Timeline for Why is adopting Russell's Axiom of Reducibility as strong as eliminating the Ramified Hierarchy?
Current License: CC BY-SA 3.0
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| Apr 29, 2014 at 3:37 | history | edited | user44143 | CC BY-SA 3.0 |
shorter first sentence
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| Apr 25, 2014 at 17:05 | comment | added | user44143 | It's a many-sorted language. A Henkin model for it has integers which range over $N$, grade-0 sets which range over some $S_0$, grade-1 sets which range over some $S_1$, ..., and some ungraded sets which range over $S_{-1}$, where all the $S_i$ are subsets of $P(N)$. (You can replace "ungraded subset" in the above with "element of S_{-1}".) I claim there are models for $2+\neg 1$ in which $S_0, S_1, ...$ are properly smaller than $P(N)$, but $S_{-1}=P(N)$. | |
| Apr 25, 2014 at 16:56 | comment | added | Keshav Srinivasan | "The proof is that 2 does not imply 1, because we can take a non-trivial model for 2, and expand it by having all subsets of N as ungraded subsets." I'm afraid I don mt understand what you're saying. Could you elaborate on this? I'm not even sure what you mean by "ungraded subsets". | |
| Apr 25, 2014 at 16:09 | history | answered | user44143 | CC BY-SA 3.0 |