Prove that $mathcal{A}preceq mathcal{B}$.












2












$begingroup$


Can someone check whether my solution is okay?




If $mathcal{A}subseteq mathcal{B}$, $mathcal{A}preceq mathcal{C}$, and $mathcal{B}preceq mathcal{C}$, prove that $mathcal{A}preceq mathcal{B}$.




Let $mathcal{A} subseteq mathcal{B}$, $mathcal{A}preceq mathcal{C}$, and $mathcal{B}preceq mathcal{C}$. Then for any $mathcal{L}$-formulas $phi$ and assignments $alpha$, $mathcal{A}models phi[alpha]$ if and only if $mathcal{C}models phi[alpha]$ if and only if $mathcal{B}models phi[alpha]$. Then $mathcal{A}models phi[alpha]$ if and only if $mathcal{B}models phi[alpha]$ and by definition of substructures, $Asubseteq B$. Then $mathcal{A}preceq mathcal{B}$.






proof-verification logic model-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can you define $preceq, mathcal L$-formula, assignment and $models$?
    $endgroup$
    – Jimmy R.
    Nov 29 '18 at 2:04








  • 4




    $begingroup$
    @Jimmy That seems unreasonable; it would make the question too long. These are basic model-theoretic concepts.
    $endgroup$
    – Andrés E. Caicedo
    Nov 29 '18 at 2:08






  • 3




    $begingroup$
    Yes, the proof is fine.
    $endgroup$
    – Andrés E. Caicedo
    Nov 29 '18 at 2:09






  • 3




    $begingroup$
    The proof makes sense if $preceq$ stands for "is an elementary substructure of". I think it would be reasonable to at least say explicitly either that this is what $preceq$ means here, or to say explicitly which kinds of things the letters $mathcal A$, $mathcal B$, $mathcal C$ range over in the quoted goal.
    $endgroup$
    – Henning Makholm
    Nov 29 '18 at 2:21












  • $begingroup$
    @HenningMakholm I see. Yes, it stands for elementary substructure, and $mathcal{A,B,C}$ are just structures for a language! I am not sure about the range...
    $endgroup$
    – numericalorange
    Nov 29 '18 at 2:23
















2












$begingroup$


Can someone check whether my solution is okay?




If $mathcal{A}subseteq mathcal{B}$, $mathcal{A}preceq mathcal{C}$, and $mathcal{B}preceq mathcal{C}$, prove that $mathcal{A}preceq mathcal{B}$.




Let $mathcal{A} subseteq mathcal{B}$, $mathcal{A}preceq mathcal{C}$, and $mathcal{B}preceq mathcal{C}$. Then for any $mathcal{L}$-formulas $phi$ and assignments $alpha$, $mathcal{A}models phi[alpha]$ if and only if $mathcal{C}models phi[alpha]$ if and only if $mathcal{B}models phi[alpha]$. Then $mathcal{A}models phi[alpha]$ if and only if $mathcal{B}models phi[alpha]$ and by definition of substructures, $Asubseteq B$. Then $mathcal{A}preceq mathcal{B}$.






proof-verification logic model-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can you define $preceq, mathcal L$-formula, assignment and $models$?
    $endgroup$
    – Jimmy R.
    Nov 29 '18 at 2:04








  • 4




    $begingroup$
    @Jimmy That seems unreasonable; it would make the question too long. These are basic model-theoretic concepts.
    $endgroup$
    – Andrés E. Caicedo
    Nov 29 '18 at 2:08






  • 3




    $begingroup$
    Yes, the proof is fine.
    $endgroup$
    – Andrés E. Caicedo
    Nov 29 '18 at 2:09






  • 3




    $begingroup$
    The proof makes sense if $preceq$ stands for "is an elementary substructure of". I think it would be reasonable to at least say explicitly either that this is what $preceq$ means here, or to say explicitly which kinds of things the letters $mathcal A$, $mathcal B$, $mathcal C$ range over in the quoted goal.
    $endgroup$
    – Henning Makholm
    Nov 29 '18 at 2:21












  • $begingroup$
    @HenningMakholm I see. Yes, it stands for elementary substructure, and $mathcal{A,B,C}$ are just structures for a language! I am not sure about the range...
    $endgroup$
    – numericalorange
    Nov 29 '18 at 2:23














2












2








2





$begingroup$


Can someone check whether my solution is okay?




If $mathcal{A}subseteq mathcal{B}$, $mathcal{A}preceq mathcal{C}$, and $mathcal{B}preceq mathcal{C}$, prove that $mathcal{A}preceq mathcal{B}$.




Let $mathcal{A} subseteq mathcal{B}$, $mathcal{A}preceq mathcal{C}$, and $mathcal{B}preceq mathcal{C}$. Then for any $mathcal{L}$-formulas $phi$ and assignments $alpha$, $mathcal{A}models phi[alpha]$ if and only if $mathcal{C}models phi[alpha]$ if and only if $mathcal{B}models phi[alpha]$. Then $mathcal{A}models phi[alpha]$ if and only if $mathcal{B}models phi[alpha]$ and by definition of substructures, $Asubseteq B$. Then $mathcal{A}preceq mathcal{B}$.






proof-verification logic model-theory







share|cite|improve this question









$endgroup$




Can someone check whether my solution is okay?




If $mathcal{A}subseteq mathcal{B}$, $mathcal{A}preceq mathcal{C}$, and $mathcal{B}preceq mathcal{C}$, prove that $mathcal{A}preceq mathcal{B}$.




Let $mathcal{A} subseteq mathcal{B}$, $mathcal{A}preceq mathcal{C}$, and $mathcal{B}preceq mathcal{C}$. Then for any $mathcal{L}$-formulas $phi$ and assignments $alpha$, $mathcal{A}models phi[alpha]$ if and only if $mathcal{C}models phi[alpha]$ if and only if $mathcal{B}models phi[alpha]$. Then $mathcal{A}models phi[alpha]$ if and only if $mathcal{B}models phi[alpha]$ and by definition of substructures, $Asubseteq B$. Then $mathcal{A}preceq mathcal{B}$.






proof-verification logic model-theory




proof-verification logic model-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '18 at 2:02









numericalorangenumericalorange

1,728311




1,728311












  • $begingroup$
    Can you define $preceq, mathcal L$-formula, assignment and $models$?
    $endgroup$
    – Jimmy R.
    Nov 29 '18 at 2:04








  • 4




    $begingroup$
    @Jimmy That seems unreasonable; it would make the question too long. These are basic model-theoretic concepts.
    $endgroup$
    – Andrés E. Caicedo
    Nov 29 '18 at 2:08






  • 3




    $begingroup$
    Yes, the proof is fine.
    $endgroup$
    – Andrés E. Caicedo
    Nov 29 '18 at 2:09






  • 3




    $begingroup$
    The proof makes sense if $preceq$ stands for "is an elementary substructure of". I think it would be reasonable to at least say explicitly either that this is what $preceq$ means here, or to say explicitly which kinds of things the letters $mathcal A$, $mathcal B$, $mathcal C$ range over in the quoted goal.
    $endgroup$
    – Henning Makholm
    Nov 29 '18 at 2:21












  • $begingroup$
    @HenningMakholm I see. Yes, it stands for elementary substructure, and $mathcal{A,B,C}$ are just structures for a language! I am not sure about the range...
    $endgroup$
    – numericalorange
    Nov 29 '18 at 2:23


















  • $begingroup$
    Can you define $preceq, mathcal L$-formula, assignment and $models$?
    $endgroup$
    – Jimmy R.
    Nov 29 '18 at 2:04








  • 4




    $begingroup$
    @Jimmy That seems unreasonable; it would make the question too long. These are basic model-theoretic concepts.
    $endgroup$
    – Andrés E. Caicedo
    Nov 29 '18 at 2:08






  • 3




    $begingroup$
    Yes, the proof is fine.
    $endgroup$
    – Andrés E. Caicedo
    Nov 29 '18 at 2:09






  • 3




    $begingroup$
    The proof makes sense if $preceq$ stands for "is an elementary substructure of". I think it would be reasonable to at least say explicitly either that this is what $preceq$ means here, or to say explicitly which kinds of things the letters $mathcal A$, $mathcal B$, $mathcal C$ range over in the quoted goal.
    $endgroup$
    – Henning Makholm
    Nov 29 '18 at 2:21












  • $begingroup$
    @HenningMakholm I see. Yes, it stands for elementary substructure, and $mathcal{A,B,C}$ are just structures for a language! I am not sure about the range...
    $endgroup$
    – numericalorange
    Nov 29 '18 at 2:23
















$begingroup$
Can you define $preceq, mathcal L$-formula, assignment and $models$?
$endgroup$
– Jimmy R.
Nov 29 '18 at 2:04






$begingroup$
Can you define $preceq, mathcal L$-formula, assignment and $models$?
$endgroup$
– Jimmy R.
Nov 29 '18 at 2:04






4




4




$begingroup$
@Jimmy That seems unreasonable; it would make the question too long. These are basic model-theoretic concepts.
$endgroup$
– Andrés E. Caicedo
Nov 29 '18 at 2:08




$begingroup$
@Jimmy That seems unreasonable; it would make the question too long. These are basic model-theoretic concepts.
$endgroup$
– Andrés E. Caicedo
Nov 29 '18 at 2:08




3




3




$begingroup$
Yes, the proof is fine.
$endgroup$
– Andrés E. Caicedo
Nov 29 '18 at 2:09




$begingroup$
Yes, the proof is fine.
$endgroup$
– Andrés E. Caicedo
Nov 29 '18 at 2:09




3




3




$begingroup$
The proof makes sense if $preceq$ stands for "is an elementary substructure of". I think it would be reasonable to at least say explicitly either that this is what $preceq$ means here, or to say explicitly which kinds of things the letters $mathcal A$, $mathcal B$, $mathcal C$ range over in the quoted goal.
$endgroup$
– Henning Makholm
Nov 29 '18 at 2:21






$begingroup$
The proof makes sense if $preceq$ stands for "is an elementary substructure of". I think it would be reasonable to at least say explicitly either that this is what $preceq$ means here, or to say explicitly which kinds of things the letters $mathcal A$, $mathcal B$, $mathcal C$ range over in the quoted goal.
$endgroup$
– Henning Makholm
Nov 29 '18 at 2:21














$begingroup$
@HenningMakholm I see. Yes, it stands for elementary substructure, and $mathcal{A,B,C}$ are just structures for a language! I am not sure about the range...
$endgroup$
– numericalorange
Nov 29 '18 at 2:23




$begingroup$
@HenningMakholm I see. Yes, it stands for elementary substructure, and $mathcal{A,B,C}$ are just structures for a language! I am not sure about the range...
$endgroup$
– numericalorange
Nov 29 '18 at 2:23










1 Answer
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$begingroup$

The proof seems okay. One point you glossed over a little bit more than you should have: when you choose $alpha$, you should specify that it is an arbitrary assignment in $A$, and later on, when you use $Bpreceq C$, you should deduce that it is therefore an assignment in $B$ (so you can apply $Bpreceq C$). Just make sure you apply the hypothesis directly, and how you apply it. That should remove all doubt about correctness.






share|cite|improve this answer









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    $begingroup$

    The proof seems okay. One point you glossed over a little bit more than you should have: when you choose $alpha$, you should specify that it is an arbitrary assignment in $A$, and later on, when you use $Bpreceq C$, you should deduce that it is therefore an assignment in $B$ (so you can apply $Bpreceq C$). Just make sure you apply the hypothesis directly, and how you apply it. That should remove all doubt about correctness.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The proof seems okay. One point you glossed over a little bit more than you should have: when you choose $alpha$, you should specify that it is an arbitrary assignment in $A$, and later on, when you use $Bpreceq C$, you should deduce that it is therefore an assignment in $B$ (so you can apply $Bpreceq C$). Just make sure you apply the hypothesis directly, and how you apply it. That should remove all doubt about correctness.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The proof seems okay. One point you glossed over a little bit more than you should have: when you choose $alpha$, you should specify that it is an arbitrary assignment in $A$, and later on, when you use $Bpreceq C$, you should deduce that it is therefore an assignment in $B$ (so you can apply $Bpreceq C$). Just make sure you apply the hypothesis directly, and how you apply it. That should remove all doubt about correctness.






        share|cite|improve this answer









        $endgroup$



        The proof seems okay. One point you glossed over a little bit more than you should have: when you choose $alpha$, you should specify that it is an arbitrary assignment in $A$, and later on, when you use $Bpreceq C$, you should deduce that it is therefore an assignment in $B$ (so you can apply $Bpreceq C$). Just make sure you apply the hypothesis directly, and how you apply it. That should remove all doubt about correctness.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 '18 at 19:40









        tomasztomasz

        23.5k23378




        23.5k23378






























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