-1
$\begingroup$

Please explain the error in my reasoning with this:

Let T be a formal theory within which Godel's first incompleteness theorem can be proved. In other words... suppose when we write the proof of godel's incompleteness theorem we are doing so within theory T. And suppose theory T is also of the kind Godel's incompleteness theorem refers to... ie: formal theory capable of doing arithmetic etc...

Can such a theory T as described above exist?

If it is possible... then by proving godel's incompleteness theorem within T... we prove that the Godel sentence for T is true... we also prove that the godel sentence is true and unprovable since the theorem applies to theory T.... a contradiction.

So my question is... about the theory within which godel's incompleteness theorem itself is proved... if the theorem is proved within a theory T... then is there something that stipulates the proof cannot itself be applied to theory T...

So when Godel's incompleteness theorem is proved... it applies to all formal theories except the one within which the theorem itself is proved?

Or is the writing of the proof of godel's incompleteness theorem done outside something referred to as a "formal theory capable of arithmetic" ?

Appreciate any help with this.

|cite|improve this question
$\endgroup$
2
$\begingroup$

There are many theories $T$ in which one can prove Gödel's first incompleteness theorem for $T$, but you have to be careful about the statement of the theorem. The form of the theorem relevant to your question says that if $T$ is consistent, then its Gödel sentence $g$ is true and unprovable in $T$. So, if $T$ is consistent, then $g$ cannot be proved in $T$, but it can be proved in the stronger theory obtained by adding to $T$ the additional hypothesis that $T$ is consistent. The conclusion to draw from this is that the second, stronger theory here is genuinely stronger; the consistency of $T$ cannot be proved in $T$. That's essentially how one proves (after filling in a lot of details) the second incompleteness theorem.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks Andreas... so it looks to me like the proof of godel's incompleteness theorem for T cannot be done from within T? $\endgroup$ – Ameet Sharma Apr 23 '15 at 22:04
  • $\begingroup$ @AmeetSharma If you formulate the theorem for $T$ as I did, including the hypothesis that $T$ is consistent, then it can be proved within $T$ when $T$ is, for example, Peano arithmetic, or Zermelo-Fraenkel set theory. But if you formulate the theorem without the hypothesis of consistency, then it cannot be proved within $T$. $\endgroup$ – Andreas Blass Apr 23 '15 at 22:06
  • $\begingroup$ So suppose T is consistent... and we prove godel's incomploeteness theorem about T inside T... So we just proved that g is true and unprovable in T... but we get a contradiction... we just proved g is true within T, and we also proved g is unproveable within T.... so our initial assumption is wrong and T must be inconsistent? $\endgroup$ – Ameet Sharma Apr 23 '15 at 22:10
  • $\begingroup$ No, you do not prove "$g$ is true but unprovable" within $T$. You prove it within the stronger theory that contains, in addition to $T$, your supposition that $T$ is consistent. $\endgroup$ – Andreas Blass Apr 23 '15 at 22:12
  • $\begingroup$ Your confusion may be due to overlooking the fact that, even if $T$ is consistent, that does not make the theory $T$ the same as the theory "$T$ and '$T$ is consistent'." $\endgroup$ – Andreas Blass Apr 23 '15 at 22:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.