$Gammamodelsphi$ if and only if $Gamma,negphimodelspsilandnegpsi$ The 2019 Stack Overflow Developer Survey Results Are InShow that $(phi rightarrow psi), (phi rightarrow neg psi) vdash neg phi$Is it correct that If $mathcal A $ is a model of $Gamma $, and if $Gamma models psi$ then $mathcal A models psi $?Show that $Gamma cup neg phi$ is satisfiable if and only if $Gammanot models phi$How to show that if $neg b = a land d$ then $a land neg b = neg b$ and $b land neg a = neg a$If $models neg phi$, then $models phi^circ$, where $phi^circ$ is the “semi-dual” of $phi$Prove that a theory $Gamma$ is consistent if and only if there is a structure $M$ so that $M$ $models$ $Gamma$.not always $A models phi$ or $A models neg phi$ exampleIf $Gamma$ is consistent and $Gammanotvdashphi$, then $Gammacupnegphi$ is also consistent. Why?Is there any way to simplify $(Aland B land C) lor (neg A land neg B land neg C)$?Show that $vdash Gamma cup psi$ implies $vdash Gamma cup psi'$ where $psi'$ is $psi$ with one of its bound variables renamed.

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$Gammamodelsphi$ if and only if $Gamma,negphimodelspsilandnegpsi$



The 2019 Stack Overflow Developer Survey Results Are InShow that $(phi rightarrow psi), (phi rightarrow neg psi) vdash neg phi$Is it correct that If $mathcal A $ is a model of $Gamma $, and if $Gamma models psi$ then $mathcal A models psi $?Show that $Gamma cup neg phi$ is satisfiable if and only if $Gammanot models phi$How to show that if $neg b = a land d$ then $a land neg b = neg b$ and $b land neg a = neg a$If $models neg phi$, then $models phi^circ$, where $phi^circ$ is the “semi-dual” of $phi$Prove that a theory $Gamma$ is consistent if and only if there is a structure $M$ so that $M$ $models$ $Gamma$.not always $A models phi$ or $A models neg phi$ exampleIf $Gamma$ is consistent and $Gammanotvdashphi$, then $Gammacupnegphi$ is also consistent. Why?Is there any way to simplify $(Aland B land C) lor (neg A land neg B land neg C)$?Show that $vdash Gamma cup psi$ implies $vdash Gamma cup psi'$ where $psi'$ is $psi$ with one of its bound variables renamed.










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Let $Gammacupphi,psisubseteq L epsilon$ then $Gammamodelspsi$ if and only if $Gamma,(negphi)models(psiland(negpsi))$. I don't seem to understand how the reverse implication goes. Can anyone help me out ? Thanks.










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    $begingroup$


    Let $Gammacupphi,psisubseteq L epsilon$ then $Gammamodelspsi$ if and only if $Gamma,(negphi)models(psiland(negpsi))$. I don't seem to understand how the reverse implication goes. Can anyone help me out ? Thanks.










    share|cite|improve this question











    $endgroup$














      1












      1








      1





      $begingroup$


      Let $Gammacupphi,psisubseteq L epsilon$ then $Gammamodelspsi$ if and only if $Gamma,(negphi)models(psiland(negpsi))$. I don't seem to understand how the reverse implication goes. Can anyone help me out ? Thanks.










      share|cite|improve this question











      $endgroup$




      Let $Gammacupphi,psisubseteq L epsilon$ then $Gammamodelspsi$ if and only if $Gamma,(negphi)models(psiland(negpsi))$. I don't seem to understand how the reverse implication goes. Can anyone help me out ? Thanks.







      logic first-order-logic






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      share|cite|improve this question













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      share|cite|improve this question








      edited Mar 30 at 11:09









      blub

      3,299929




      3,299929










      asked Mar 30 at 10:50









      Pedro SantosPedro Santos

      16810




      16810




















          1 Answer
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          $begingroup$

          Written like this, it makes no sense. I assume you wanted to write



          $$Gammamodelsphitext iff Gamma,negphimodelspsilandnegpsi$$




          To prove this, it is helpful to note that for $Deltacuppsisubseteqmathcal L_FO$, $Deltanotmodelspsilandnegpsi$ iff $Delta$ is satisfiable, as then there is an interpretation $mathcal I$ s.t. $mathcal ImodelsDelta$, and naturally $mathcal Inotmodelspsilandnegpsi$.



          Now on to proving the equivalence. Let $Gammacupphi,psisubseteqmathcal L_FO$.



          From left to right, assume $Gammamodelsphi$, i.e. for every interpretation $mathcal I$: $mathcal ImodelsGamma$ implies $mathcal Imodelsphi$. Thus, no interpretation $mathcal I$ models $Gamma,negphi$ and thus for every interpretation $mathcal I$: $mathcal ImodelsGamma,negphi$ implies $mathcal Imodelspsilandnegpsi$.



          From right to left, assume $Gammanotmodelsphi$, i.e. there is an interpretation $mathcal I$ s.t. $mathcal ImodelsGamma$ but $mathcal Inotmodelsphi$. The latter implies $mathcal Imodelsnegphi$. Thus $mathcal ImodelsGamma,negphi$, i.e. $Gamma,negphi$ is satisfiable and thus $Gamma,negphinotmodelspsilandnegpsi$.






          share|cite|improve this answer











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          • $begingroup$
            Ah yes thats what i meant , yes Thank you my friend !
            $endgroup$
            – Pedro Santos
            Mar 30 at 11:00











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          $begingroup$

          Written like this, it makes no sense. I assume you wanted to write



          $$Gammamodelsphitext iff Gamma,negphimodelspsilandnegpsi$$




          To prove this, it is helpful to note that for $Deltacuppsisubseteqmathcal L_FO$, $Deltanotmodelspsilandnegpsi$ iff $Delta$ is satisfiable, as then there is an interpretation $mathcal I$ s.t. $mathcal ImodelsDelta$, and naturally $mathcal Inotmodelspsilandnegpsi$.



          Now on to proving the equivalence. Let $Gammacupphi,psisubseteqmathcal L_FO$.



          From left to right, assume $Gammamodelsphi$, i.e. for every interpretation $mathcal I$: $mathcal ImodelsGamma$ implies $mathcal Imodelsphi$. Thus, no interpretation $mathcal I$ models $Gamma,negphi$ and thus for every interpretation $mathcal I$: $mathcal ImodelsGamma,negphi$ implies $mathcal Imodelspsilandnegpsi$.



          From right to left, assume $Gammanotmodelsphi$, i.e. there is an interpretation $mathcal I$ s.t. $mathcal ImodelsGamma$ but $mathcal Inotmodelsphi$. The latter implies $mathcal Imodelsnegphi$. Thus $mathcal ImodelsGamma,negphi$, i.e. $Gamma,negphi$ is satisfiable and thus $Gamma,negphinotmodelspsilandnegpsi$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Ah yes thats what i meant , yes Thank you my friend !
            $endgroup$
            – Pedro Santos
            Mar 30 at 11:00















          1












          $begingroup$

          Written like this, it makes no sense. I assume you wanted to write



          $$Gammamodelsphitext iff Gamma,negphimodelspsilandnegpsi$$




          To prove this, it is helpful to note that for $Deltacuppsisubseteqmathcal L_FO$, $Deltanotmodelspsilandnegpsi$ iff $Delta$ is satisfiable, as then there is an interpretation $mathcal I$ s.t. $mathcal ImodelsDelta$, and naturally $mathcal Inotmodelspsilandnegpsi$.



          Now on to proving the equivalence. Let $Gammacupphi,psisubseteqmathcal L_FO$.



          From left to right, assume $Gammamodelsphi$, i.e. for every interpretation $mathcal I$: $mathcal ImodelsGamma$ implies $mathcal Imodelsphi$. Thus, no interpretation $mathcal I$ models $Gamma,negphi$ and thus for every interpretation $mathcal I$: $mathcal ImodelsGamma,negphi$ implies $mathcal Imodelspsilandnegpsi$.



          From right to left, assume $Gammanotmodelsphi$, i.e. there is an interpretation $mathcal I$ s.t. $mathcal ImodelsGamma$ but $mathcal Inotmodelsphi$. The latter implies $mathcal Imodelsnegphi$. Thus $mathcal ImodelsGamma,negphi$, i.e. $Gamma,negphi$ is satisfiable and thus $Gamma,negphinotmodelspsilandnegpsi$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Ah yes thats what i meant , yes Thank you my friend !
            $endgroup$
            – Pedro Santos
            Mar 30 at 11:00













          1












          1








          1





          $begingroup$

          Written like this, it makes no sense. I assume you wanted to write



          $$Gammamodelsphitext iff Gamma,negphimodelspsilandnegpsi$$




          To prove this, it is helpful to note that for $Deltacuppsisubseteqmathcal L_FO$, $Deltanotmodelspsilandnegpsi$ iff $Delta$ is satisfiable, as then there is an interpretation $mathcal I$ s.t. $mathcal ImodelsDelta$, and naturally $mathcal Inotmodelspsilandnegpsi$.



          Now on to proving the equivalence. Let $Gammacupphi,psisubseteqmathcal L_FO$.



          From left to right, assume $Gammamodelsphi$, i.e. for every interpretation $mathcal I$: $mathcal ImodelsGamma$ implies $mathcal Imodelsphi$. Thus, no interpretation $mathcal I$ models $Gamma,negphi$ and thus for every interpretation $mathcal I$: $mathcal ImodelsGamma,negphi$ implies $mathcal Imodelspsilandnegpsi$.



          From right to left, assume $Gammanotmodelsphi$, i.e. there is an interpretation $mathcal I$ s.t. $mathcal ImodelsGamma$ but $mathcal Inotmodelsphi$. The latter implies $mathcal Imodelsnegphi$. Thus $mathcal ImodelsGamma,negphi$, i.e. $Gamma,negphi$ is satisfiable and thus $Gamma,negphinotmodelspsilandnegpsi$.






          share|cite|improve this answer











          $endgroup$



          Written like this, it makes no sense. I assume you wanted to write



          $$Gammamodelsphitext iff Gamma,negphimodelspsilandnegpsi$$




          To prove this, it is helpful to note that for $Deltacuppsisubseteqmathcal L_FO$, $Deltanotmodelspsilandnegpsi$ iff $Delta$ is satisfiable, as then there is an interpretation $mathcal I$ s.t. $mathcal ImodelsDelta$, and naturally $mathcal Inotmodelspsilandnegpsi$.



          Now on to proving the equivalence. Let $Gammacupphi,psisubseteqmathcal L_FO$.



          From left to right, assume $Gammamodelsphi$, i.e. for every interpretation $mathcal I$: $mathcal ImodelsGamma$ implies $mathcal Imodelsphi$. Thus, no interpretation $mathcal I$ models $Gamma,negphi$ and thus for every interpretation $mathcal I$: $mathcal ImodelsGamma,negphi$ implies $mathcal Imodelspsilandnegpsi$.



          From right to left, assume $Gammanotmodelsphi$, i.e. there is an interpretation $mathcal I$ s.t. $mathcal ImodelsGamma$ but $mathcal Inotmodelsphi$. The latter implies $mathcal Imodelsnegphi$. Thus $mathcal ImodelsGamma,negphi$, i.e. $Gamma,negphi$ is satisfiable and thus $Gamma,negphinotmodelspsilandnegpsi$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 30 at 11:02

























          answered Mar 30 at 10:57









          blubblub

          3,299929




          3,299929











          • $begingroup$
            Ah yes thats what i meant , yes Thank you my friend !
            $endgroup$
            – Pedro Santos
            Mar 30 at 11:00
















          • $begingroup$
            Ah yes thats what i meant , yes Thank you my friend !
            $endgroup$
            – Pedro Santos
            Mar 30 at 11:00















          $begingroup$
          Ah yes thats what i meant , yes Thank you my friend !
          $endgroup$
          – Pedro Santos
          Mar 30 at 11:00




          $begingroup$
          Ah yes thats what i meant , yes Thank you my friend !
          $endgroup$
          – Pedro Santos
          Mar 30 at 11:00

















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