Is metatheory, providing proof of the incompleteness theorem, consistent? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-19T23:04:46Zhttp://mathoverflow.net/feeds/question/119144http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/119144/is-metatheory-providing-proof-of-the-incompleteness-theorem-consistentIs metatheory, providing proof of the incompleteness theorem, consistent?Max2013-01-17T07:43:42Z2013-01-17T15:46:13Z
<p>Is metatheory, considered as a formal system, that used to prove the First Incompleteness Theorem, consistent?</p>
http://mathoverflow.net/questions/119144/is-metatheory-providing-proof-of-the-incompleteness-theorem-consistent/119158#119158Answer by Steven Landsburg for Is metatheory, providing proof of the incompleteness theorem, consistent?Steven Landsburg2013-01-17T10:59:08Z2013-01-17T15:46:13Z<p>Peano Arithmetic suffices to prove Godel's Theorem, and Peano Arithmetic is consistent, so yes.</p>
<p>If you don't believe Peano Arithmetic is consistent, then the question becomes: What <em>do</em> you believe is consistent? As you can see in the accepted answer to <a href="http://mathoverflow.net/questions/118183/what-axioms-are-used-to-prove-godels-incompleteness-theorems" rel="nofollow">this question</a>, you can prove Godel's Theorem in <a href="http://en.wikipedia.org/wiki/Primitive_recursive_arithmetic" rel="nofollow">Primitive Recursive Arithmetic</a>, or even a bit less. If you doubt that PRA is consistent, you'll be forced to doubt so much that any skepticism you might have about Godel's Theorem is probably among the least of your worries.</p>
http://mathoverflow.net/questions/119144/is-metatheory-providing-proof-of-the-incompleteness-theorem-consistent/119159#119159Answer by Emil Jeřábek for Is metatheory, providing proof of the incompleteness theorem, consistent?Emil Jeřábek2013-01-17T11:23:50Z2013-01-17T11:23:50Z<p>As already pointed out by Steven Landsburg, there are plenty of such theories if you stick to conventional mathematics. If you are some sort of an ultrafinitist,
the incompleteness theorem is provable in weak fragments of bounded arithmetic, such as $PV$ or $S^1_2$. These theories are interpretable in Robinson’s arithmetic $Q$, so as long as you accept that $Q$ is consistent, there is such a theory. Do you believe that for every natural numbers $n,m$, the number $\underbrace{2^{2^{\cdot^{\cdot^{\cdot^{2^m}}}}}}_n$ exists? Then $Q$ is consistent.</p>