (t_1 = t_2) land (t_3 ne t_4)$. Now suppose in the terms we have some constants from $A$. Then neither $varphi$ nor $neg varphi$ is in $T cup operatorname{Diag}(mathfrak A)$ as it is not in $operatorname{Diag}(mathfrak A)$ for it is not atomic, nor is it in $T$ as it is a statement over the enrichted language $L(A)$, but not over $L$. Could someone please explain the above statement (and what I oversee here...)?










      share|cite|improve this question









      $endgroup$




      Let $T$ be a theory in first order logic over some language $L$. Let $mathfrak A$ be some structure over $L$ with $mathfrak A models T$ and with $A$ be its universe. Then consider every $a in A$ as a constant and look at the enriched language $L(A) = L cup A$ with the $L(A)$-structure $mathfrak A_A = (mathfrak A, a)_{ain A}$. A formula over $L(A)$ is called basic if it is an atomic formula. The set
      $$
      operatorname{Diag}(mathfrak A) = { varphi mbox{ is a basic $L(A)$-sentence } mid mathfrak A_A models varphi }
      $$

      is called the atomic diagram of $mathfrak A$.



      A theory $T$ is called model-complete if every substructure relation between two models is actually an elementary embedding.




      Then $T$ is model-complete if and only if for any $mathfrak A models T$ the theory $T cup operatorname{Diag}(mathfrak A)$ is complete.




      These definitions are from A course in model theory by K.Tent/M.Zeigler.



      I do not understand the quoted statement. For let $t_1 = t_2$ and $t_3 ne t_4$ be two atomic sentences for terms $t_1, t_2,t_3,t_4$ in $L(A)$. Then Set $varphi = (t_1 = t_2) land (t_3 ne t_4)$. Now suppose in the terms we have some constants from $A$. Then neither $varphi$ nor $neg varphi$ is in $T cup operatorname{Diag}(mathfrak A)$ as it is not in $operatorname{Diag}(mathfrak A)$ for it is not atomic, nor is it in $T$ as it is a statement over the enrichted language $L(A)$, but not over $L$. Could someone please explain the above statement (and what I oversee here...)?







      logic first-order-logic model-theory






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      asked Dec 14 '18 at 17:34









      StefanHStefanH

      8,12152366




      8,12152366






















          2 Answers
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          2












          $begingroup$

          Complete means for any sentence $varphi,$ either $Tvdash varphi$ or $Tvdash lnot varphi,$ not $varphiin T$ or $lnotvarphiin T.$



          If $mathfrak A,mathfrak Bmodels T$ and $mathfrak Asubseteq mathfrak B,$ then both are models of $Tcup Diag(mathfrak A).$ If $Tcup Diag(mathfrak A)$ is complete, then they must agree on all $L(A)$-sentences, and hence the embedding is elementary. On the other hand if $Tcup Diag(mathfrak A)$ is not complete for some $mathfrak Amodels T,$ then we can find a $mathfrak Bmodels T$ with $mathfrak Asubseteq mathfrak B$ that differs from $mathfrak A$ on some $L(A)$-sentence (just let it be a model of the negation of some $varphi$ that is true in $mathfrak A$ but that $Tcup Diag(mathfrak A)$ does not decide). Hence this embedding is not elementary.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What makes you sure that such a model $mathfrak B$ of $T cup Diag(mathfrak A) cup {neg varphi}$ exists that contains $mathfrak A$ as an $L$-structure?
            $endgroup$
            – StefanH
            Dec 15 '18 at 17:58












          • $begingroup$
            @StefanH The whole point of the diagram of $A$ is that any model of it contains an isomorphic copy of $A$ as a substructure (via the interpretation of the constant symbols).
            $endgroup$
            – Alex Kruckman
            Dec 15 '18 at 18:12










          • $begingroup$
            Got it, just needs additionally that completeness of a theory is equivalent with the property that all models are elementary equivalent, then surely non-completeness lets us choose such a model.
            $endgroup$
            – StefanH
            Dec 16 '18 at 21:09





















          4












          $begingroup$

          You have two errors here:




          1. "$T$ is complete" means that for every sentence $varphi$, $Tmodels varphi$ or $Tmodels lnotvarphi$. It does not mean that $varphiin T$ or $lnot varphiin T$.


          2. A basic formula is not just an atomic formula. It is an atomic formula or a negation of an atomic formula.



          In your example, if $mathfrak{A}models varphi$, then $t_1 = t_2in text{Diag}(mathfrak{A})$ and $t_3neq t_4in text{Diag}(mathfrak{A})$, so $Tcup text{Diag}(mathfrak{A})models varphi$. Otherwise, if $mathfrak{A}notmodels varphi$, then either $t_1neq t_2in text{Diag}(mathfrak{A})$ or $t_3 = t_4in text{Diag}(mathfrak{A})$, and in either case $Tcup text{Diag}(mathfrak{A})models lnot varphi$.



          It's an easy exercise to show that for any structure $mathfrak{A}$ and any quantifier-free $L(A)$-sentence $varphi$, either $text{Diag}(mathfrak{A})models varphi$ or $text{Diag}(mathfrak{A})models lnot varphi$. The interesting fact is that $T$ is model complete if and only if $Tcuptext{Diag}(mathfrak{A})$ decides the truth of all $L(A)$-formulas, even those with quantifiers. The answer by spaceisdarkgreen explains the easy proof of this fact.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Guess you mean $T cup operatorname{Diag}(mathfrak A)$ decides the truth of all $L(A)$-formulas, not just $operatorname{Diag}(mathfrak A)$?
            $endgroup$
            – StefanH
            Dec 15 '18 at 17:52






          • 1




            $begingroup$
            @StefanH "All extensions of $mathfrak A$ are elementary" means exactly the same thing as "$Diag(mathfrak A)$ is complete." If this holds and $mathfrak Amodels T,$ then all the extensions are models of $T$ as well. "$T$ is model complete" means all models of $T$ have this property.
            $endgroup$
            – spaceisdarkgreen
            Dec 15 '18 at 23:54










          • $begingroup$
            @StefanH Yes, that was a typo! I misunderstood what you were referring to at first. Fixed.
            $endgroup$
            – Alex Kruckman
            Dec 16 '18 at 1:52











          Your Answer





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          2 Answers
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          2 Answers
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          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          2












          $begingroup$

          Complete means for any sentence $varphi,$ either $Tvdash varphi$ or $Tvdash lnot varphi,$ not $varphiin T$ or $lnotvarphiin T.$



          If $mathfrak A,mathfrak Bmodels T$ and $mathfrak Asubseteq mathfrak B,$ then both are models of $Tcup Diag(mathfrak A).$ If $Tcup Diag(mathfrak A)$ is complete, then they must agree on all $L(A)$-sentences, and hence the embedding is elementary. On the other hand if $Tcup Diag(mathfrak A)$ is not complete for some $mathfrak Amodels T,$ then we can find a $mathfrak Bmodels T$ with $mathfrak Asubseteq mathfrak B$ that differs from $mathfrak A$ on some $L(A)$-sentence (just let it be a model of the negation of some $varphi$ that is true in $mathfrak A$ but that $Tcup Diag(mathfrak A)$ does not decide). Hence this embedding is not elementary.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What makes you sure that such a model $mathfrak B$ of $T cup Diag(mathfrak A) cup {neg varphi}$ exists that contains $mathfrak A$ as an $L$-structure?
            $endgroup$
            – StefanH
            Dec 15 '18 at 17:58












          • $begingroup$
            @StefanH The whole point of the diagram of $A$ is that any model of it contains an isomorphic copy of $A$ as a substructure (via the interpretation of the constant symbols).
            $endgroup$
            – Alex Kruckman
            Dec 15 '18 at 18:12










          • $begingroup$
            Got it, just needs additionally that completeness of a theory is equivalent with the property that all models are elementary equivalent, then surely non-completeness lets us choose such a model.
            $endgroup$
            – StefanH
            Dec 16 '18 at 21:09


















          2












          $begingroup$

          Complete means for any sentence $varphi,$ either $Tvdash varphi$ or $Tvdash lnot varphi,$ not $varphiin T$ or $lnotvarphiin T.$



          If $mathfrak A,mathfrak Bmodels T$ and $mathfrak Asubseteq mathfrak B,$ then both are models of $Tcup Diag(mathfrak A).$ If $Tcup Diag(mathfrak A)$ is complete, then they must agree on all $L(A)$-sentences, and hence the embedding is elementary. On the other hand if $Tcup Diag(mathfrak A)$ is not complete for some $mathfrak Amodels T,$ then we can find a $mathfrak Bmodels T$ with $mathfrak Asubseteq mathfrak B$ that differs from $mathfrak A$ on some $L(A)$-sentence (just let it be a model of the negation of some $varphi$ that is true in $mathfrak A$ but that $Tcup Diag(mathfrak A)$ does not decide). Hence this embedding is not elementary.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What makes you sure that such a model $mathfrak B$ of $T cup Diag(mathfrak A) cup {neg varphi}$ exists that contains $mathfrak A$ as an $L$-structure?
            $endgroup$
            – StefanH
            Dec 15 '18 at 17:58












          • $begingroup$
            @StefanH The whole point of the diagram of $A$ is that any model of it contains an isomorphic copy of $A$ as a substructure (via the interpretation of the constant symbols).
            $endgroup$
            – Alex Kruckman
            Dec 15 '18 at 18:12










          • $begingroup$
            Got it, just needs additionally that completeness of a theory is equivalent with the property that all models are elementary equivalent, then surely non-completeness lets us choose such a model.
            $endgroup$
            – StefanH
            Dec 16 '18 at 21:09
















          2












          2








          2





          $begingroup$

          Complete means for any sentence $varphi,$ either $Tvdash varphi$ or $Tvdash lnot varphi,$ not $varphiin T$ or $lnotvarphiin T.$



          If $mathfrak A,mathfrak Bmodels T$ and $mathfrak Asubseteq mathfrak B,$ then both are models of $Tcup Diag(mathfrak A).$ If $Tcup Diag(mathfrak A)$ is complete, then they must agree on all $L(A)$-sentences, and hence the embedding is elementary. On the other hand if $Tcup Diag(mathfrak A)$ is not complete for some $mathfrak Amodels T,$ then we can find a $mathfrak Bmodels T$ with $mathfrak Asubseteq mathfrak B$ that differs from $mathfrak A$ on some $L(A)$-sentence (just let it be a model of the negation of some $varphi$ that is true in $mathfrak A$ but that $Tcup Diag(mathfrak A)$ does not decide). Hence this embedding is not elementary.






          share|cite|improve this answer









          $endgroup$



          Complete means for any sentence $varphi,$ either $Tvdash varphi$ or $Tvdash lnot varphi,$ not $varphiin T$ or $lnotvarphiin T.$



          If $mathfrak A,mathfrak Bmodels T$ and $mathfrak Asubseteq mathfrak B,$ then both are models of $Tcup Diag(mathfrak A).$ If $Tcup Diag(mathfrak A)$ is complete, then they must agree on all $L(A)$-sentences, and hence the embedding is elementary. On the other hand if $Tcup Diag(mathfrak A)$ is not complete for some $mathfrak Amodels T,$ then we can find a $mathfrak Bmodels T$ with $mathfrak Asubseteq mathfrak B$ that differs from $mathfrak A$ on some $L(A)$-sentence (just let it be a model of the negation of some $varphi$ that is true in $mathfrak A$ but that $Tcup Diag(mathfrak A)$ does not decide). Hence this embedding is not elementary.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 18:36









          spaceisdarkgreenspaceisdarkgreen

          33k21753




          33k21753












          • $begingroup$
            What makes you sure that such a model $mathfrak B$ of $T cup Diag(mathfrak A) cup {neg varphi}$ exists that contains $mathfrak A$ as an $L$-structure?
            $endgroup$
            – StefanH
            Dec 15 '18 at 17:58












          • $begingroup$
            @StefanH The whole point of the diagram of $A$ is that any model of it contains an isomorphic copy of $A$ as a substructure (via the interpretation of the constant symbols).
            $endgroup$
            – Alex Kruckman
            Dec 15 '18 at 18:12










          • $begingroup$
            Got it, just needs additionally that completeness of a theory is equivalent with the property that all models are elementary equivalent, then surely non-completeness lets us choose such a model.
            $endgroup$
            – StefanH
            Dec 16 '18 at 21:09




















          • $begingroup$
            What makes you sure that such a model $mathfrak B$ of $T cup Diag(mathfrak A) cup {neg varphi}$ exists that contains $mathfrak A$ as an $L$-structure?
            $endgroup$
            – StefanH
            Dec 15 '18 at 17:58












          • $begingroup$
            @StefanH The whole point of the diagram of $A$ is that any model of it contains an isomorphic copy of $A$ as a substructure (via the interpretation of the constant symbols).
            $endgroup$
            – Alex Kruckman
            Dec 15 '18 at 18:12










          • $begingroup$
            Got it, just needs additionally that completeness of a theory is equivalent with the property that all models are elementary equivalent, then surely non-completeness lets us choose such a model.
            $endgroup$
            – StefanH
            Dec 16 '18 at 21:09


















          $begingroup$
          What makes you sure that such a model $mathfrak B$ of $T cup Diag(mathfrak A) cup {neg varphi}$ exists that contains $mathfrak A$ as an $L$-structure?
          $endgroup$
          – StefanH
          Dec 15 '18 at 17:58






          $begingroup$
          What makes you sure that such a model $mathfrak B$ of $T cup Diag(mathfrak A) cup {neg varphi}$ exists that contains $mathfrak A$ as an $L$-structure?
          $endgroup$
          – StefanH
          Dec 15 '18 at 17:58














          $begingroup$
          @StefanH The whole point of the diagram of $A$ is that any model of it contains an isomorphic copy of $A$ as a substructure (via the interpretation of the constant symbols).
          $endgroup$
          – Alex Kruckman
          Dec 15 '18 at 18:12




          $begingroup$
          @StefanH The whole point of the diagram of $A$ is that any model of it contains an isomorphic copy of $A$ as a substructure (via the interpretation of the constant symbols).
          $endgroup$
          – Alex Kruckman
          Dec 15 '18 at 18:12












          $begingroup$
          Got it, just needs additionally that completeness of a theory is equivalent with the property that all models are elementary equivalent, then surely non-completeness lets us choose such a model.
          $endgroup$
          – StefanH
          Dec 16 '18 at 21:09






          $begingroup$
          Got it, just needs additionally that completeness of a theory is equivalent with the property that all models are elementary equivalent, then surely non-completeness lets us choose such a model.
          $endgroup$
          – StefanH
          Dec 16 '18 at 21:09













          4












          $begingroup$

          You have two errors here:




          1. "$T$ is complete" means that for every sentence $varphi$, $Tmodels varphi$ or $Tmodels lnotvarphi$. It does not mean that $varphiin T$ or $lnot varphiin T$.


          2. A basic formula is not just an atomic formula. It is an atomic formula or a negation of an atomic formula.



          In your example, if $mathfrak{A}models varphi$, then $t_1 = t_2in text{Diag}(mathfrak{A})$ and $t_3neq t_4in text{Diag}(mathfrak{A})$, so $Tcup text{Diag}(mathfrak{A})models varphi$. Otherwise, if $mathfrak{A}notmodels varphi$, then either $t_1neq t_2in text{Diag}(mathfrak{A})$ or $t_3 = t_4in text{Diag}(mathfrak{A})$, and in either case $Tcup text{Diag}(mathfrak{A})models lnot varphi$.



          It's an easy exercise to show that for any structure $mathfrak{A}$ and any quantifier-free $L(A)$-sentence $varphi$, either $text{Diag}(mathfrak{A})models varphi$ or $text{Diag}(mathfrak{A})models lnot varphi$. The interesting fact is that $T$ is model complete if and only if $Tcuptext{Diag}(mathfrak{A})$ decides the truth of all $L(A)$-formulas, even those with quantifiers. The answer by spaceisdarkgreen explains the easy proof of this fact.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Guess you mean $T cup operatorname{Diag}(mathfrak A)$ decides the truth of all $L(A)$-formulas, not just $operatorname{Diag}(mathfrak A)$?
            $endgroup$
            – StefanH
            Dec 15 '18 at 17:52






          • 1




            $begingroup$
            @StefanH "All extensions of $mathfrak A$ are elementary" means exactly the same thing as "$Diag(mathfrak A)$ is complete." If this holds and $mathfrak Amodels T,$ then all the extensions are models of $T$ as well. "$T$ is model complete" means all models of $T$ have this property.
            $endgroup$
            – spaceisdarkgreen
            Dec 15 '18 at 23:54










          • $begingroup$
            @StefanH Yes, that was a typo! I misunderstood what you were referring to at first. Fixed.
            $endgroup$
            – Alex Kruckman
            Dec 16 '18 at 1:52
















          4












          $begingroup$

          You have two errors here:




          1. "$T$ is complete" means that for every sentence $varphi$, $Tmodels varphi$ or $Tmodels lnotvarphi$. It does not mean that $varphiin T$ or $lnot varphiin T$.


          2. A basic formula is not just an atomic formula. It is an atomic formula or a negation of an atomic formula.



          In your example, if $mathfrak{A}models varphi$, then $t_1 = t_2in text{Diag}(mathfrak{A})$ and $t_3neq t_4in text{Diag}(mathfrak{A})$, so $Tcup text{Diag}(mathfrak{A})models varphi$. Otherwise, if $mathfrak{A}notmodels varphi$, then either $t_1neq t_2in text{Diag}(mathfrak{A})$ or $t_3 = t_4in text{Diag}(mathfrak{A})$, and in either case $Tcup text{Diag}(mathfrak{A})models lnot varphi$.



          It's an easy exercise to show that for any structure $mathfrak{A}$ and any quantifier-free $L(A)$-sentence $varphi$, either $text{Diag}(mathfrak{A})models varphi$ or $text{Diag}(mathfrak{A})models lnot varphi$. The interesting fact is that $T$ is model complete if and only if $Tcuptext{Diag}(mathfrak{A})$ decides the truth of all $L(A)$-formulas, even those with quantifiers. The answer by spaceisdarkgreen explains the easy proof of this fact.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Guess you mean $T cup operatorname{Diag}(mathfrak A)$ decides the truth of all $L(A)$-formulas, not just $operatorname{Diag}(mathfrak A)$?
            $endgroup$
            – StefanH
            Dec 15 '18 at 17:52






          • 1




            $begingroup$
            @StefanH "All extensions of $mathfrak A$ are elementary" means exactly the same thing as "$Diag(mathfrak A)$ is complete." If this holds and $mathfrak Amodels T,$ then all the extensions are models of $T$ as well. "$T$ is model complete" means all models of $T$ have this property.
            $endgroup$
            – spaceisdarkgreen
            Dec 15 '18 at 23:54










          • $begingroup$
            @StefanH Yes, that was a typo! I misunderstood what you were referring to at first. Fixed.
            $endgroup$
            – Alex Kruckman
            Dec 16 '18 at 1:52














          4












          4








          4





          $begingroup$

          You have two errors here:




          1. "$T$ is complete" means that for every sentence $varphi$, $Tmodels varphi$ or $Tmodels lnotvarphi$. It does not mean that $varphiin T$ or $lnot varphiin T$.


          2. A basic formula is not just an atomic formula. It is an atomic formula or a negation of an atomic formula.



          In your example, if $mathfrak{A}models varphi$, then $t_1 = t_2in text{Diag}(mathfrak{A})$ and $t_3neq t_4in text{Diag}(mathfrak{A})$, so $Tcup text{Diag}(mathfrak{A})models varphi$. Otherwise, if $mathfrak{A}notmodels varphi$, then either $t_1neq t_2in text{Diag}(mathfrak{A})$ or $t_3 = t_4in text{Diag}(mathfrak{A})$, and in either case $Tcup text{Diag}(mathfrak{A})models lnot varphi$.



          It's an easy exercise to show that for any structure $mathfrak{A}$ and any quantifier-free $L(A)$-sentence $varphi$, either $text{Diag}(mathfrak{A})models varphi$ or $text{Diag}(mathfrak{A})models lnot varphi$. The interesting fact is that $T$ is model complete if and only if $Tcuptext{Diag}(mathfrak{A})$ decides the truth of all $L(A)$-formulas, even those with quantifiers. The answer by spaceisdarkgreen explains the easy proof of this fact.






          share|cite|improve this answer











          $endgroup$



          You have two errors here:




          1. "$T$ is complete" means that for every sentence $varphi$, $Tmodels varphi$ or $Tmodels lnotvarphi$. It does not mean that $varphiin T$ or $lnot varphiin T$.


          2. A basic formula is not just an atomic formula. It is an atomic formula or a negation of an atomic formula.



          In your example, if $mathfrak{A}models varphi$, then $t_1 = t_2in text{Diag}(mathfrak{A})$ and $t_3neq t_4in text{Diag}(mathfrak{A})$, so $Tcup text{Diag}(mathfrak{A})models varphi$. Otherwise, if $mathfrak{A}notmodels varphi$, then either $t_1neq t_2in text{Diag}(mathfrak{A})$ or $t_3 = t_4in text{Diag}(mathfrak{A})$, and in either case $Tcup text{Diag}(mathfrak{A})models lnot varphi$.



          It's an easy exercise to show that for any structure $mathfrak{A}$ and any quantifier-free $L(A)$-sentence $varphi$, either $text{Diag}(mathfrak{A})models varphi$ or $text{Diag}(mathfrak{A})models lnot varphi$. The interesting fact is that $T$ is model complete if and only if $Tcuptext{Diag}(mathfrak{A})$ decides the truth of all $L(A)$-formulas, even those with quantifiers. The answer by spaceisdarkgreen explains the easy proof of this fact.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 16 '18 at 1:51

























          answered Dec 14 '18 at 18:52









          Alex KruckmanAlex Kruckman

          27.4k32657




          27.4k32657












          • $begingroup$
            Guess you mean $T cup operatorname{Diag}(mathfrak A)$ decides the truth of all $L(A)$-formulas, not just $operatorname{Diag}(mathfrak A)$?
            $endgroup$
            – StefanH
            Dec 15 '18 at 17:52






          • 1




            $begingroup$
            @StefanH "All extensions of $mathfrak A$ are elementary" means exactly the same thing as "$Diag(mathfrak A)$ is complete." If this holds and $mathfrak Amodels T,$ then all the extensions are models of $T$ as well. "$T$ is model complete" means all models of $T$ have this property.
            $endgroup$
            – spaceisdarkgreen
            Dec 15 '18 at 23:54










          • $begingroup$
            @StefanH Yes, that was a typo! I misunderstood what you were referring to at first. Fixed.
            $endgroup$
            – Alex Kruckman
            Dec 16 '18 at 1:52


















          • $begingroup$
            Guess you mean $T cup operatorname{Diag}(mathfrak A)$ decides the truth of all $L(A)$-formulas, not just $operatorname{Diag}(mathfrak A)$?
            $endgroup$
            – StefanH
            Dec 15 '18 at 17:52






          • 1




            $begingroup$
            @StefanH "All extensions of $mathfrak A$ are elementary" means exactly the same thing as "$Diag(mathfrak A)$ is complete." If this holds and $mathfrak Amodels T,$ then all the extensions are models of $T$ as well. "$T$ is model complete" means all models of $T$ have this property.
            $endgroup$
            – spaceisdarkgreen
            Dec 15 '18 at 23:54










          • $begingroup$
            @StefanH Yes, that was a typo! I misunderstood what you were referring to at first. Fixed.
            $endgroup$
            – Alex Kruckman
            Dec 16 '18 at 1:52
















          $begingroup$
          Guess you mean $T cup operatorname{Diag}(mathfrak A)$ decides the truth of all $L(A)$-formulas, not just $operatorname{Diag}(mathfrak A)$?
          $endgroup$
          – StefanH
          Dec 15 '18 at 17:52




          $begingroup$
          Guess you mean $T cup operatorname{Diag}(mathfrak A)$ decides the truth of all $L(A)$-formulas, not just $operatorname{Diag}(mathfrak A)$?
          $endgroup$
          – StefanH
          Dec 15 '18 at 17:52




          1




          1




          $begingroup$
          @StefanH "All extensions of $mathfrak A$ are elementary" means exactly the same thing as "$Diag(mathfrak A)$ is complete." If this holds and $mathfrak Amodels T,$ then all the extensions are models of $T$ as well. "$T$ is model complete" means all models of $T$ have this property.
          $endgroup$
          – spaceisdarkgreen
          Dec 15 '18 at 23:54




          $begingroup$
          @StefanH "All extensions of $mathfrak A$ are elementary" means exactly the same thing as "$Diag(mathfrak A)$ is complete." If this holds and $mathfrak Amodels T,$ then all the extensions are models of $T$ as well. "$T$ is model complete" means all models of $T$ have this property.
          $endgroup$
          – spaceisdarkgreen
          Dec 15 '18 at 23:54












          $begingroup$
          @StefanH Yes, that was a typo! I misunderstood what you were referring to at first. Fixed.
          $endgroup$
          – Alex Kruckman
          Dec 16 '18 at 1:52




          $begingroup$
          @StefanH Yes, that was a typo! I misunderstood what you were referring to at first. Fixed.
          $endgroup$
          – Alex Kruckman
          Dec 16 '18 at 1:52


















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