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Science Forum Index » Logic Forum » Question about the wiki article about gödels theorems...
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| Bjarne... |
 Posted: Fri Jun 13, 2008 6:40 pm |
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Hey guys
There is something that I don't understand in this article:
http://en.wikipedia.org/wiki/G%C3%B6del's_incompleteness_theorems
In the part about the second incompleteness theorem you can find this:
"Gödel's second incompleteness theorem also implies that a theory T1
satisfying the technical conditions outlined above can't prove the
consistency of any theory T2 which proves the consistency of T1. This
is because then T1 can prove that if T2 proves the consistency of T1,
then T1 is in fact consistent. For the claim that T1 is consistent has
form "for all numbers n, n has the decidable property of not being a
code for a proof of contradiction in T1". If T1 were in fact
inconsistent, then T2 would prove for some n that n is the code of a
contradiction in T1"
I don't understand the last part. I understand that if T1 is
inconsistentent there must be an n, such that n is the Gödel number
for a proof of a contradiction, but how come that the existence of
this n is provable in T2? |
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| Rupert... |
| Posted: Sun Jun 15, 2008 1:06 pm |
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On Jun 15, 11:44 pm, Bjarne <n... at (no spam) no.no> wrote:
Quote: On 2008-06-14, in sci.logic, Rupert wrote:
On Jun 13, 4:40 pm, Bjarne <n... at (no spam) no.no> wrote, quoting Wikipedia:
"Gödel's second incompleteness theorem also implies that a theory T1
satisfying the technical conditions outlined above can't prove the
consistency of any theory T2 which proves the consistency of T1. This
is because then T1 can prove that if T2 proves the consistency of T1,
then T1 is in fact consistent.
This is definitely wrong.
Well, trivially so (and also somewhat confusingly phrased): T1 can
certainly prove the consistency of the theory with "T1 is consistent"
as its sole axiom. What is missing in the above is the requirement
that T2 be T1-provably Sigma-1 -complete.
Thanks, but this all seems a bit complicated. I'm just trying to
understand how the 2nd incompleteness theorem can be used as evidence
for that there is no finitary proof for the consistency of arithmetic
as in Hilbert's 2nd problem.
Can't you just argue, that primitive recursive arithmetic is the
formalization of Hilbert's finitism and that PRA is formalizable in
PA. So any proof of constistency of PA in PRA would also be a proof
in PA, which is impossible. Is this wrong?
No, that's fine. |
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| Rupert... |
| Posted: Mon Jun 16, 2008 3:26 pm |
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On Jun 17, 5:38 am, Bjarne <n... at (no spam) no.no> wrote:
Quote: The complications that perturbed you show that we
can't escape the conclusion e.g. by denying that PRA is the correct
formalisation of finitism or by some such evasive manoeuvre.
I'm not sure what you're saying here... Do you mean, that it is proven
that there is no finitistic proof of consistency of arithmetic? I
thought that there was really no consensus about this?
So long as you grant that PRA is the correct formalisation of
finitism, then it is certainly proven. And I think that is generally
agreed on today. |
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| Bjarne... |
| Posted: Mon Jun 16, 2008 4:38 pm |
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Guest
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Quote: The complications that perturbed you show that we
can't escape the conclusion e.g. by denying that PRA is the correct
formalisation of finitism or by some such evasive manoeuvre.
I'm not sure what you're saying here... Do you mean, that it is proven
that there is no finitistic proof of consistency of arithmetic? I
thought that there was really no consensus about this? |
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