Timeline for Is every true statement independent of $PA$ equivalent to some consistency statement?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Feb 20, 2018 at 8:57 | comment | added | user103227 | I've edited my answer to reflect Will Sawin's observation. | |
| Feb 20, 2018 at 8:56 | history | edited | user103227 | CC BY-SA 3.0 |
Incorporating insights from the comments.
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| Feb 18, 2018 at 21:20 | vote | accept | Christopher King | ||
| Feb 18, 2018 at 21:01 | comment | added | user103227 | Excellent point. If you post that as an answer I'll be happy to upvote yours and remove mine. | |
| Feb 18, 2018 at 16:45 | comment | added | Will Sawin | Right, so the answer is "yes" for plain PA, because it is stronger than $EA + Con(EA)$. | |
| Feb 18, 2018 at 16:41 | comment | added | user103227 | It is safe to replace PA with elementary arithmetic EA throughout my answer. See Visser's Faith and Falsity in Annals of Pure and Applied Logic, 131(1-3), 2005, pp. 103-131 for a proof. | |
| Feb 18, 2018 at 16:31 | comment | added | Will Sawin | Does that theorem really need the full strength of $PA$ to be proved? f we can prove that $T + Con(T)$ prove $\pi$ is equivalent to $Con(T+ Psi)$ for some theory $T$ weaker than $PA$, then as long as $PA$ proves consistency of $T$, $PA$ will prove the same theorem. | |
| Feb 18, 2018 at 9:54 | history | answered | user103227 | CC BY-SA 3.0 |