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From: joes <noreply@example.org>
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Subject: Re: Title: A Structural Analysis of the Standard Halting Problem
 Proof
Date: Thu, 24 Jul 2025 21:24:16 -0000 (UTC)
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Am Thu, 24 Jul 2025 09:32:45 -0500 schrieb olcott:
> On 7/24/2025 5:03 AM, Fred. Zwarts wrote:
>> Op 23.jul.2025 om 17:29 schreef olcott:

>>> The directly executed HHH does reach its final halt state. DDD
>>> correctly simulated by HHH cannot possibly reach its final halt state
>>> no matter what HHH does because it remains stuck in recursive
>>> simulation.
>> 
>> Indeed, but irrelevant. The simulating HHH does not do a correct
>> simulation, it aborts prematurely.

> _DDD()
> [00002192] 55             push ebp [00002193] 8bec           mov ebp,esp
> [00002195] 6892210000     push 00002192  // push DDD [0000219a]
> e833f4ffff     call 000015d2  // call HHH [0000219f] 83c404         add
> esp,+04 [000021a2] 5d             pop ebp [000021a3] c3             ret
> Size in bytes:(0018) [000021a3]
> 
> Aborting prematurely literally means that after N instructions of DDD
> are correctly emulated by HHH that this emulated DDD would reach its own
> emulated "ret" instruction final halt state.
> What value of N are you proposing?

Let's see: the call to HHH is #4, [waves hands], then another 4 inside
the next level of simulation, and after another 4 the first simulated
HHH (the one called by the input, not the outermost simulator. We are
now 3 levels in) decides that enough is enough and aborts, returning
to the outermost level which takes 3 more instructions to halt,
whereupon our treasured HHH returns that DDD halts. So, 4+4+4+3=15?

Of course the crux is that changing "HHH" changes the input, so HHH
can never do it.

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.