$exists^infty$-elimination and model companion.











up vote
2
down vote

favorite
2












In the book Ziegler, Tent: A course in model theory it states




Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.




I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise




Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
$T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.




I define the theory



$(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$



To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?










share|cite|improve this question


























    up vote
    2
    down vote

    favorite
    2












    In the book Ziegler, Tent: A course in model theory it states




    Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
    languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.




    I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise




    Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
    $T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
    elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.




    I define the theory



    $(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
    elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$



    To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?










    share|cite|improve this question
























      up vote
      2
      down vote

      favorite
      2









      up vote
      2
      down vote

      favorite
      2






      2





      In the book Ziegler, Tent: A course in model theory it states




      Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
      languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.




      I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise




      Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
      $T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
      elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.




      I define the theory



      $(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
      elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$



      To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?










      share|cite|improve this question













      In the book Ziegler, Tent: A course in model theory it states




      Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
      languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.




      I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise




      Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
      $T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
      elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.




      I define the theory



      $(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
      elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$



      To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?







      model-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 14 at 6:10









      Zikrunumea

      438




      438






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          2
          down vote



          accepted
          +50










          Since $T_1$ and $T_2$ are model complete, they are $forallexists$-axiomatizable, and $T_1cup T_2$ is also $forallexists$-axiomatizable. So to show the model companion exists, you need to axiomatize the existentially closed models of $T_1cup T_2$.



          So suppose $Msubseteq M'$, both models of $T_1cup T_2$. We want to write down a sufficient condition for $M$ to be existentially closed in $M'$. Let $psi(x)$ be a quantifier-free formula with parameters from $M$ such that $M'models exists x, psi(x)$. You want to observe the following things:




          1. It suffices to assume that $psi(x)$ is a conjunction of atomic and negated atomic $(L_1cup L_2)$-formulas.


          2. So if $L_1$ and $L_2$ are relational, $psi(x)$ is actually a conjunction $varphi_1(x)land varphi_2(x)$, where $varphi_1$ is an $L_1$-formula and $varphi_2$ is an $L_2$-formula. If the languages aren't relational, there's an issue that $psi$ could mention terms formed using function symbols from both $L_1$ and $L_2$. Then you have to think about replacing $psi$ with another formula obtained by "unnesting" all terms. It might be better to work out the details in the relational case first and then come back to this complication.


          3. It suffices to assume that there is a witness $M'models psi(a')$ such that each element of $a'$ is in $M'setminus M$ and the elements of $a'$ are all distinct.



          This leads us to the following axiomatization: Let $varphi_1(x,y)$ be a quantifier-free $L_1$-formula, and let $varphi_2(x,z)$ be a quantifier-free $L_2$-formula (here $x$, $y$, $z$ are tuples of variables). Let $theta_1(y)$ and $theta_2(z)$ be the formulas provided by Exercise 5.5.6. for $varphi_1$ and $varphi_2$, respectively. Let $theta'_1(y)$ be the conjunction of $theta_1(y)$ and inequations $y_ineq y_j$ for $ineq j$, and similarly for $theta'_2(z)$. Then look at the following sentence: $$forall y, forall z, ((theta'_1(y)land theta'_2(z))rightarrow exists x, (varphi_1(x,y)land varphi_2(x,z))).$$



          The model companion of $T_1cup T_2$ is axiomatized by $T_1cup T_2$ together with all sentences of the above form. The discussion above can be viewed as an extended hint that every model of this theory is an existentially closed model of $T_1cup T_2$. You also need to show the converse: that every existentially closed model of $T_1cup T_2$ satisfies these extra axioms. Explicitly, given $varphi_1(x,y)$ and $varphi_2(x,z)$, if $Mmodels theta_1'(b)land theta_2'(c)$, then you can embed $M$ in a model $M'$ of $T_1cup T_2$ such that $M'models varphi_1(a',b)land varphi_2(a',c)$ for some $a'in M'$, and use the fact that $M$ is existentially closed to find a witness in $M$.



          Aside: The statement of the exercise is a result from Peter Winkler's 1975 PhD thesis. It's a nice coincidence that you posted this problem now. Minh Tran, Erik Walsberg, and I are working on a project that we call "Interpolative Fusions" which at its heart is a generalization of this result of Winkler's. We just posted the first paper from this project to the arXiv this week: https://arxiv.org/abs/1811.06108






          share|cite|improve this answer























            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997862%2fexists-infty-elimination-and-model-companion%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted
            +50










            Since $T_1$ and $T_2$ are model complete, they are $forallexists$-axiomatizable, and $T_1cup T_2$ is also $forallexists$-axiomatizable. So to show the model companion exists, you need to axiomatize the existentially closed models of $T_1cup T_2$.



            So suppose $Msubseteq M'$, both models of $T_1cup T_2$. We want to write down a sufficient condition for $M$ to be existentially closed in $M'$. Let $psi(x)$ be a quantifier-free formula with parameters from $M$ such that $M'models exists x, psi(x)$. You want to observe the following things:




            1. It suffices to assume that $psi(x)$ is a conjunction of atomic and negated atomic $(L_1cup L_2)$-formulas.


            2. So if $L_1$ and $L_2$ are relational, $psi(x)$ is actually a conjunction $varphi_1(x)land varphi_2(x)$, where $varphi_1$ is an $L_1$-formula and $varphi_2$ is an $L_2$-formula. If the languages aren't relational, there's an issue that $psi$ could mention terms formed using function symbols from both $L_1$ and $L_2$. Then you have to think about replacing $psi$ with another formula obtained by "unnesting" all terms. It might be better to work out the details in the relational case first and then come back to this complication.


            3. It suffices to assume that there is a witness $M'models psi(a')$ such that each element of $a'$ is in $M'setminus M$ and the elements of $a'$ are all distinct.



            This leads us to the following axiomatization: Let $varphi_1(x,y)$ be a quantifier-free $L_1$-formula, and let $varphi_2(x,z)$ be a quantifier-free $L_2$-formula (here $x$, $y$, $z$ are tuples of variables). Let $theta_1(y)$ and $theta_2(z)$ be the formulas provided by Exercise 5.5.6. for $varphi_1$ and $varphi_2$, respectively. Let $theta'_1(y)$ be the conjunction of $theta_1(y)$ and inequations $y_ineq y_j$ for $ineq j$, and similarly for $theta'_2(z)$. Then look at the following sentence: $$forall y, forall z, ((theta'_1(y)land theta'_2(z))rightarrow exists x, (varphi_1(x,y)land varphi_2(x,z))).$$



            The model companion of $T_1cup T_2$ is axiomatized by $T_1cup T_2$ together with all sentences of the above form. The discussion above can be viewed as an extended hint that every model of this theory is an existentially closed model of $T_1cup T_2$. You also need to show the converse: that every existentially closed model of $T_1cup T_2$ satisfies these extra axioms. Explicitly, given $varphi_1(x,y)$ and $varphi_2(x,z)$, if $Mmodels theta_1'(b)land theta_2'(c)$, then you can embed $M$ in a model $M'$ of $T_1cup T_2$ such that $M'models varphi_1(a',b)land varphi_2(a',c)$ for some $a'in M'$, and use the fact that $M$ is existentially closed to find a witness in $M$.



            Aside: The statement of the exercise is a result from Peter Winkler's 1975 PhD thesis. It's a nice coincidence that you posted this problem now. Minh Tran, Erik Walsberg, and I are working on a project that we call "Interpolative Fusions" which at its heart is a generalization of this result of Winkler's. We just posted the first paper from this project to the arXiv this week: https://arxiv.org/abs/1811.06108






            share|cite|improve this answer



























              up vote
              2
              down vote



              accepted
              +50










              Since $T_1$ and $T_2$ are model complete, they are $forallexists$-axiomatizable, and $T_1cup T_2$ is also $forallexists$-axiomatizable. So to show the model companion exists, you need to axiomatize the existentially closed models of $T_1cup T_2$.



              So suppose $Msubseteq M'$, both models of $T_1cup T_2$. We want to write down a sufficient condition for $M$ to be existentially closed in $M'$. Let $psi(x)$ be a quantifier-free formula with parameters from $M$ such that $M'models exists x, psi(x)$. You want to observe the following things:




              1. It suffices to assume that $psi(x)$ is a conjunction of atomic and negated atomic $(L_1cup L_2)$-formulas.


              2. So if $L_1$ and $L_2$ are relational, $psi(x)$ is actually a conjunction $varphi_1(x)land varphi_2(x)$, where $varphi_1$ is an $L_1$-formula and $varphi_2$ is an $L_2$-formula. If the languages aren't relational, there's an issue that $psi$ could mention terms formed using function symbols from both $L_1$ and $L_2$. Then you have to think about replacing $psi$ with another formula obtained by "unnesting" all terms. It might be better to work out the details in the relational case first and then come back to this complication.


              3. It suffices to assume that there is a witness $M'models psi(a')$ such that each element of $a'$ is in $M'setminus M$ and the elements of $a'$ are all distinct.



              This leads us to the following axiomatization: Let $varphi_1(x,y)$ be a quantifier-free $L_1$-formula, and let $varphi_2(x,z)$ be a quantifier-free $L_2$-formula (here $x$, $y$, $z$ are tuples of variables). Let $theta_1(y)$ and $theta_2(z)$ be the formulas provided by Exercise 5.5.6. for $varphi_1$ and $varphi_2$, respectively. Let $theta'_1(y)$ be the conjunction of $theta_1(y)$ and inequations $y_ineq y_j$ for $ineq j$, and similarly for $theta'_2(z)$. Then look at the following sentence: $$forall y, forall z, ((theta'_1(y)land theta'_2(z))rightarrow exists x, (varphi_1(x,y)land varphi_2(x,z))).$$



              The model companion of $T_1cup T_2$ is axiomatized by $T_1cup T_2$ together with all sentences of the above form. The discussion above can be viewed as an extended hint that every model of this theory is an existentially closed model of $T_1cup T_2$. You also need to show the converse: that every existentially closed model of $T_1cup T_2$ satisfies these extra axioms. Explicitly, given $varphi_1(x,y)$ and $varphi_2(x,z)$, if $Mmodels theta_1'(b)land theta_2'(c)$, then you can embed $M$ in a model $M'$ of $T_1cup T_2$ such that $M'models varphi_1(a',b)land varphi_2(a',c)$ for some $a'in M'$, and use the fact that $M$ is existentially closed to find a witness in $M$.



              Aside: The statement of the exercise is a result from Peter Winkler's 1975 PhD thesis. It's a nice coincidence that you posted this problem now. Minh Tran, Erik Walsberg, and I are working on a project that we call "Interpolative Fusions" which at its heart is a generalization of this result of Winkler's. We just posted the first paper from this project to the arXiv this week: https://arxiv.org/abs/1811.06108






              share|cite|improve this answer

























                up vote
                2
                down vote



                accepted
                +50







                up vote
                2
                down vote



                accepted
                +50




                +50




                Since $T_1$ and $T_2$ are model complete, they are $forallexists$-axiomatizable, and $T_1cup T_2$ is also $forallexists$-axiomatizable. So to show the model companion exists, you need to axiomatize the existentially closed models of $T_1cup T_2$.



                So suppose $Msubseteq M'$, both models of $T_1cup T_2$. We want to write down a sufficient condition for $M$ to be existentially closed in $M'$. Let $psi(x)$ be a quantifier-free formula with parameters from $M$ such that $M'models exists x, psi(x)$. You want to observe the following things:




                1. It suffices to assume that $psi(x)$ is a conjunction of atomic and negated atomic $(L_1cup L_2)$-formulas.


                2. So if $L_1$ and $L_2$ are relational, $psi(x)$ is actually a conjunction $varphi_1(x)land varphi_2(x)$, where $varphi_1$ is an $L_1$-formula and $varphi_2$ is an $L_2$-formula. If the languages aren't relational, there's an issue that $psi$ could mention terms formed using function symbols from both $L_1$ and $L_2$. Then you have to think about replacing $psi$ with another formula obtained by "unnesting" all terms. It might be better to work out the details in the relational case first and then come back to this complication.


                3. It suffices to assume that there is a witness $M'models psi(a')$ such that each element of $a'$ is in $M'setminus M$ and the elements of $a'$ are all distinct.



                This leads us to the following axiomatization: Let $varphi_1(x,y)$ be a quantifier-free $L_1$-formula, and let $varphi_2(x,z)$ be a quantifier-free $L_2$-formula (here $x$, $y$, $z$ are tuples of variables). Let $theta_1(y)$ and $theta_2(z)$ be the formulas provided by Exercise 5.5.6. for $varphi_1$ and $varphi_2$, respectively. Let $theta'_1(y)$ be the conjunction of $theta_1(y)$ and inequations $y_ineq y_j$ for $ineq j$, and similarly for $theta'_2(z)$. Then look at the following sentence: $$forall y, forall z, ((theta'_1(y)land theta'_2(z))rightarrow exists x, (varphi_1(x,y)land varphi_2(x,z))).$$



                The model companion of $T_1cup T_2$ is axiomatized by $T_1cup T_2$ together with all sentences of the above form. The discussion above can be viewed as an extended hint that every model of this theory is an existentially closed model of $T_1cup T_2$. You also need to show the converse: that every existentially closed model of $T_1cup T_2$ satisfies these extra axioms. Explicitly, given $varphi_1(x,y)$ and $varphi_2(x,z)$, if $Mmodels theta_1'(b)land theta_2'(c)$, then you can embed $M$ in a model $M'$ of $T_1cup T_2$ such that $M'models varphi_1(a',b)land varphi_2(a',c)$ for some $a'in M'$, and use the fact that $M$ is existentially closed to find a witness in $M$.



                Aside: The statement of the exercise is a result from Peter Winkler's 1975 PhD thesis. It's a nice coincidence that you posted this problem now. Minh Tran, Erik Walsberg, and I are working on a project that we call "Interpolative Fusions" which at its heart is a generalization of this result of Winkler's. We just posted the first paper from this project to the arXiv this week: https://arxiv.org/abs/1811.06108






                share|cite|improve this answer














                Since $T_1$ and $T_2$ are model complete, they are $forallexists$-axiomatizable, and $T_1cup T_2$ is also $forallexists$-axiomatizable. So to show the model companion exists, you need to axiomatize the existentially closed models of $T_1cup T_2$.



                So suppose $Msubseteq M'$, both models of $T_1cup T_2$. We want to write down a sufficient condition for $M$ to be existentially closed in $M'$. Let $psi(x)$ be a quantifier-free formula with parameters from $M$ such that $M'models exists x, psi(x)$. You want to observe the following things:




                1. It suffices to assume that $psi(x)$ is a conjunction of atomic and negated atomic $(L_1cup L_2)$-formulas.


                2. So if $L_1$ and $L_2$ are relational, $psi(x)$ is actually a conjunction $varphi_1(x)land varphi_2(x)$, where $varphi_1$ is an $L_1$-formula and $varphi_2$ is an $L_2$-formula. If the languages aren't relational, there's an issue that $psi$ could mention terms formed using function symbols from both $L_1$ and $L_2$. Then you have to think about replacing $psi$ with another formula obtained by "unnesting" all terms. It might be better to work out the details in the relational case first and then come back to this complication.


                3. It suffices to assume that there is a witness $M'models psi(a')$ such that each element of $a'$ is in $M'setminus M$ and the elements of $a'$ are all distinct.



                This leads us to the following axiomatization: Let $varphi_1(x,y)$ be a quantifier-free $L_1$-formula, and let $varphi_2(x,z)$ be a quantifier-free $L_2$-formula (here $x$, $y$, $z$ are tuples of variables). Let $theta_1(y)$ and $theta_2(z)$ be the formulas provided by Exercise 5.5.6. for $varphi_1$ and $varphi_2$, respectively. Let $theta'_1(y)$ be the conjunction of $theta_1(y)$ and inequations $y_ineq y_j$ for $ineq j$, and similarly for $theta'_2(z)$. Then look at the following sentence: $$forall y, forall z, ((theta'_1(y)land theta'_2(z))rightarrow exists x, (varphi_1(x,y)land varphi_2(x,z))).$$



                The model companion of $T_1cup T_2$ is axiomatized by $T_1cup T_2$ together with all sentences of the above form. The discussion above can be viewed as an extended hint that every model of this theory is an existentially closed model of $T_1cup T_2$. You also need to show the converse: that every existentially closed model of $T_1cup T_2$ satisfies these extra axioms. Explicitly, given $varphi_1(x,y)$ and $varphi_2(x,z)$, if $Mmodels theta_1'(b)land theta_2'(c)$, then you can embed $M$ in a model $M'$ of $T_1cup T_2$ such that $M'models varphi_1(a',b)land varphi_2(a',c)$ for some $a'in M'$, and use the fact that $M$ is existentially closed to find a witness in $M$.



                Aside: The statement of the exercise is a result from Peter Winkler's 1975 PhD thesis. It's a nice coincidence that you posted this problem now. Minh Tran, Erik Walsberg, and I are working on a project that we call "Interpolative Fusions" which at its heart is a generalization of this result of Winkler's. We just posted the first paper from this project to the arXiv this week: https://arxiv.org/abs/1811.06108







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 18 at 11:15

























                answered Nov 18 at 9:29









                Alex Kruckman

                25.7k22455




                25.7k22455






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997862%2fexists-infty-elimination-and-model-companion%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    AnyDesk - Fatal Program Failure

                    How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

                    QoS: MAC-Priority for clients behind a repeater