Prove that $mathcal{A}preceq mathcal{B}$.
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
asked Nov 29 at 2:02
numericalorange
1,719311
|
show 1 more comment
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
asked Nov 29 at 2:02
numericalorange
1,719311
Can you define $preceq, mathcal L$-formula, assignment and $models$?
– Jimmy R.
Nov 29 at 2:04
4
@Jimmy That seems unreasonable; it would make the question too long. These are basic model-theoretic concepts.
– Andrés E. Caicedo
Nov 29 at 2:08
3
Yes, the proof is fine.
– Andrés E. Caicedo
Nov 29 at 2:09
3
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.
– Henning Makholm
Nov 29 at 2:21
@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...
– numericalorange
Nov 29 at 2:23
|
show 1 more comment
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
asked Nov 29 at 2:02
numericalorange
1,719311
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
asked Nov 29 at 2:02
numericalorange
1,719311
asked Nov 29 at 2:02
numericalorange
1,719311
asked Nov 29 at 2:02
numericalorange
1,719311
asked Nov 29 at 2:02
numericalorange
1,719311
asked Nov 29 at 2:02
numericalorange
1,719311
1,719311
Can you define $preceq, mathcal L$-formula, assignment and $models$?
– Jimmy R.
Nov 29 at 2:04
4
@Jimmy That seems unreasonable; it would make the question too long. These are basic model-theoretic concepts.
– Andrés E. Caicedo
Nov 29 at 2:08
3
Yes, the proof is fine.
– Andrés E. Caicedo
Nov 29 at 2:09
3
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.
– Henning Makholm
Nov 29 at 2:21
@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...
– numericalorange
Nov 29 at 2:23
|
show 1 more comment
Can you define $preceq, mathcal L$-formula, assignment and $models$?
– Jimmy R.
Nov 29 at 2:04
4
@Jimmy That seems unreasonable; it would make the question too long. These are basic model-theoretic concepts.
– Andrés E. Caicedo
Nov 29 at 2:08
3
Yes, the proof is fine.
– Andrés E. Caicedo
Nov 29 at 2:09
3
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.
– Henning Makholm
Nov 29 at 2:21
@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...
– numericalorange
Nov 29 at 2:23
Can you define $preceq, mathcal L$-formula, assignment and $models$?
– Jimmy R.
Nov 29 at 2:04
Can you define $preceq, mathcal L$-formula, assignment and $models$?
– Jimmy R.
Nov 29 at 2:04
4
4
@Jimmy That seems unreasonable; it would make the question too long. These are basic model-theoretic concepts.
– Andrés E. Caicedo
Nov 29 at 2:08
@Jimmy That seems unreasonable; it would make the question too long. These are basic model-theoretic concepts.
– Andrés E. Caicedo
Nov 29 at 2:08
3
3
Yes, the proof is fine.
– Andrés E. Caicedo
Nov 29 at 2:09
Yes, the proof is fine.
– Andrés E. Caicedo
Nov 29 at 2:09
3
3
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.
– Henning Makholm
Nov 29 at 2:21
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.
– Henning Makholm
Nov 29 at 2:21
@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...
– numericalorange
Nov 29 at 2:23
@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...
– numericalorange
Nov 29 at 2:23
|
show 1 more comment
1 Answer
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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.
answered Dec 1 at 19:40
tomasz
23.4k23178
add a comment |
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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.
answered Dec 1 at 19:40
tomasz
23.4k23178
add a comment |
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.
answered Dec 1 at 19:40
tomasz
23.4k23178
add a comment |
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.
answered Dec 1 at 19:40
tomasz
23.4k23178
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.
answered Dec 1 at 19:40
tomasz
23.4k23178
answered Dec 1 at 19:40
tomasz
23.4k23178
answered Dec 1 at 19:40
tomasz
23.4k23178
answered Dec 1 at 19:40
tomasz
23.4k23178
23.4k23178
add a comment |
add a comment |
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Can you define $preceq, mathcal L$-formula, assignment and $models$?
– Jimmy R.
Nov 29 at 2:04
4
@Jimmy That seems unreasonable; it would make the question too long. These are basic model-theoretic concepts.
– Andrés E. Caicedo
Nov 29 at 2:08
3
Yes, the proof is fine.
– Andrés E. Caicedo
Nov 29 at 2:09
3
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.
– Henning Makholm
Nov 29 at 2:21
@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...
– numericalorange
Nov 29 at 2:23