Re: Computable Functions --- finite string transformation rules

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Sujet : Re: Computable Functions --- finite string transformation rules
De : F.Zwarts (at) *nospam* HetNet.nl (Fred. Zwarts)
Groupes : comp.theory
Date : 26. Apr 2025, 09:58:23
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vui77f$217h6$2@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
User-Agent : Mozilla Thunderbird
Op 25.apr.2025 om 23:21 schreef olcott:
On 4/25/2025 8:56 AM, joes wrote:
Am Thu, 24 Apr 2025 19:03:34 -0500 schrieb olcott:
On 4/24/2025 6:10 PM, Richard Damon wrote:
On 4/24/25 5:01 PM, olcott wrote:
On 4/24/2025 2:59 PM, Fred. Zwarts wrote:
Op 24.apr.2025 om 21:41 schreef olcott:
On 4/24/2025 2:12 PM, Fred. Zwarts wrote:
Op 24.apr.2025 om 19:13 schreef olcott:
>
HHH correctly determines through mathematical induction that DD
emulated by HHH (according to the finite string transformations
specified by the x86 language) cannot possibly reach its final
halt state in an infinite number of steps.
>
No, HHH has a bug which makes that it fails to see that there is
only a finite recursion,
>
When the finite string transformation rules of the x86 language are
applied to the input to HHH(DD)
THIS DD CANNOT POSSIBLY REACH ITS FINAL HALT STATE not even after an
infinite number of emulated steps.
>
Again a lot of text, but no rebuttal.
>
When the finite string transformation rules of the x86 language are
applied to the input to HHH(DD)
THIS DD CANNOT POSSIBLY REACH ITS FINAL HALT STATE not even after an
infinite number of emulated steps.
>
No, HHH just stops performing those before it get to the end.
The transformation, which by definition of the x86 language, don't just
stop in the middle, continue to the point where the emulated HHH aborts
its emulation and returns 0 to the emulated DD which the halts.
>
Mathematical induction proves that DD emulated by HHH cannot possibly
reach its own final state in an infinite number of steps and it does
this with one recursive emulation.
There is a repeating pattern that every C programmer can see.
 
Like Fred wrote months ago, that has nothing to do with the contradictory
part of DD,
 Sure it does. The contradictory part of DD has always
been unreachable thus only a ruse.
 
only with it being simulated by the same simulator it calls.
 That <is> the Halting Problem counter-example input.
 
The program EE(){ HHH(EE); } also halts and cannot be simulated by HHH.
>
 HHH cannot possibly do this without violating the rules of
the x86 language.
HHH already violates the rules of the x86 language by prematurely aborting the halting program.

Date Sujet#  Auteur
22 Jul 25 o 

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