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Article #431Re: A question in Recursion Theory -- OOPS!
From: Mark Fulk
Date: Wed, 30 Sep 1992 14:21
Date: Wed, 30 Sep 1992 14:21
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2750 bytes
In article <1992Sep30.140356.9401@cs.rochester.edu> fulk@cs.rochester.edu (Mark Fulk) writes: >My previous post on this matter was poorly worded, and contained some >mistakes in the presentation of the proof. Herewith an improvement: Wherein I still left a gap. >Let f be the left projection function for some pairing function. >Let J be the set of natural numbers. > >We construct I by deleting elements from N; note that this makes I co-r.e. >In the construction, we will refer to certain requirements R_i: > >R_i: range phi_i is finite or there is y in N such that phi_i(y) not in I. > >We will also restrain some numbers into I for requirements R_i. These >restraints can be broken for higher-priority (lower-numbered) requirements, >but the standard finite-injury argument will establish that every requirement >gets met. The point of these restraints is that the range of f restricted >to I (f|I) will still be N; this forces any section of f|I to have an >infinite range, since a section must be in injection. > >Stage s: > > Search for s steps for all i such that: requirement i is not met, > and there are x,y such that phi_i(x) = y, f(y)=x, and y is not > restrained into I for any requirement i'<i. > > Let i0 be the least such i found; let x0 and y0 be corresponding > values of x and y. > > Delete y0 from I. > > find y'>y0 such that y' has not yet been deleted from I, and such that > f(y') = f(y0) = x0, and restrain y' into I for requirement i0. > > Mark requirement i0 as being met. > > For all requirements i>i0 that are marked as met, remove the mark > (that is, start treating them as unmet). > >End stage s. > >I will leave the verification up to the reader. The basic structure of >the proof is that every requirement will be met, either by default >(phi_i has finite range) or by being met in the construction. The >latter will happen to each requirement at most a finite number of times >(2^(i-1) will do it). Then note that meeting every requirement is enough >to establish that f|I has no section. > >Mark -- Mark A. Fulk University of Rochester Computer Science Department fulk@cs.rochester.edu
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Newsgroups: sci.logic Path: arc1-ca.pugleaf.net!sparky!uunet!zaphod.mps.ohio-state.edu!rpi!batcomputer!cornell!rochester!fulk From: fulk@cs.rochester.edu (Mark Fulk) Subject: Re: A question in Recursion Theory -- OOPS! Message-ID: <1992Sep30.142159.10200@cs.rochester.edu> Keywords: recursion Organization: Computer Science Department University of Rochester References: <1992Sep29.161016.19533@ghost.dsi.unimi.it> <1992Sep29.175554.12758@cs.rochester.edu> <1992Sep30.140356.9401@cs.rochester.edu> Date: Wed, 30 Sep 1992 14:21:59 GMT Lines: 56