Timeline for Extensions of $PA+\neg Con(PA)$ with large consistency strength
Current License: CC BY-SA 4.0
13 events
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| Jan 19 at 16:29 | comment | added | Joel David Hamkins | @Gro-Tsen My argument wouldn't work if you did it that way. We know how to control exactly the level of the theory $\text{PA}_k$ that becomes inconsistent in the interpreted model, but we have much less control over the size of the proofs of contradiction (which are necessarily much larger). I'm not sure how to analyze your theory $T'$, which is closer to the OP's question, and this is why in my answer I moved to $\text{PA}_k$ instead. | |
| Jan 19 at 9:08 | comment | added | Gro-Tsen | Does it matter that you defined $k$ as the smallest such that $\mathsf{PA}_k$ is inconsistent, or would the argument work just as well with the length of the smallest proof of inconsistency in $\mathsf{PA}$? (In other words, what can be said about the theory $T'$ which says “$\mathsf{PA}$ is inconsistent and no proof of a contradiction in $\mathsf{ZFC}$ comes before the earliest proof of a contradiction in $\mathsf{PA}$”? I can make this into a separate question if you think the answer is too long for a comment.) | |
| Dec 3, 2023 at 2:50 | comment | added | Tom Bouley | Thank you for your excellent and prompt answer! My delay in accepting it lay only in my intending to mull it over and then promptly moving on to other things. | |
| Dec 3, 2023 at 2:24 | vote | accept | Tom Bouley | ||
| Nov 6, 2023 at 18:52 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
Got the equiconsistency.
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| Nov 6, 2023 at 17:58 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
Fixed upper bound consistency hypothesis
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| Nov 6, 2023 at 16:58 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Nov 6, 2023 at 2:42 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Nov 6, 2023 at 2:03 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Nov 6, 2023 at 1:17 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Nov 6, 2023 at 1:15 | comment | added | Joel David Hamkins | This argument doesn't seem to show consistency of your statement $P$, since the proof of a contradiction in PA in the end-extension will be much bigger than $k$ and indeed in the end-extension above the original nonstandard PA model, with it's log therefore also outside, so we can't seem to know that ZFC isn't refuted by that stage. | |
| Nov 6, 2023 at 1:04 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Nov 6, 2023 at 0:59 | history | answered | Joel David Hamkins | CC BY-SA 4.0 |