Timeline for Seemingly ill-founded recursion and the recursion theorem
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 30, 2013 at 20:32 | answer | added | Andreas Blass | timeline score: 4 | |
| Dec 30, 2013 at 18:37 | answer | added | Will Sawin | timeline score: 2 | |
| Dec 30, 2013 at 18:20 | answer | added | Andrej Bauer | timeline score: 3 | |
| Dec 30, 2013 at 18:12 | comment | added | Andrej Bauer | Don't careful people write $\lambda \leftrightarrow \lnot \mathrm{Prov}_{PA}(\ulcorner \lambda \urcorner)$? | |
| Dec 30, 2013 at 17:47 | answer | added | Bjørn Kjos-Hanssen | timeline score: 2 | |
| Dec 30, 2013 at 17:05 | review | First posts | |||
| Dec 30, 2013 at 17:05 | |||||
| Dec 30, 2013 at 17:04 | comment | added | Anonymous | Yes, the clarification IS the question. We don't bat an eye when we say, "There is a $\lambda$ such that $PA$ proves $\lambda\leftrightarrow \neg\mathrm{Prov}_{PA}(\lambda)$, even though your complaint applies there just as well. My question is, can the latter two definitions in my OP be justified in a similar way to this liar's paradox... As far as first-order-second-order issues, within the predicate, $S_j$ should not be considered as set, but as formula. | |
| Dec 30, 2013 at 17:00 | comment | added | Joel David Hamkins | Could you clarify what you mean precisely by including $S_j$ inside the provability predicate? You seem to be asking whether a given equivalence determines a unique set, which is essentially a second order matter, but in this case, it makes no sense to refer to that set inside the provability predicate, which I take you to refer to first-order provability in PA. Your question seems to want clarification. | |
| Dec 30, 2013 at 16:56 | history | edited | Anonymous | CC BY-SA 3.0 |
added 4 characters in body
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| Dec 30, 2013 at 16:55 | comment | added | Anonymous | I said so because membership in S_i depends on membership in S_i (circularly). But you're right, technically, so I'll change it to $n\in S_i$ iff $n=2i$ or $\exists j$ such that $n\in S_j$. Then to check whether $1\in S_0$, we are instructed (among other things) to check whether $1\in S_0$, which in turn means checking whether $1\in S_0$, etc. | |
| Dec 30, 2013 at 16:52 | comment | added | Joel David Hamkins | Your second displayed line does seem to define a unique family of sets, namely, $S_i=\mathbb{N}$ for all $i$, which is the only family satisfying the property. Why do you say that it does not? | |
| Dec 30, 2013 at 16:48 | history | asked | Anonymous | CC BY-SA 3.0 |