$Gammamodelsphi$ 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.
$begingroup$
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.
logic first-order-logic
edited Mar 30 at 11:09
blub
3,299929
asked Mar 30 at 10:50
Pedro SantosPedro Santos
16810
$endgroup$
add a comment |
$begingroup$
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.
logic first-order-logic
edited Mar 30 at 11:09
blub
3,299929
asked Mar 30 at 10:50
Pedro SantosPedro Santos
16810
$endgroup$
add a comment |
$begingroup$
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.
logic first-order-logic
edited Mar 30 at 11:09
blub
3,299929
asked Mar 30 at 10:50
Pedro SantosPedro Santos
16810
$endgroup$
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
logic first-order-logic
edited Mar 30 at 11:09
blub
3,299929
asked Mar 30 at 10:50
Pedro SantosPedro Santos
16810
edited Mar 30 at 11:09
blub
3,299929
asked Mar 30 at 10:50
Pedro SantosPedro Santos
16810
edited Mar 30 at 11:09
blub
3,299929
edited Mar 30 at 11:09
blub
3,299929
edited Mar 30 at 11:09
blub
3,299929
3,299929
asked Mar 30 at 10:50
Pedro SantosPedro Santos
16810
asked Mar 30 at 10:50
Pedro SantosPedro Santos
16810
asked Mar 30 at 10:50
Pedro SantosPedro Santos
16810
16810
add a comment |
add a comment |
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 $Gamma,negphinotmodelspsilandnegpsi$.
answered Mar 30 at 10:57
blubblub
3,299929
$endgroup$
$begingroup$
Ah yes thats what i meant , yes Thank you my friend !
$endgroup$
– Pedro Santos
Mar 30 at 11:00
add a comment |
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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 $Gamma,negphinotmodelspsilandnegpsi$.
answered Mar 30 at 10:57
blubblub
3,299929
$endgroup$
$begingroup$
Ah yes thats what i meant , yes Thank you my friend !
$endgroup$
– Pedro Santos
Mar 30 at 11:00
add a comment |
$begingroup$
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 $Gamma,negphinotmodelspsilandnegpsi$.
answered Mar 30 at 10:57
blubblub
3,299929
$endgroup$
$begingroup$
Ah yes thats what i meant , yes Thank you my friend !
$endgroup$
– Pedro Santos
Mar 30 at 11:00
add a comment |
$begingroup$
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 $Gamma,negphinotmodelspsilandnegpsi$.
answered Mar 30 at 10:57
blubblub
3,299929
$endgroup$
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 $Gamma,negphinotmodelspsilandnegpsi$.
answered Mar 30 at 10:57
blubblub
3,299929
edited Mar 30 at 11:02
answered Mar 30 at 10:57
blubblub
3,299929
answered Mar 30 at 10:57
blubblub
3,299929
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
add a comment |
$begingroup$
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
add a comment |
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