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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 19:50:43 -0500
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On 8/19/2024 7:46 PM, Richard Damon wrote:
> On 8/19/24 8:06 PM, olcott wrote:
>> 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.
>>
> 
> But not EXPLICITLY stated, so that is a lie.
> 

If you want to pay head games you can play them by yourself.

> And what is WRONG with running the code on a rubber ducky, it might be 
> powered by Pentium.
> 
> And, if they ARE in the same memory space, then it is DDDn not DDD, as 
> there are each different by the memory that came with them.
> 
> Sorry, you are just caught out in your lie and stupdity.
> 
> You just don't knunderstand what you are talking about.
> 
> 
>>> 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