What is the Turing degree of Truth?
$begingroup$
First of all by Truth I mean the set $T$ of the Gôdel numbers of the true formulas of first order arithmetic.
First order arithmetic is not decidable and $T$ is not decidable, additionally
the undefinability theorem says that $T$ can not be expressed in first order logic, so $T$ does not belong to the arithmetical hierarchy and his Turing degree is at least $omega$.
What is then the Turing degree of $T$?
Is there a good textbook that somebody can suggest where this kind of subjects are explained thoroughly?
logic arithmetic computability foundations
edited Jan 11 at 20:54
Bernard
119k740113
asked Jan 11 at 20:47
holmesholmes
664
$endgroup$
add a comment |
$begingroup$
First of all by Truth I mean the set $T$ of the Gôdel numbers of the true formulas of first order arithmetic.
First order arithmetic is not decidable and $T$ is not decidable, additionally
the undefinability theorem says that $T$ can not be expressed in first order logic, so $T$ does not belong to the arithmetical hierarchy and his Turing degree is at least $omega$.
What is then the Turing degree of $T$?
Is there a good textbook that somebody can suggest where this kind of subjects are explained thoroughly?
logic arithmetic computability foundations
edited Jan 11 at 20:54
Bernard
119k740113
asked Jan 11 at 20:47
holmesholmes
664
$endgroup$
$begingroup$
See Turing degree for bibliography.
$endgroup$
– Mauro ALLEGRANZA
Jan 11 at 21:16
$begingroup$
Indeed its degree is $0^{(omega)}$, the Turing join of the $0^{(n)}$.
$endgroup$
– Andrés E. Caicedo
Jan 11 at 22:07
add a comment |
$begingroup$
First of all by Truth I mean the set $T$ of the Gôdel numbers of the true formulas of first order arithmetic.
First order arithmetic is not decidable and $T$ is not decidable, additionally
the undefinability theorem says that $T$ can not be expressed in first order logic, so $T$ does not belong to the arithmetical hierarchy and his Turing degree is at least $omega$.
What is then the Turing degree of $T$?
Is there a good textbook that somebody can suggest where this kind of subjects are explained thoroughly?
logic arithmetic computability foundations
edited Jan 11 at 20:54
Bernard
119k740113
asked Jan 11 at 20:47
holmesholmes
664
$endgroup$
First of all by Truth I mean the set $T$ of the Gôdel numbers of the true formulas of first order arithmetic.
First order arithmetic is not decidable and $T$ is not decidable, additionally
the undefinability theorem says that $T$ can not be expressed in first order logic, so $T$ does not belong to the arithmetical hierarchy and his Turing degree is at least $omega$.
What is then the Turing degree of $T$?
Is there a good textbook that somebody can suggest where this kind of subjects are explained thoroughly?
logic arithmetic computability foundations
logic arithmetic computability foundations
edited Jan 11 at 20:54
Bernard
119k740113
asked Jan 11 at 20:47
holmesholmes
664
edited Jan 11 at 20:54
Bernard
119k740113
asked Jan 11 at 20:47
holmesholmes
664
edited Jan 11 at 20:54
Bernard
119k740113
edited Jan 11 at 20:54
Bernard
119k740113
edited Jan 11 at 20:54
Bernard
119k740113
119k740113
asked Jan 11 at 20:47
holmesholmes
664
asked Jan 11 at 20:47
holmesholmes
664
asked Jan 11 at 20:47
holmesholmes
664
664
$begingroup$
See Turing degree for bibliography.
$endgroup$
– Mauro ALLEGRANZA
Jan 11 at 21:16
$begingroup$
Indeed its degree is $0^{(omega)}$, the Turing join of the $0^{(n)}$.
$endgroup$
– Andrés E. Caicedo
Jan 11 at 22:07
add a comment |
$begingroup$
See Turing degree for bibliography.
$endgroup$
– Mauro ALLEGRANZA
Jan 11 at 21:16
$begingroup$
Indeed its degree is $0^{(omega)}$, the Turing join of the $0^{(n)}$.
$endgroup$
– Andrés E. Caicedo
Jan 11 at 22:07
$begingroup$
See Turing degree for bibliography.
$endgroup$
– Mauro ALLEGRANZA
Jan 11 at 21:16
$begingroup$
See Turing degree for bibliography.
$endgroup$
– Mauro ALLEGRANZA
Jan 11 at 21:16
$begingroup$
Indeed its degree is $0^{(omega)}$, the Turing join of the $0^{(n)}$.
$endgroup$
– Andrés E. Caicedo
Jan 11 at 22:07
$begingroup$
Indeed its degree is $0^{(omega)}$, the Turing join of the $0^{(n)}$.
$endgroup$
– Andrés E. Caicedo
Jan 11 at 22:07
add a comment |
1 Answer
1
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oldest
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$begingroup$
$T$ does not belong to the arithmetical hierarchy and his Turing degree is at least $omega$.
A couple quick comments:
I assume "$omega$" should be "$0^{(omega)}$" - in which case that's correct - since $omega$ isn't a Turing degree.
More substantively, there's a serious issue in your reasoning. You're right that $T$ can't be arithmetic. However, that is not sufficient to conclude that $Tge_T0^{(omega)}$: there are Turing degrees incomparable with $0^{(omega)}$ (and hence not arithmetic or $ge_T0^{(omega)}$), and even Turing degrees above every arithmetic set which are incomparable with $0^{(omega)}$ (this second fact is an instance of the more general exact pair theorem).
To show that $Tge_T0^{(omega)}$, you need to argue that given $langle a,brangleinmathbb{N}^2$ you can find - computably - a sentence $varphi$ which is true iff $bin 0^{(a)}$. You can do that via Kleene's $T$-predicate. Conversely, to show $0^{(omega)}ge_TT$ you just show that uniformly the $Sigma_n$-theory of $mathbb{N}$ has Turing degree $0^{(n)}$. Soare's book is a good source on this material.
answered Jan 11 at 22:12
Noah SchweberNoah Schweber
123k10150285
$endgroup$
add a comment |
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$begingroup$
$T$ does not belong to the arithmetical hierarchy and his Turing degree is at least $omega$.
A couple quick comments:
I assume "$omega$" should be "$0^{(omega)}$" - in which case that's correct - since $omega$ isn't a Turing degree.
More substantively, there's a serious issue in your reasoning. You're right that $T$ can't be arithmetic. However, that is not sufficient to conclude that $Tge_T0^{(omega)}$: there are Turing degrees incomparable with $0^{(omega)}$ (and hence not arithmetic or $ge_T0^{(omega)}$), and even Turing degrees above every arithmetic set which are incomparable with $0^{(omega)}$ (this second fact is an instance of the more general exact pair theorem).
To show that $Tge_T0^{(omega)}$, you need to argue that given $langle a,brangleinmathbb{N}^2$ you can find - computably - a sentence $varphi$ which is true iff $bin 0^{(a)}$. You can do that via Kleene's $T$-predicate. Conversely, to show $0^{(omega)}ge_TT$ you just show that uniformly the $Sigma_n$-theory of $mathbb{N}$ has Turing degree $0^{(n)}$. Soare's book is a good source on this material.
answered Jan 11 at 22:12
Noah SchweberNoah Schweber
123k10150285
$endgroup$
add a comment |
$begingroup$
$T$ does not belong to the arithmetical hierarchy and his Turing degree is at least $omega$.
A couple quick comments:
I assume "$omega$" should be "$0^{(omega)}$" - in which case that's correct - since $omega$ isn't a Turing degree.
More substantively, there's a serious issue in your reasoning. You're right that $T$ can't be arithmetic. However, that is not sufficient to conclude that $Tge_T0^{(omega)}$: there are Turing degrees incomparable with $0^{(omega)}$ (and hence not arithmetic or $ge_T0^{(omega)}$), and even Turing degrees above every arithmetic set which are incomparable with $0^{(omega)}$ (this second fact is an instance of the more general exact pair theorem).
To show that $Tge_T0^{(omega)}$, you need to argue that given $langle a,brangleinmathbb{N}^2$ you can find - computably - a sentence $varphi$ which is true iff $bin 0^{(a)}$. You can do that via Kleene's $T$-predicate. Conversely, to show $0^{(omega)}ge_TT$ you just show that uniformly the $Sigma_n$-theory of $mathbb{N}$ has Turing degree $0^{(n)}$. Soare's book is a good source on this material.
answered Jan 11 at 22:12
Noah SchweberNoah Schweber
123k10150285
$endgroup$
add a comment |
$begingroup$
$T$ does not belong to the arithmetical hierarchy and his Turing degree is at least $omega$.
A couple quick comments:
I assume "$omega$" should be "$0^{(omega)}$" - in which case that's correct - since $omega$ isn't a Turing degree.
More substantively, there's a serious issue in your reasoning. You're right that $T$ can't be arithmetic. However, that is not sufficient to conclude that $Tge_T0^{(omega)}$: there are Turing degrees incomparable with $0^{(omega)}$ (and hence not arithmetic or $ge_T0^{(omega)}$), and even Turing degrees above every arithmetic set which are incomparable with $0^{(omega)}$ (this second fact is an instance of the more general exact pair theorem).
To show that $Tge_T0^{(omega)}$, you need to argue that given $langle a,brangleinmathbb{N}^2$ you can find - computably - a sentence $varphi$ which is true iff $bin 0^{(a)}$. You can do that via Kleene's $T$-predicate. Conversely, to show $0^{(omega)}ge_TT$ you just show that uniformly the $Sigma_n$-theory of $mathbb{N}$ has Turing degree $0^{(n)}$. Soare's book is a good source on this material.
answered Jan 11 at 22:12
Noah SchweberNoah Schweber
123k10150285
$endgroup$
$T$ does not belong to the arithmetical hierarchy and his Turing degree is at least $omega$.
A couple quick comments:
I assume "$omega$" should be "$0^{(omega)}$" - in which case that's correct - since $omega$ isn't a Turing degree.
More substantively, there's a serious issue in your reasoning. You're right that $T$ can't be arithmetic. However, that is not sufficient to conclude that $Tge_T0^{(omega)}$: there are Turing degrees incomparable with $0^{(omega)}$ (and hence not arithmetic or $ge_T0^{(omega)}$), and even Turing degrees above every arithmetic set which are incomparable with $0^{(omega)}$ (this second fact is an instance of the more general exact pair theorem).
To show that $Tge_T0^{(omega)}$, you need to argue that given $langle a,brangleinmathbb{N}^2$ you can find - computably - a sentence $varphi$ which is true iff $bin 0^{(a)}$. You can do that via Kleene's $T$-predicate. Conversely, to show $0^{(omega)}ge_TT$ you just show that uniformly the $Sigma_n$-theory of $mathbb{N}$ has Turing degree $0^{(n)}$. Soare's book is a good source on this material.
answered Jan 11 at 22:12
Noah SchweberNoah Schweber
123k10150285
answered Jan 11 at 22:12
Noah SchweberNoah Schweber
123k10150285
answered Jan 11 at 22:12
Noah SchweberNoah Schweber
123k10150285
answered Jan 11 at 22:12
Noah SchweberNoah Schweber
123k10150285
123k10150285
add a comment |
add a comment |
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$begingroup$
See Turing degree for bibliography.
$endgroup$
– Mauro ALLEGRANZA
Jan 11 at 21:16
$begingroup$
Indeed its degree is $0^{(omega)}$, the Turing join of the $0^{(n)}$.
$endgroup$
– Andrés E. Caicedo
Jan 11 at 22:07