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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Anyone that disagrees with this is not telling the truth --- V5
Date: Tue, 20 Aug 2024 08:29:39 -0500
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On 8/20/2024 4:32 AM, Mikko wrote:
> On 2024-08-20 02:47:49 +0000, olcott said:
> 
>> *Everything that is not expressly stated below is*
>> *specified as unspecified*
>>
>> void DDD()
>> {
>>    HHH(DDD);
>>    return;
>> }
>>
>> _DDD()
>> [00002172] 55         push ebp      ; housekeeping
>> [00002173] 8bec       mov ebp,esp   ; housekeeping
>> [00002175] 6872210000 push 00002172 ; push DDD
>> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
>> [0000217f] 83c404     add esp,+04
>> [00002182] 5d         pop ebp
>> [00002183] c3         ret
>> Size in bytes:(0018) [00002183]
>>
>> *It is a basic fact that DDD emulated by HHH according to*
>> *the semantics of the x86 language cannot possibly stop*
>> *running unless aborted* (out of memory error excluded)
>>
>> X = DDD emulated by HHH∞ according to the semantics of the x86 language
>> Y = HHH∞ never aborts its emulation of DDD
>> Z = DDD never stops running
>>
>> The above claim boils down to this: (X ∧ Y) ↔ Z
> 
> No, it does not.

https://en.wikipedia.org/wiki/Stipulative_definition
It is stipulated that the above claim boils down to this:  (X ∧ Y) ↔ Z

>  Above X is not a truth bearer (no verb).
It breaks the line by adding one more word

X = DDD is emulated by HHH∞ according to the semantics of the x86 language

> Y and Z are distinct propositions abot different things,

They are a pair of premises if you know what premises
are then you know that can can't have any objection to
a pair of them.

> Y about HHH∞ and Z about DDD with no obvious connection.

HHH∞(DDD) never aborts its emulation of DDD

> In addition your "basic fact" menstions HHH but not HHH∞.
> It is unspecifend whether HHH or HHH∞ ever emulate or
> ever abort their emulation or have aborted their emulation

The hypothetical HHHn aborts The hypothetical HHH∞ never aborts.
HHH can be either one of these.

> and therefore whether "DDD emulated by HHH" and "DDD
> emulated by HHH∞" denote anything. As DDD only calls HHH
> but not HHH∞ there is no connection between X and Z.
> 
>> x86utm takes the compiled Halt7.obj file of this c program
>> https://github.com/plolcott/x86utm/blob/master/Halt7.c
>> Thus making all of the code of HHH directly available to
>> DDD and itself. HHH emulates itself emulating DDD.
>>
>> void EEE()
>> {
>>    HERE: goto HERE;
>> }
>>
>> HHHn correctly predicts the behavior of DDD the same
>> way that HHHn correctly predicts the behavior of EEE.
> 
> It cannot correctly predicted the same if the behavours
> of DDD and EEE are different.
> 

Sure it can. the actual implemented HHH determines the
correct halt status of each.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer