How to show that $phi vdash psi$ implies $phi^{mathbf{M}} vdash psi^{mathbf{M}}$












0














Given a class $mathbf{M} neq emptyset$ and a formula $phi$ from the language of set theory, let $phi^{mathbf{M}}$ denote the relativizatization of $phi$ to $mathbf{M}$. I want to show the if $phi vdash psi$, then $phi^{mathbf{M}} vdash psi^{mathbf{M}}$. I tried proving that if $vdash phi$, then $vdash phi^{mathbf{M}}$, but I'm getting stucked in the induction steps - and one of the induction steps seems to be exactly what I want to prove. Anyone can shed some light?






logic set-theory







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  • 3




    Show that your inference rules relativize, and then mechanically relativize the proof.
    – Asaf Karagila♦
    Dec 26 at 8:26
















0














Given a class $mathbf{M} neq emptyset$ and a formula $phi$ from the language of set theory, let $phi^{mathbf{M}}$ denote the relativizatization of $phi$ to $mathbf{M}$. I want to show the if $phi vdash psi$, then $phi^{mathbf{M}} vdash psi^{mathbf{M}}$. I tried proving that if $vdash phi$, then $vdash phi^{mathbf{M}}$, but I'm getting stucked in the induction steps - and one of the induction steps seems to be exactly what I want to prove. Anyone can shed some light?






logic set-theory







share|cite|improve this question


















  • 3




    Show that your inference rules relativize, and then mechanically relativize the proof.
    – Asaf Karagila♦
    Dec 26 at 8:26














0












0








0







Given a class $mathbf{M} neq emptyset$ and a formula $phi$ from the language of set theory, let $phi^{mathbf{M}}$ denote the relativizatization of $phi$ to $mathbf{M}$. I want to show the if $phi vdash psi$, then $phi^{mathbf{M}} vdash psi^{mathbf{M}}$. I tried proving that if $vdash phi$, then $vdash phi^{mathbf{M}}$, but I'm getting stucked in the induction steps - and one of the induction steps seems to be exactly what I want to prove. Anyone can shed some light?






logic set-theory







share|cite|improve this question













Given a class $mathbf{M} neq emptyset$ and a formula $phi$ from the language of set theory, let $phi^{mathbf{M}}$ denote the relativizatization of $phi$ to $mathbf{M}$. I want to show the if $phi vdash psi$, then $phi^{mathbf{M}} vdash psi^{mathbf{M}}$. I tried proving that if $vdash phi$, then $vdash phi^{mathbf{M}}$, but I'm getting stucked in the induction steps - and one of the induction steps seems to be exactly what I want to prove. Anyone can shed some light?






logic set-theory




logic set-theory






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share|cite|improve this question











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asked Dec 26 at 6:32









Nuntractatuses Amável

61812




61812








  • 3




    Show that your inference rules relativize, and then mechanically relativize the proof.
    – Asaf Karagila♦
    Dec 26 at 8:26














  • 3




    Show that your inference rules relativize, and then mechanically relativize the proof.
    – Asaf Karagila♦
    Dec 26 at 8:26








3




3




Show that your inference rules relativize, and then mechanically relativize the proof.
– Asaf Karagila♦
Dec 26 at 8:26




Show that your inference rules relativize, and then mechanically relativize the proof.
– Asaf Karagila♦
Dec 26 at 8:26










1 Answer
1






active

oldest

votes


















2














For any structure $A$ in the language of set theory, the class $mathbf{M}$ can be interpreted as a subset $mathbf{M}^A$ of $A$, which we view as a substructure of $A$ by restricting $in$ to $mathbf{M}^A$.



Then given a formula $varphi$, the defining property of the formula $varphi^mathbf{M}$ is that for any structure $A$, we have $Amodels varphi^mathbf{M}$ if and only if $mathbf{M}^Amodels varphi$.



Now suppose $varphivdash psi$. By soundness and completeness, we have $varphimodels psi$, and it suffices to show $varphi^mathbf{M}models psi^mathbf{M}$. So let $A$ be any model of $varphi^mathbf{M}$. Then $mathbf{M}^Amodels varphi$. Since $varphimodels psi$, we have $mathbf{M}^Amodels psi$, so $Amodels psi^mathbf{M}$.






share|cite|improve this answer





















  • nice. I was trying to do it without using $vDash$ - only the formal deductions. I would prefer to do it in this way, but it is just personal preference.
    – Nuntractatuses Amável
    Dec 26 at 6:47










  • I tried proving the defining property you mentioned, and I encountered some dificulties. Mainly, the point that if $phi$ is the sentence $exists x psi$, then $psi$ is not necessarily a sentence - in which case I am not sure what the defining property means. I then tried proving for arbitrary formulas, with an assignment of values for the variables: then it struck me that, for this to make any sense, the variables can only assume values in $mathbf{M}^A$. Is this the case? I've seen relativization only in the context of set theory - in Kunen's book. I'm slightly confused.
    – Nuntractatuses Amável
    Dec 26 at 8:07








  • 2




    Yes, to prove the "defining property", you need to do an induction over formulas. Here the statement you want to prove by induction is: For any formula $varphi(x)$, any structure $A$, and any tuple $ain (mathbf{M}^A)^x$, we have $Amodels varphi^{mathbf{M}}(a)$ if and only if $mathbf{M}^Amodels varphi(a)$. This restricts to the property I wrote above when $varphi$ is a sentence.
    – Alex Kruckman
    Dec 26 at 15:50








  • 2




    If you prefer a purely syntactic proof, you can follow Asaf's hint above.
    – Alex Kruckman
    Dec 26 at 15:52










  • I managed to prove it (the "defining property"), thank you :). I'll try to do it in a purely syntactical way, now.
    – Nuntractatuses Amável
    yesterday











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For any structure $A$ in the language of set theory, the class $mathbf{M}$ can be interpreted as a subset $mathbf{M}^A$ of $A$, which we view as a substructure of $A$ by restricting $in$ to $mathbf{M}^A$.



Then given a formula $varphi$, the defining property of the formula $varphi^mathbf{M}$ is that for any structure $A$, we have $Amodels varphi^mathbf{M}$ if and only if $mathbf{M}^Amodels varphi$.



Now suppose $varphivdash psi$. By soundness and completeness, we have $varphimodels psi$, and it suffices to show $varphi^mathbf{M}models psi^mathbf{M}$. So let $A$ be any model of $varphi^mathbf{M}$. Then $mathbf{M}^Amodels varphi$. Since $varphimodels psi$, we have $mathbf{M}^Amodels psi$, so $Amodels psi^mathbf{M}$.






share|cite|improve this answer





















  • nice. I was trying to do it without using $vDash$ - only the formal deductions. I would prefer to do it in this way, but it is just personal preference.
    – Nuntractatuses Amável
    Dec 26 at 6:47










  • I tried proving the defining property you mentioned, and I encountered some dificulties. Mainly, the point that if $phi$ is the sentence $exists x psi$, then $psi$ is not necessarily a sentence - in which case I am not sure what the defining property means. I then tried proving for arbitrary formulas, with an assignment of values for the variables: then it struck me that, for this to make any sense, the variables can only assume values in $mathbf{M}^A$. Is this the case? I've seen relativization only in the context of set theory - in Kunen's book. I'm slightly confused.
    – Nuntractatuses Amável
    Dec 26 at 8:07








  • 2




    Yes, to prove the "defining property", you need to do an induction over formulas. Here the statement you want to prove by induction is: For any formula $varphi(x)$, any structure $A$, and any tuple $ain (mathbf{M}^A)^x$, we have $Amodels varphi^{mathbf{M}}(a)$ if and only if $mathbf{M}^Amodels varphi(a)$. This restricts to the property I wrote above when $varphi$ is a sentence.
    – Alex Kruckman
    Dec 26 at 15:50








  • 2




    If you prefer a purely syntactic proof, you can follow Asaf's hint above.
    – Alex Kruckman
    Dec 26 at 15:52










  • I managed to prove it (the "defining property"), thank you :). I'll try to do it in a purely syntactical way, now.
    – Nuntractatuses Amável
    yesterday
















2














For any structure $A$ in the language of set theory, the class $mathbf{M}$ can be interpreted as a subset $mathbf{M}^A$ of $A$, which we view as a substructure of $A$ by restricting $in$ to $mathbf{M}^A$.



Then given a formula $varphi$, the defining property of the formula $varphi^mathbf{M}$ is that for any structure $A$, we have $Amodels varphi^mathbf{M}$ if and only if $mathbf{M}^Amodels varphi$.



Now suppose $varphivdash psi$. By soundness and completeness, we have $varphimodels psi$, and it suffices to show $varphi^mathbf{M}models psi^mathbf{M}$. So let $A$ be any model of $varphi^mathbf{M}$. Then $mathbf{M}^Amodels varphi$. Since $varphimodels psi$, we have $mathbf{M}^Amodels psi$, so $Amodels psi^mathbf{M}$.






share|cite|improve this answer





















  • nice. I was trying to do it without using $vDash$ - only the formal deductions. I would prefer to do it in this way, but it is just personal preference.
    – Nuntractatuses Amável
    Dec 26 at 6:47










  • I tried proving the defining property you mentioned, and I encountered some dificulties. Mainly, the point that if $phi$ is the sentence $exists x psi$, then $psi$ is not necessarily a sentence - in which case I am not sure what the defining property means. I then tried proving for arbitrary formulas, with an assignment of values for the variables: then it struck me that, for this to make any sense, the variables can only assume values in $mathbf{M}^A$. Is this the case? I've seen relativization only in the context of set theory - in Kunen's book. I'm slightly confused.
    – Nuntractatuses Amável
    Dec 26 at 8:07








  • 2




    Yes, to prove the "defining property", you need to do an induction over formulas. Here the statement you want to prove by induction is: For any formula $varphi(x)$, any structure $A$, and any tuple $ain (mathbf{M}^A)^x$, we have $Amodels varphi^{mathbf{M}}(a)$ if and only if $mathbf{M}^Amodels varphi(a)$. This restricts to the property I wrote above when $varphi$ is a sentence.
    – Alex Kruckman
    Dec 26 at 15:50








  • 2




    If you prefer a purely syntactic proof, you can follow Asaf's hint above.
    – Alex Kruckman
    Dec 26 at 15:52










  • I managed to prove it (the "defining property"), thank you :). I'll try to do it in a purely syntactical way, now.
    – Nuntractatuses Amável
    yesterday














2












2








2






For any structure $A$ in the language of set theory, the class $mathbf{M}$ can be interpreted as a subset $mathbf{M}^A$ of $A$, which we view as a substructure of $A$ by restricting $in$ to $mathbf{M}^A$.



Then given a formula $varphi$, the defining property of the formula $varphi^mathbf{M}$ is that for any structure $A$, we have $Amodels varphi^mathbf{M}$ if and only if $mathbf{M}^Amodels varphi$.



Now suppose $varphivdash psi$. By soundness and completeness, we have $varphimodels psi$, and it suffices to show $varphi^mathbf{M}models psi^mathbf{M}$. So let $A$ be any model of $varphi^mathbf{M}$. Then $mathbf{M}^Amodels varphi$. Since $varphimodels psi$, we have $mathbf{M}^Amodels psi$, so $Amodels psi^mathbf{M}$.






share|cite|improve this answer












For any structure $A$ in the language of set theory, the class $mathbf{M}$ can be interpreted as a subset $mathbf{M}^A$ of $A$, which we view as a substructure of $A$ by restricting $in$ to $mathbf{M}^A$.



Then given a formula $varphi$, the defining property of the formula $varphi^mathbf{M}$ is that for any structure $A$, we have $Amodels varphi^mathbf{M}$ if and only if $mathbf{M}^Amodels varphi$.



Now suppose $varphivdash psi$. By soundness and completeness, we have $varphimodels psi$, and it suffices to show $varphi^mathbf{M}models psi^mathbf{M}$. So let $A$ be any model of $varphi^mathbf{M}$. Then $mathbf{M}^Amodels varphi$. Since $varphimodels psi$, we have $mathbf{M}^Amodels psi$, so $Amodels psi^mathbf{M}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 26 at 6:43









Alex Kruckman

26.4k22555




26.4k22555












  • nice. I was trying to do it without using $vDash$ - only the formal deductions. I would prefer to do it in this way, but it is just personal preference.
    – Nuntractatuses Amável
    Dec 26 at 6:47










  • I tried proving the defining property you mentioned, and I encountered some dificulties. Mainly, the point that if $phi$ is the sentence $exists x psi$, then $psi$ is not necessarily a sentence - in which case I am not sure what the defining property means. I then tried proving for arbitrary formulas, with an assignment of values for the variables: then it struck me that, for this to make any sense, the variables can only assume values in $mathbf{M}^A$. Is this the case? I've seen relativization only in the context of set theory - in Kunen's book. I'm slightly confused.
    – Nuntractatuses Amável
    Dec 26 at 8:07








  • 2




    Yes, to prove the "defining property", you need to do an induction over formulas. Here the statement you want to prove by induction is: For any formula $varphi(x)$, any structure $A$, and any tuple $ain (mathbf{M}^A)^x$, we have $Amodels varphi^{mathbf{M}}(a)$ if and only if $mathbf{M}^Amodels varphi(a)$. This restricts to the property I wrote above when $varphi$ is a sentence.
    – Alex Kruckman
    Dec 26 at 15:50








  • 2




    If you prefer a purely syntactic proof, you can follow Asaf's hint above.
    – Alex Kruckman
    Dec 26 at 15:52










  • I managed to prove it (the "defining property"), thank you :). I'll try to do it in a purely syntactical way, now.
    – Nuntractatuses Amável
    yesterday


















  • nice. I was trying to do it without using $vDash$ - only the formal deductions. I would prefer to do it in this way, but it is just personal preference.
    – Nuntractatuses Amável
    Dec 26 at 6:47










  • I tried proving the defining property you mentioned, and I encountered some dificulties. Mainly, the point that if $phi$ is the sentence $exists x psi$, then $psi$ is not necessarily a sentence - in which case I am not sure what the defining property means. I then tried proving for arbitrary formulas, with an assignment of values for the variables: then it struck me that, for this to make any sense, the variables can only assume values in $mathbf{M}^A$. Is this the case? I've seen relativization only in the context of set theory - in Kunen's book. I'm slightly confused.
    – Nuntractatuses Amável
    Dec 26 at 8:07








  • 2




    Yes, to prove the "defining property", you need to do an induction over formulas. Here the statement you want to prove by induction is: For any formula $varphi(x)$, any structure $A$, and any tuple $ain (mathbf{M}^A)^x$, we have $Amodels varphi^{mathbf{M}}(a)$ if and only if $mathbf{M}^Amodels varphi(a)$. This restricts to the property I wrote above when $varphi$ is a sentence.
    – Alex Kruckman
    Dec 26 at 15:50








  • 2




    If you prefer a purely syntactic proof, you can follow Asaf's hint above.
    – Alex Kruckman
    Dec 26 at 15:52










  • I managed to prove it (the "defining property"), thank you :). I'll try to do it in a purely syntactical way, now.
    – Nuntractatuses Amável
    yesterday
















nice. I was trying to do it without using $vDash$ - only the formal deductions. I would prefer to do it in this way, but it is just personal preference.
– Nuntractatuses Amável
Dec 26 at 6:47




nice. I was trying to do it without using $vDash$ - only the formal deductions. I would prefer to do it in this way, but it is just personal preference.
– Nuntractatuses Amável
Dec 26 at 6:47












I tried proving the defining property you mentioned, and I encountered some dificulties. Mainly, the point that if $phi$ is the sentence $exists x psi$, then $psi$ is not necessarily a sentence - in which case I am not sure what the defining property means. I then tried proving for arbitrary formulas, with an assignment of values for the variables: then it struck me that, for this to make any sense, the variables can only assume values in $mathbf{M}^A$. Is this the case? I've seen relativization only in the context of set theory - in Kunen's book. I'm slightly confused.
– Nuntractatuses Amável
Dec 26 at 8:07






I tried proving the defining property you mentioned, and I encountered some dificulties. Mainly, the point that if $phi$ is the sentence $exists x psi$, then $psi$ is not necessarily a sentence - in which case I am not sure what the defining property means. I then tried proving for arbitrary formulas, with an assignment of values for the variables: then it struck me that, for this to make any sense, the variables can only assume values in $mathbf{M}^A$. Is this the case? I've seen relativization only in the context of set theory - in Kunen's book. I'm slightly confused.
– Nuntractatuses Amável
Dec 26 at 8:07






2




2




Yes, to prove the "defining property", you need to do an induction over formulas. Here the statement you want to prove by induction is: For any formula $varphi(x)$, any structure $A$, and any tuple $ain (mathbf{M}^A)^x$, we have $Amodels varphi^{mathbf{M}}(a)$ if and only if $mathbf{M}^Amodels varphi(a)$. This restricts to the property I wrote above when $varphi$ is a sentence.
– Alex Kruckman
Dec 26 at 15:50






Yes, to prove the "defining property", you need to do an induction over formulas. Here the statement you want to prove by induction is: For any formula $varphi(x)$, any structure $A$, and any tuple $ain (mathbf{M}^A)^x$, we have $Amodels varphi^{mathbf{M}}(a)$ if and only if $mathbf{M}^Amodels varphi(a)$. This restricts to the property I wrote above when $varphi$ is a sentence.
– Alex Kruckman
Dec 26 at 15:50






2




2




If you prefer a purely syntactic proof, you can follow Asaf's hint above.
– Alex Kruckman
Dec 26 at 15:52




If you prefer a purely syntactic proof, you can follow Asaf's hint above.
– Alex Kruckman
Dec 26 at 15:52












I managed to prove it (the "defining property"), thank you :). I'll try to do it in a purely syntactical way, now.
– Nuntractatuses Amável
yesterday




I managed to prove it (the "defining property"), thank you :). I'll try to do it in a purely syntactical way, now.
– Nuntractatuses Amável
yesterday


















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