Timeline for A property of < in Primitive recursive arithmetic
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 14 at 11:03 | comment | added | Joel David Hamkins | I see, I hadn't realized that it included $I\Delta_0$. Thanks! (And apologies for all my poorly informed comments on this thread.) | |
| Jul 14 at 10:06 | comment | added | Ali Enayat | @JoelDavidHamkins PRA can be viewed as a definitional expansion of the theory obtained by augementing the theort $\mathrm{I}\Delta_0$ with axioms $\phi_n$ (for $n\in\omega$), such that $\varphi_0$ asserts the totality of exponentiation, $\varphi_1$ asserts the totality of superexponentiation (tetration), etc. | |
| Jul 13 at 15:27 | comment | added | Joel David Hamkins | This answer seems to claim that PRA includes $\text{IE}_1$ and has no computable nonstandard models. mathoverflow.net/a/121252/1946 | |
| Jul 13 at 15:24 | comment | added | Joel David Hamkins | For example, does the Tennenbaum theorem hold for nonstandard models of PRA? That would be decisive negative answer to my inquiry. | |
| Jul 13 at 12:26 | comment | added | Joel David Hamkins | @NoahSchweber Oh, I had thought that PRA specifically did not have induction. Isn't that the point of it, that it is weak that way? I thought we just have the recursive definitions of the primitive recursive functions, and that's it. Of course if you have induction then the whole question here is easy. (Meanwhile, having one point at infinity that is self-successor does not violate injectivity of successor, but you can't make a model of PRA this way, since the parity function won't work.) But how easily can we make nonstandard models of PRA? | |
| Jul 13 at 5:12 | comment | added | Noah Schweber | @JoelDavidHamkins Since PRA has induction + injectivity of successor, that shouldn't work. (More substantively, see Francois Dorais' answer at the MSE version.) | |
| Jul 12 at 17:55 | comment | added | Joel David Hamkins | Could someone tell me whether we can make badly behaved nonstandard models of PRA as easily as for, say, Robinson's Q? In that theory you can just adjoin a point or two at infinity and make things strange. Does this method work in PRA as well? If so, the answer here will be negative. | |
| Jul 12 at 7:46 | review | Close votes | |||
| Jul 17 at 3:06 | |||||
| Jul 12 at 2:00 | comment | added | Gerald Edgar | Generally I recommend not cross-posting until at least a week has passed with no answer. | |
| Jul 12 at 0:53 | comment | added | Noah Schweber | Crossposted: math.stackexchange.com/q/4944796/28111 | |
| S Jul 12 at 0:43 | review | First questions | |||
| Jul 12 at 1:20 | |||||
| S Jul 12 at 0:43 | history | asked | user532222 | CC BY-SA 4.0 |