Are there any statements $phi$ where “$vdash phi$ or $vdash neg phi$” was shown nonconstructively?
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I am wondering if there is an example of a statement $phi$ in the language of some formal system $T$ satisfying (1-4):
$phi$ was shown to not be independent of $T$ (i.e. it was proved that $T vdash phi$ or $T vdash neg phi$)- The proof in (1) was nonconstructive: it was initially (or still) unknown which of {$phi, neg phi$} was a theorem of $T$.
$phi$ was not clearly $Delta_0$ (any bounded statement like "there is no odd perfect number less than 10↑↑↑10" has a (possibly long) proof/disproof).
$phi$ was not part of some class of statements $C$ that were shown by some algorithm to be decidable. (For instance, any statement in Presburger arithmetic or ACF/RCF which we know to be decidable by a quantifier elimination algorithm).
$T$ is r.e.
logic examples-counterexamples proof-theory
asked Nov 16 at 19:58
Morgan Sinclaire
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up vote
0
down vote
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I am wondering if there is an example of a statement $phi$ in the language of some formal system $T$ satisfying (1-4):
$phi$ was shown to not be independent of $T$ (i.e. it was proved that $T vdash phi$ or $T vdash neg phi$)- The proof in (1) was nonconstructive: it was initially (or still) unknown which of {$phi, neg phi$} was a theorem of $T$.
$phi$ was not clearly $Delta_0$ (any bounded statement like "there is no odd perfect number less than 10↑↑↑10" has a (possibly long) proof/disproof).
$phi$ was not part of some class of statements $C$ that were shown by some algorithm to be decidable. (For instance, any statement in Presburger arithmetic or ACF/RCF which we know to be decidable by a quantifier elimination algorithm).
$T$ is r.e.
logic examples-counterexamples proof-theory
asked Nov 16 at 19:58
Morgan Sinclaire
163
1
As stated, the answer is yes and here's a rather silly example: Let $T$ be the theory of the constructible universe $L$ (in some fixed universe $V$ such that $T$ exists) and let $phi$ be Riemann's hypothesis. Since $T$ is complete, we have $T vdash phi$ or $T vdash neg phi$. But we don't yet know which one it is.
– Stefan Mesken
Nov 16 at 20:04
Yes, I thought I forgot a non-triviality condition: $T$ should be r.e. I think that fixes that class of examples?
– Morgan Sinclaire
Nov 16 at 20:35
1
Not really. We know that there is a finite subtheory $T'$ of $T$ that decides Riemann's hypothesis.
– Stefan Mesken
Nov 16 at 20:36
Yeah, but for that finite subtheory, we know exactly which way it decides RH. We want an example of a theory $T$ where we don't know which way $T$ decides $phi$, just that it does. Unless we have access to the theory of $L$ I don't see how to recursively enumerate any $T'$ as you've described it. Unless I'm misunderstanding?
– Morgan Sinclaire
Nov 16 at 20:56
The point is we don't know how $T'$ decides RH (only that it decides RH in the same way as $T$). And we do have access to $T$ and $T'$. The latter, being finite, is trivially r.e.
– Stefan Mesken
Nov 16 at 21:00
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am wondering if there is an example of a statement $phi$ in the language of some formal system $T$ satisfying (1-4):
$phi$ was shown to not be independent of $T$ (i.e. it was proved that $T vdash phi$ or $T vdash neg phi$)- The proof in (1) was nonconstructive: it was initially (or still) unknown which of {$phi, neg phi$} was a theorem of $T$.
$phi$ was not clearly $Delta_0$ (any bounded statement like "there is no odd perfect number less than 10↑↑↑10" has a (possibly long) proof/disproof).
$phi$ was not part of some class of statements $C$ that were shown by some algorithm to be decidable. (For instance, any statement in Presburger arithmetic or ACF/RCF which we know to be decidable by a quantifier elimination algorithm).
$T$ is r.e.
logic examples-counterexamples proof-theory
asked Nov 16 at 19:58
Morgan Sinclaire
163
I am wondering if there is an example of a statement $phi$ in the language of some formal system $T$ satisfying (1-4):
$phi$ was shown to not be independent of $T$ (i.e. it was proved that $T vdash phi$ or $T vdash neg phi$)- The proof in (1) was nonconstructive: it was initially (or still) unknown which of {$phi, neg phi$} was a theorem of $T$.
$phi$ was not clearly $Delta_0$ (any bounded statement like "there is no odd perfect number less than 10↑↑↑10" has a (possibly long) proof/disproof).
$phi$ was not part of some class of statements $C$ that were shown by some algorithm to be decidable. (For instance, any statement in Presburger arithmetic or ACF/RCF which we know to be decidable by a quantifier elimination algorithm).
$T$ is r.e.
logic examples-counterexamples proof-theory
logic examples-counterexamples proof-theory
asked Nov 16 at 19:58
Morgan Sinclaire
163
asked Nov 16 at 19:58
Morgan Sinclaire
163
edited Nov 16 at 22:25
asked Nov 16 at 19:58
Morgan Sinclaire
163
asked Nov 16 at 19:58
Morgan Sinclaire
163
asked Nov 16 at 19:58
Morgan Sinclaire
163
163
1
As stated, the answer is yes and here's a rather silly example: Let $T$ be the theory of the constructible universe $L$ (in some fixed universe $V$ such that $T$ exists) and let $phi$ be Riemann's hypothesis. Since $T$ is complete, we have $T vdash phi$ or $T vdash neg phi$. But we don't yet know which one it is.
– Stefan Mesken
Nov 16 at 20:04
Yes, I thought I forgot a non-triviality condition: $T$ should be r.e. I think that fixes that class of examples?
– Morgan Sinclaire
Nov 16 at 20:35
1
Not really. We know that there is a finite subtheory $T'$ of $T$ that decides Riemann's hypothesis.
– Stefan Mesken
Nov 16 at 20:36
Yeah, but for that finite subtheory, we know exactly which way it decides RH. We want an example of a theory $T$ where we don't know which way $T$ decides $phi$, just that it does. Unless we have access to the theory of $L$ I don't see how to recursively enumerate any $T'$ as you've described it. Unless I'm misunderstanding?
– Morgan Sinclaire
Nov 16 at 20:56
The point is we don't know how $T'$ decides RH (only that it decides RH in the same way as $T$). And we do have access to $T$ and $T'$. The latter, being finite, is trivially r.e.
– Stefan Mesken
Nov 16 at 21:00
|
show 2 more comments
1
As stated, the answer is yes and here's a rather silly example: Let $T$ be the theory of the constructible universe $L$ (in some fixed universe $V$ such that $T$ exists) and let $phi$ be Riemann's hypothesis. Since $T$ is complete, we have $T vdash phi$ or $T vdash neg phi$. But we don't yet know which one it is.
– Stefan Mesken
Nov 16 at 20:04
Yes, I thought I forgot a non-triviality condition: $T$ should be r.e. I think that fixes that class of examples?
– Morgan Sinclaire
Nov 16 at 20:35
1
Not really. We know that there is a finite subtheory $T'$ of $T$ that decides Riemann's hypothesis.
– Stefan Mesken
Nov 16 at 20:36
Yeah, but for that finite subtheory, we know exactly which way it decides RH. We want an example of a theory $T$ where we don't know which way $T$ decides $phi$, just that it does. Unless we have access to the theory of $L$ I don't see how to recursively enumerate any $T'$ as you've described it. Unless I'm misunderstanding?
– Morgan Sinclaire
Nov 16 at 20:56
The point is we don't know how $T'$ decides RH (only that it decides RH in the same way as $T$). And we do have access to $T$ and $T'$. The latter, being finite, is trivially r.e.
– Stefan Mesken
Nov 16 at 21:00
1
1
As stated, the answer is yes and here's a rather silly example: Let $T$ be the theory of the constructible universe $L$ (in some fixed universe $V$ such that $T$ exists) and let $phi$ be Riemann's hypothesis. Since $T$ is complete, we have $T vdash phi$ or $T vdash neg phi$. But we don't yet know which one it is.
– Stefan Mesken
Nov 16 at 20:04
As stated, the answer is yes and here's a rather silly example: Let $T$ be the theory of the constructible universe $L$ (in some fixed universe $V$ such that $T$ exists) and let $phi$ be Riemann's hypothesis. Since $T$ is complete, we have $T vdash phi$ or $T vdash neg phi$. But we don't yet know which one it is.
– Stefan Mesken
Nov 16 at 20:04
Yes, I thought I forgot a non-triviality condition: $T$ should be r.e. I think that fixes that class of examples?
– Morgan Sinclaire
Nov 16 at 20:35
Yes, I thought I forgot a non-triviality condition: $T$ should be r.e. I think that fixes that class of examples?
– Morgan Sinclaire
Nov 16 at 20:35
1
1
Not really. We know that there is a finite subtheory $T'$ of $T$ that decides Riemann's hypothesis.
– Stefan Mesken
Nov 16 at 20:36
Not really. We know that there is a finite subtheory $T'$ of $T$ that decides Riemann's hypothesis.
– Stefan Mesken
Nov 16 at 20:36
Yeah, but for that finite subtheory, we know exactly which way it decides RH. We want an example of a theory $T$ where we don't know which way $T$ decides $phi$, just that it does. Unless we have access to the theory of $L$ I don't see how to recursively enumerate any $T'$ as you've described it. Unless I'm misunderstanding?
– Morgan Sinclaire
Nov 16 at 20:56
Yeah, but for that finite subtheory, we know exactly which way it decides RH. We want an example of a theory $T$ where we don't know which way $T$ decides $phi$, just that it does. Unless we have access to the theory of $L$ I don't see how to recursively enumerate any $T'$ as you've described it. Unless I'm misunderstanding?
– Morgan Sinclaire
Nov 16 at 20:56
The point is we don't know how $T'$ decides RH (only that it decides RH in the same way as $T$). And we do have access to $T$ and $T'$. The latter, being finite, is trivially r.e.
– Stefan Mesken
Nov 16 at 21:00
The point is we don't know how $T'$ decides RH (only that it decides RH in the same way as $T$). And we do have access to $T$ and $T'$. The latter, being finite, is trivially r.e.
– Stefan Mesken
Nov 16 at 21:00
|
show 2 more comments
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1
As stated, the answer is yes and here's a rather silly example: Let $T$ be the theory of the constructible universe $L$ (in some fixed universe $V$ such that $T$ exists) and let $phi$ be Riemann's hypothesis. Since $T$ is complete, we have $T vdash phi$ or $T vdash neg phi$. But we don't yet know which one it is.
– Stefan Mesken
Nov 16 at 20:04
Yes, I thought I forgot a non-triviality condition: $T$ should be r.e. I think that fixes that class of examples?
– Morgan Sinclaire
Nov 16 at 20:35
1
Not really. We know that there is a finite subtheory $T'$ of $T$ that decides Riemann's hypothesis.
– Stefan Mesken
Nov 16 at 20:36
Yeah, but for that finite subtheory, we know exactly which way it decides RH. We want an example of a theory $T$ where we don't know which way $T$ decides $phi$, just that it does. Unless we have access to the theory of $L$ I don't see how to recursively enumerate any $T'$ as you've described it. Unless I'm misunderstanding?
– Morgan Sinclaire
Nov 16 at 20:56
The point is we don't know how $T'$ decides RH (only that it decides RH in the same way as $T$). And we do have access to $T$ and $T'$. The latter, being finite, is trivially r.e.
– Stefan Mesken
Nov 16 at 21:00