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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Anyone that disagrees with this is not telling the truth ---- V4
Date: Mon, 19 Aug 2024 19:06:11 -0500
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On 8/19/2024 6:08 PM, Richard Damon wrote:
> On 8/19/24 8:14 AM, olcott wrote:
>> On 8/19/2024 5:17 AM, Fred. Zwarts wrote:
>>
>> *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
>>
>> My claim boils down to this: (X ∧ Y) ↔ Z
>>
>> void EEE()
>> {
>>    HERE: goto HERE;
>> }
>>
>> HHHn correctly predicts the behavior of DDD the same
>> way that HHHn correctly predicts the behavior of EEE.
>>
>>>>
>>>
>>> Which proves that the simulation failed to reach the end. This makes 
>>> the simulation incomplete and therefore incorrect.
>>> The simulating HHH is programmed to abort and halt. The simulated HHH 
>>> should behave exactly in the same way, so no cheating with the Root 
>>> variable is allowed.
>>> The the simulating HHH aborts when the simulated HHH has only one 
>>> cycle to go, after which it would also abort and halt, but the 
>>> simulating HHH failed to reach this end.
>>
>> I made my claim more precise.
>>
> 
> Remember, you said: Everything that is not expressly stated below is*
> specified as unspecified
> 
> Therefore HHHn can NOT correctly emulate DDD past the call HHH 
> instruction, because it doesn't HAVE the instruciton of the PROGRAM DDD 
> (which is what you emulate) since it doesn't have the instruction at 
> 000015D2.
> 

That they are in the same memory space is entailed
in the same way that the x86 code is not being run
on a rubber ducky is entailed.

> The contents of the memory at 000015D2 can not be accessable to HHHn, as 
> the input is described as DDD and not DDDn, so the input doesn't change 
> between instances, and thus CAN'T contain that memory that changes, and 
> thus is not valid to be part of the input.
> 
> Thus we also have that HHH∞ can not exist, so both your premises just 
> fail to be possible.
> 
> Sorry, you are just repeating your error because apparently you just 
> can't learn.
> 
> 


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