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  • $\begingroup$ 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. $\endgroup$
    – Joel David Hamkins
    Commented Nov 6, 2023 at 1:15
  • 1
    $\begingroup$ 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. $\endgroup$
    – Tom Bouley
    Commented Dec 3, 2023 at 2:50
  • $\begingroup$ 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.) $\endgroup$
    – Gro-Tsen
    Commented Jan 19 at 9:08
  • $\begingroup$ @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. $\endgroup$
    – Joel David Hamkins
    Commented Jan 19 at 16:29