proof theoretic ordinal for Robinson's arithmetic
$begingroup$
Does a theory like Robinson's arithmetic have a proof-theoretic ordinal? If so, what is it?
logic ordinal-analysis
edited Dec 6 '18 at 4:32
Andrés E. Caicedo
65.3k8158247
asked Oct 19 '16 at 21:03
Cecil BurrowCecil Burrow
362
$endgroup$
add a comment |
$begingroup$
Does a theory like Robinson's arithmetic have a proof-theoretic ordinal? If so, what is it?
logic ordinal-analysis
edited Dec 6 '18 at 4:32
Andrés E. Caicedo
65.3k8158247
asked Oct 19 '16 at 21:03
Cecil BurrowCecil Burrow
362
$endgroup$
add a comment |
$begingroup$
Does a theory like Robinson's arithmetic have a proof-theoretic ordinal? If so, what is it?
logic ordinal-analysis
edited Dec 6 '18 at 4:32
Andrés E. Caicedo
65.3k8158247
asked Oct 19 '16 at 21:03
Cecil BurrowCecil Burrow
362
$endgroup$
Does a theory like Robinson's arithmetic have a proof-theoretic ordinal? If so, what is it?
logic ordinal-analysis
logic ordinal-analysis
edited Dec 6 '18 at 4:32
Andrés E. Caicedo
65.3k8158247
asked Oct 19 '16 at 21:03
Cecil BurrowCecil Burrow
362
edited Dec 6 '18 at 4:32
Andrés E. Caicedo
65.3k8158247
asked Oct 19 '16 at 21:03
Cecil BurrowCecil Burrow
362
edited Dec 6 '18 at 4:32
Andrés E. Caicedo
65.3k8158247
edited Dec 6 '18 at 4:32
Andrés E. Caicedo
65.3k8158247
edited Dec 6 '18 at 4:32
Andrés E. Caicedo
65.3k8158247
65.3k8158247
asked Oct 19 '16 at 21:03
Cecil BurrowCecil Burrow
362
asked Oct 19 '16 at 21:03
Cecil BurrowCecil Burrow
362
asked Oct 19 '16 at 21:03
Cecil BurrowCecil Burrow
362
362
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Well, since $Q$ doesn't even have induction along $mathbb{N}$, I'm dubious that the notion of proof-theoretic ordinal makes sense for it; but if forced, I'd say the answer has to be $omega$. Induction along finite orderings is trivial, so $omega$ is the first ordinal for which it makes sense to ask "Does $Q$ prove induction along this ordinal?," and $Q$ doesn't.
answered Oct 20 '16 at 0:43
Noah SchweberNoah Schweber
123k10150286
$endgroup$
$begingroup$
Well, I guess I was hoping for some ordinal $alpha$ such that the principlw of transfinite induction up to $alpha$ plus some minimal theory (e.g., PRA) proves Con(Q).
$endgroup$
– Cecil Burrow
Oct 20 '16 at 3:10
$begingroup$
@CecilBurrow I believe PRA already proves Con(Q), but I could be wrong. Note that, interestingly, EFA does not prove Con(Q) (see section 3 of this message on the FOM mailing list); Friedman cites Paris and Wilkie for that.
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:19
$begingroup$
I don't know of any proof that PRA proves Con(Q) - a reference would be wonderful!
$endgroup$
– Cecil Burrow
Oct 20 '16 at 3:38
$begingroup$
@CecilBurrow Hm, I'm having trouble finding a source. Section 3H of Moschovakis' notes is where I remember the fact cropping up, but it's not there (although it may still be interesting to you). I'll keep looking . . .
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:44
$begingroup$
On the other hand, the proof theoretic ordinal of PRA is $omega^omega$, while the proof theoretic ordinal of EFA is just $omega^3$. So that suggests that PRA proves Con(EFA) and hence Con(Q) . . .
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:46
add a comment |
$begingroup$
Noah is correct that PRA $vdash$ Con(Q): On p. 139 of the Handbook of Proof Theory, Chapter 2 (available here), Buss shows that $IDelta_0$ + "tetration is total" $vdash$ Con($IDelta_0$), so it certainly proves Con(Q). Since $IDelta_0 subseteq$ PRA and PRA proves tetration total, the result follows.
(This is in response to a side question that came up in the comment thread of Noah's answer, but I don't have the karma to post this as a comment).
answered Sep 24 '18 at 4:11
Morgan SinclaireMorgan Sinclaire
385
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
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$begingroup$
Well, since $Q$ doesn't even have induction along $mathbb{N}$, I'm dubious that the notion of proof-theoretic ordinal makes sense for it; but if forced, I'd say the answer has to be $omega$. Induction along finite orderings is trivial, so $omega$ is the first ordinal for which it makes sense to ask "Does $Q$ prove induction along this ordinal?," and $Q$ doesn't.
answered Oct 20 '16 at 0:43
Noah SchweberNoah Schweber
123k10150286
$endgroup$
$begingroup$
Well, I guess I was hoping for some ordinal $alpha$ such that the principlw of transfinite induction up to $alpha$ plus some minimal theory (e.g., PRA) proves Con(Q).
$endgroup$
– Cecil Burrow
Oct 20 '16 at 3:10
$begingroup$
@CecilBurrow I believe PRA already proves Con(Q), but I could be wrong. Note that, interestingly, EFA does not prove Con(Q) (see section 3 of this message on the FOM mailing list); Friedman cites Paris and Wilkie for that.
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:19
$begingroup$
I don't know of any proof that PRA proves Con(Q) - a reference would be wonderful!
$endgroup$
– Cecil Burrow
Oct 20 '16 at 3:38
$begingroup$
@CecilBurrow Hm, I'm having trouble finding a source. Section 3H of Moschovakis' notes is where I remember the fact cropping up, but it's not there (although it may still be interesting to you). I'll keep looking . . .
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:44
$begingroup$
On the other hand, the proof theoretic ordinal of PRA is $omega^omega$, while the proof theoretic ordinal of EFA is just $omega^3$. So that suggests that PRA proves Con(EFA) and hence Con(Q) . . .
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:46
add a comment |
$begingroup$
Well, since $Q$ doesn't even have induction along $mathbb{N}$, I'm dubious that the notion of proof-theoretic ordinal makes sense for it; but if forced, I'd say the answer has to be $omega$. Induction along finite orderings is trivial, so $omega$ is the first ordinal for which it makes sense to ask "Does $Q$ prove induction along this ordinal?," and $Q$ doesn't.
answered Oct 20 '16 at 0:43
Noah SchweberNoah Schweber
123k10150286
$endgroup$
$begingroup$
Well, I guess I was hoping for some ordinal $alpha$ such that the principlw of transfinite induction up to $alpha$ plus some minimal theory (e.g., PRA) proves Con(Q).
$endgroup$
– Cecil Burrow
Oct 20 '16 at 3:10
$begingroup$
@CecilBurrow I believe PRA already proves Con(Q), but I could be wrong. Note that, interestingly, EFA does not prove Con(Q) (see section 3 of this message on the FOM mailing list); Friedman cites Paris and Wilkie for that.
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:19
$begingroup$
I don't know of any proof that PRA proves Con(Q) - a reference would be wonderful!
$endgroup$
– Cecil Burrow
Oct 20 '16 at 3:38
$begingroup$
@CecilBurrow Hm, I'm having trouble finding a source. Section 3H of Moschovakis' notes is where I remember the fact cropping up, but it's not there (although it may still be interesting to you). I'll keep looking . . .
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:44
$begingroup$
On the other hand, the proof theoretic ordinal of PRA is $omega^omega$, while the proof theoretic ordinal of EFA is just $omega^3$. So that suggests that PRA proves Con(EFA) and hence Con(Q) . . .
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:46
add a comment |
$begingroup$
Well, since $Q$ doesn't even have induction along $mathbb{N}$, I'm dubious that the notion of proof-theoretic ordinal makes sense for it; but if forced, I'd say the answer has to be $omega$. Induction along finite orderings is trivial, so $omega$ is the first ordinal for which it makes sense to ask "Does $Q$ prove induction along this ordinal?," and $Q$ doesn't.
answered Oct 20 '16 at 0:43
Noah SchweberNoah Schweber
123k10150286
$endgroup$
Well, since $Q$ doesn't even have induction along $mathbb{N}$, I'm dubious that the notion of proof-theoretic ordinal makes sense for it; but if forced, I'd say the answer has to be $omega$. Induction along finite orderings is trivial, so $omega$ is the first ordinal for which it makes sense to ask "Does $Q$ prove induction along this ordinal?," and $Q$ doesn't.
answered Oct 20 '16 at 0:43
Noah SchweberNoah Schweber
123k10150286
answered Oct 20 '16 at 0:43
Noah SchweberNoah Schweber
123k10150286
answered Oct 20 '16 at 0:43
Noah SchweberNoah Schweber
123k10150286
answered Oct 20 '16 at 0:43
Noah SchweberNoah Schweber
123k10150286
123k10150286
$begingroup$
Well, I guess I was hoping for some ordinal $alpha$ such that the principlw of transfinite induction up to $alpha$ plus some minimal theory (e.g., PRA) proves Con(Q).
$endgroup$
– Cecil Burrow
Oct 20 '16 at 3:10
$begingroup$
@CecilBurrow I believe PRA already proves Con(Q), but I could be wrong. Note that, interestingly, EFA does not prove Con(Q) (see section 3 of this message on the FOM mailing list); Friedman cites Paris and Wilkie for that.
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:19
$begingroup$
I don't know of any proof that PRA proves Con(Q) - a reference would be wonderful!
$endgroup$
– Cecil Burrow
Oct 20 '16 at 3:38
$begingroup$
@CecilBurrow Hm, I'm having trouble finding a source. Section 3H of Moschovakis' notes is where I remember the fact cropping up, but it's not there (although it may still be interesting to you). I'll keep looking . . .
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:44
$begingroup$
On the other hand, the proof theoretic ordinal of PRA is $omega^omega$, while the proof theoretic ordinal of EFA is just $omega^3$. So that suggests that PRA proves Con(EFA) and hence Con(Q) . . .
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:46
add a comment |
$begingroup$
Well, I guess I was hoping for some ordinal $alpha$ such that the principlw of transfinite induction up to $alpha$ plus some minimal theory (e.g., PRA) proves Con(Q).
$endgroup$
– Cecil Burrow
Oct 20 '16 at 3:10
$begingroup$
@CecilBurrow I believe PRA already proves Con(Q), but I could be wrong. Note that, interestingly, EFA does not prove Con(Q) (see section 3 of this message on the FOM mailing list); Friedman cites Paris and Wilkie for that.
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:19
$begingroup$
I don't know of any proof that PRA proves Con(Q) - a reference would be wonderful!
$endgroup$
– Cecil Burrow
Oct 20 '16 at 3:38
$begingroup$
@CecilBurrow Hm, I'm having trouble finding a source. Section 3H of Moschovakis' notes is where I remember the fact cropping up, but it's not there (although it may still be interesting to you). I'll keep looking . . .
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:44
$begingroup$
On the other hand, the proof theoretic ordinal of PRA is $omega^omega$, while the proof theoretic ordinal of EFA is just $omega^3$. So that suggests that PRA proves Con(EFA) and hence Con(Q) . . .
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:46
$begingroup$
Well, I guess I was hoping for some ordinal $alpha$ such that the principlw of transfinite induction up to $alpha$ plus some minimal theory (e.g., PRA) proves Con(Q).
$endgroup$
– Cecil Burrow
Oct 20 '16 at 3:10
$begingroup$
Well, I guess I was hoping for some ordinal $alpha$ such that the principlw of transfinite induction up to $alpha$ plus some minimal theory (e.g., PRA) proves Con(Q).
$endgroup$
– Cecil Burrow
Oct 20 '16 at 3:10
$begingroup$
@CecilBurrow I believe PRA already proves Con(Q), but I could be wrong. Note that, interestingly, EFA does not prove Con(Q) (see section 3 of this message on the FOM mailing list); Friedman cites Paris and Wilkie for that.
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:19
$begingroup$
@CecilBurrow I believe PRA already proves Con(Q), but I could be wrong. Note that, interestingly, EFA does not prove Con(Q) (see section 3 of this message on the FOM mailing list); Friedman cites Paris and Wilkie for that.
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:19
$begingroup$
I don't know of any proof that PRA proves Con(Q) - a reference would be wonderful!
$endgroup$
– Cecil Burrow
Oct 20 '16 at 3:38
$begingroup$
I don't know of any proof that PRA proves Con(Q) - a reference would be wonderful!
$endgroup$
– Cecil Burrow
Oct 20 '16 at 3:38
$begingroup$
@CecilBurrow Hm, I'm having trouble finding a source. Section 3H of Moschovakis' notes is where I remember the fact cropping up, but it's not there (although it may still be interesting to you). I'll keep looking . . .
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:44
$begingroup$
@CecilBurrow Hm, I'm having trouble finding a source. Section 3H of Moschovakis' notes is where I remember the fact cropping up, but it's not there (although it may still be interesting to you). I'll keep looking . . .
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:44
$begingroup$
On the other hand, the proof theoretic ordinal of PRA is $omega^omega$, while the proof theoretic ordinal of EFA is just $omega^3$. So that suggests that PRA proves Con(EFA) and hence Con(Q) . . .
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:46
$begingroup$
On the other hand, the proof theoretic ordinal of PRA is $omega^omega$, while the proof theoretic ordinal of EFA is just $omega^3$. So that suggests that PRA proves Con(EFA) and hence Con(Q) . . .
$endgroup$
– Noah Schweber
Oct 20 '16 at 3:46
add a comment |
$begingroup$
Noah is correct that PRA $vdash$ Con(Q): On p. 139 of the Handbook of Proof Theory, Chapter 2 (available here), Buss shows that $IDelta_0$ + "tetration is total" $vdash$ Con($IDelta_0$), so it certainly proves Con(Q). Since $IDelta_0 subseteq$ PRA and PRA proves tetration total, the result follows.
(This is in response to a side question that came up in the comment thread of Noah's answer, but I don't have the karma to post this as a comment).
answered Sep 24 '18 at 4:11
Morgan SinclaireMorgan Sinclaire
385
$endgroup$
add a comment |
$begingroup$
Noah is correct that PRA $vdash$ Con(Q): On p. 139 of the Handbook of Proof Theory, Chapter 2 (available here), Buss shows that $IDelta_0$ + "tetration is total" $vdash$ Con($IDelta_0$), so it certainly proves Con(Q). Since $IDelta_0 subseteq$ PRA and PRA proves tetration total, the result follows.
(This is in response to a side question that came up in the comment thread of Noah's answer, but I don't have the karma to post this as a comment).
answered Sep 24 '18 at 4:11
Morgan SinclaireMorgan Sinclaire
385
$endgroup$
add a comment |
$begingroup$
Noah is correct that PRA $vdash$ Con(Q): On p. 139 of the Handbook of Proof Theory, Chapter 2 (available here), Buss shows that $IDelta_0$ + "tetration is total" $vdash$ Con($IDelta_0$), so it certainly proves Con(Q). Since $IDelta_0 subseteq$ PRA and PRA proves tetration total, the result follows.
(This is in response to a side question that came up in the comment thread of Noah's answer, but I don't have the karma to post this as a comment).
answered Sep 24 '18 at 4:11
Morgan SinclaireMorgan Sinclaire
385
$endgroup$
Noah is correct that PRA $vdash$ Con(Q): On p. 139 of the Handbook of Proof Theory, Chapter 2 (available here), Buss shows that $IDelta_0$ + "tetration is total" $vdash$ Con($IDelta_0$), so it certainly proves Con(Q). Since $IDelta_0 subseteq$ PRA and PRA proves tetration total, the result follows.
(This is in response to a side question that came up in the comment thread of Noah's answer, but I don't have the karma to post this as a comment).
answered Sep 24 '18 at 4:11
Morgan SinclaireMorgan Sinclaire
385
answered Sep 24 '18 at 4:11
Morgan SinclaireMorgan Sinclaire
385
answered Sep 24 '18 at 4:11
Morgan SinclaireMorgan Sinclaire
385
answered Sep 24 '18 at 4:11
Morgan SinclaireMorgan Sinclaire
385
385
add a comment |
add a comment |
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