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Apr 13, 2017 at 12:19 history edited CommunityBot
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Nov 6, 2013 at 12:36 review Suggested edits
Nov 6, 2013 at 12:44
Aug 15, 2013 at 15:49 comment added Emil Jeřábek KM contradicts a $\Pi^0_1$ truth if it proves the negation of said truth. The negation of a $\Pi^0_1$ sentence is $\Sigma^0_1$, so in other words, KM proves a false $\Sigma^0_1$ sentence. Saying that this does not happen thus amounts to all $\Sigma^0_1$ sentences provable in KM being true, i.e., that KM is $\Sigma^0_1$-sound.
Aug 15, 2013 at 14:55 comment added Keshav Srinivasan @EmilJerabek Doesn't Con(KM + all Pi_1 truths) assert that no theorems of KM contradict a Pi _1 truth? How is that different from saying that KM is Pi_1 sound?
Aug 15, 2013 at 11:12 comment added Emil Jeřábek @Keshav Srinivasan: Con(KM+True $\Pi^0_1$) says that KM is $\Sigma^0_1$-sound (or equivalently, $\Pi^0_2$-sound) rather than $\Pi^0_1$-sound.
Aug 15, 2013 at 5:02 comment added Andrés E. Caicedo "if @JoelDavidHamkins is right and we can't define a truth predicate within NBG" This is not a matter of opinion (and he's right).
Aug 15, 2013 at 4:12 comment added Keshav Srinivasan @AndresCaicedo If it's not too much trouble, could you tell me what part to read? projecteuclid.org/euclid.pl/1235421926
Aug 15, 2013 at 3:59 comment added Andrés E. Caicedo Read the reference, it may be more productive.
Aug 15, 2013 at 3:47 comment added Keshav Srinivasan @AndresCaicedo Thanks, I may read it. But could you tell me what's wrong with the reasoning in my last comment?
Aug 15, 2013 at 3:32 comment added Andrés E. Caicedo Each new edit makes the question more and more confused. And you are mistaken in your last comment. You may want to read the relevant sections of Metamathematics of first-order arithmetic, where most of this is covered carefully.
Aug 15, 2013 at 2:08 comment added Keshav Srinivasan @JoelDavidHamkins As far as I can tell, Con(KM + all true Pi_1 statements) is equivalent to the statement that KM is Pi_1 sound. But if KM were not Pi_1 sound, then it would prove some false Pi_1 statement, and since that statement is Pi_1, KM would also disprove it, so KM would be inconsistent. So if I'm not mistaken, Con(KM + all true Pi_1 statements) is equivalent to Con(KM), and Con(KM) is Pi_1. Where am I going wrong?
Aug 15, 2013 at 1:51 comment added Joel David Hamkins As I mention in my comments on my answer, Con(KM+True $\Pi^0_1$) is such a statement. (And it seems to me that KM proves "ZF is sound".)
Aug 15, 2013 at 1:44 comment added Keshav Srinivasan @JoelDavidHamkins (cont'd) Perhaps a good preliminary question would be, how would you construct a statement of first-order arithmetic that is independent of ZF + (All Pi_1 truths), and which does not follow in MK from the assumption that ZF is sound?
Aug 15, 2013 at 1:39 comment added Keshav Srinivasan @JoelDavidHamkins It's fine if it's equivalent to a statement of arithmetic, I just don't want it to be one. If there's something like a large cardinal axiom which turns out to be equivalent to a statement in the language of first-order arithmetic, that would be fine with me.
Aug 15, 2013 at 1:29 comment added Joel David Hamkins Regarding your third edit, how can you say that you want X to be not arithmetic, given that it must be provably equivalent to P?
Aug 15, 2013 at 1:22 comment added Asaf Karagila Are you sure that you know what sort of answer you are looking for?
Aug 15, 2013 at 0:15 history edited Keshav Srinivasan CC BY-SA 3.0
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Aug 14, 2013 at 19:48 answer added Noah Schweber timeline score: 2
Aug 14, 2013 at 18:36 history edited Keshav Srinivasan CC BY-SA 3.0
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Aug 14, 2013 at 4:27 answer added Andreas Blass timeline score: 6
Aug 14, 2013 at 3:23 answer added Joel David Hamkins timeline score: 5
Aug 14, 2013 at 3:23 history edited Keshav Srinivasan CC BY-SA 3.0
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Aug 14, 2013 at 3:19 comment added Asaf Karagila Amazing. Two hours of waiting between cross posting your question. You should at least point out that this was posted on m.SE and provide a link.
Aug 14, 2013 at 3:15 history asked Keshav Srinivasan CC BY-SA 3.0