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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Who here is too stupid to know that DDD correctly simulated by
 HHH cannot possibly reach its own return instruction?
Date: Sat, 3 Aug 2024 12:12:04 -0400
Organization: i2pn2 (i2pn.org)
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On 8/3/24 12:03 PM, olcott wrote:
> On 8/3/2024 10:33 AM, Richard Damon wrote:
>> On 8/3/24 10:26 AM, olcott wrote:
>>> On 8/3/2024 9:04 AM, Fred. Zwarts wrote:
>>>> Op 03.aug.2024 om 15:50 schreef olcott:
>>>>> On 8/3/2024 3:14 AM, Fred. Zwarts wrote:
>>>>>> Op 02.aug.2024 om 22:57 schreef olcott:
>>>>>>> Who here is too stupid to know that DDD correctly simulated
>>>>>>> by HHH cannot possibly reach its own return instruction?
>>>>>>>
>>>>>>> void DDD()
>>>>>>> {
>>>>>>>    HHH(DDD);
>>>>>>>    return;
>>>>>>> }
>>>>>>>
>>>>>>
>>>>>> Which proves that the simulation is incorrect.
>>>>>
>>>>> When are you going to understand that you are not allowed
>>>>> to disagree with the semantics of the x86 language?
>>>>>
>>>>>
>>>>
>>>> I do not disagree.
>>>> When are you going to understand that it is a deviation of the 
>>>> semantics of the x86 language to skip instructions of a halting 
>>>> program,
>>>
>>> HHH(DDD) simulates DDD that calls HHH(DDD) to repeat the process.
>>>
>>> If it does this an infinite number of times the simulated DDD
>>> never reaches its own return instruction.
>>>
>>> If it does this a googolplex number of times the simulated DDD
>>> never reaches its own return instruction.
>>
>> Nope, the PARTIAL SIMULATION of DDD never reaches the return instruction.
>>
> 
> For N = 0; while N <= googolplex; N++
> N instructions of DDD correctly emulated by HHH[N] never
> reach their own "return" instruction final state.
> 
> ∞ instructions of DDD correctly emulated by HHH[∞] never
> reach their own "return" instruction final state.
> 
> Thus any HHH that takes a wild guess that DDD emulated
> by itself never halts is always correct.
> 

The SIMULATION of DDD never reaches the return instruction.

The actual Program DDD that HHH simulates, WILL reach its final 
instruction for ANY FINITE N, as it will call that HHH, which will 
simulate for the finite time to emulate N instructions and return to DDD 
which will return.

Any HHH that returns, had to have done that, so the correct answer for 
any DDD based on an HHH that returns is that the DDD Halts.

That you are so stuck on your error of confusing the emulation of DDD by 
HHH with the actual behavior of DDD just proves you are nothing but a 
MORON, and a LIAR,

Sorry, just repeating you lies just adds proof to that conclusion.

You just don't know what you are talking about, and have proved in in an 
uncounted multiple of ways.