If the UTM Theorem states there's a u such that u(e,x) = f(x) where both
are defined and both are undefined when either is undefined, is that
interesting or surprising to anybody?
Replace "there's a u" by "there is a u such that for all f there is an e"
The identity function is valid for u and forces that e = f. Job done.
* u is a computable function;
* f is a computable function;
* e is the G�del number of a Turing machine that computes f.
How can e and f be equal ?
--
pa at panix dot com
| Date | Sujet | # | Auteur | |
| 21 Oct 25 |  UTM Theorem vs the identity function | 12 | Tristan Wibberley | |
| 21 Oct 25 |  Re: UTM Theorem vs the identity function | 8 | Pierre Asselin | |
| 22 Oct 25 |   Re: UTM Theorem vs the identity function | 7 | Tristan Wibberley | |
| 22 Oct 25 |  Re: UTM Theorem vs the identity function | 3 | olcott | |
| 22 Oct 25 | Re: UTM Theorem vs the identity function | 2 | Tristan Wibberley | |
| 22 Oct 25 |  Re: UTM Theorem vs the identity function --- Gödel | 1 | olcott | |
| 22 Oct 25 | Re: UTM Theorem vs the identity function | 3 | Pierre Asselin | |
| 23 Oct 25 | Re: UTM Theorem vs the identity function | 2 | Tristan Wibberley | |
| 23 Oct 25 | Re: UTM Theorem vs the identity function | 1 | Pierre Asselin | |
| 22 Oct 25 | Re: UTM Theorem vs the identity function | 3 | Mikko | |
| 22 Oct 25 | Re: UTM Theorem vs the identity function | 2 | Tristan Wibberley | |
| 23 Oct 25 | Re: UTM Theorem vs the identity function | 1 | Mikko |
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