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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1
Date: Sat, 29 Mar 2025 20:01:14 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Sat, 29 Mar 2025 10:28:15 -0500 schrieb olcott:
> On 3/29/2025 4:56 AM, Mikko wrote:
>> On 2025-03-28 22:41:12 +0000, olcott said:
>>> On 3/28/2025 4:58 PM, Richard Damon wrote:
>>>> On 3/28/25 2:13 PM, olcott wrote:
>>>>> On 3/28/2025 8:50 AM, Richard Damon wrote:
>>>>>> On 3/27/25 10:11 PM, olcott wrote:
>>>>>>> On 3/27/2025 9:02 PM, Richard Damon wrote:
>>>>>>>> On 3/27/25 9:10 PM, olcott wrote:
>>>>>>>>> On 3/27/2025 7:47 PM, Richard Damon wrote:
>>>>>>>>>> On 3/27/25 8:11 PM, olcott wrote:
>>>>>>>>>>> On 3/27/2025 4:56 PM, joes wrote:

>>>>>>>>>>>> Yes, HHH is not a correct simulator.
>>>>>>>>>>>>
>>>>>>>>>>> You say that it is not a correct simulator on the basis of
>>>>>>>>>>> your ignorance of the x86 language that conclusively proves
>>>>>>>>>>> that HHH does correctly simulate the first four instructions
>>>>>>>>>>> of DDD and correctly simulates itself simulating the first
>>>>>>>>>>> four instructions of DDD.
The x86 language or my supposed ignorance thereof doesn't prove shit.
HHH does not simulate the infinite stack of recursive simulations,
for obvious reasons.

>>>>>>>>>> It isn't a correct simulator,
>>>>>>>>>
>>>>>>>>> You know that you are lying about this or you would show how DDD
>>>>>>>>> emulated by HHH would reach its final state ACCORDING TO THE
>>>>>>>>> SEMANTICS OF THE X86 LANGUAGE.
>>>>>>>>>
>>>>>>>> It can't be, because your HHH doesn't meet your requirement.
>>>>>>>>
>>>>>>> You cannot show that because you know you are lying about that.
One cannot show something impossible.

>>>>>> Sure we can, make a main that directly calls HHH and then DDD, then
>>>>>> call HHH1(DDD)
>>>>>> That HHH will return 0, saying that DDD is non-halting, but the DDD
>>>>>> wll return, showing that DDD is halting.
>>>>>> Look at the trace that HHH generates, and that HHH1 generates,
>>>>>> HHH's will be a subset of the trace that HHH1 generates, showing
>>>>>> that it is NOT proof that this program is non-halting as that exact
>>>>>> same initial segment halts.
>>>>>> Your argument about changing HHH shows that it doesn't halt is just
>>>>>> invalid, as then you either changed the input, or demonstrated that
>>>>>> you input was a class error as it didn't contain the COMPLETE
>>>>>> representation of the code of DDD.
>>>>>
>>>>> I can't understand how that confused mess addresses the point of
>>>>> this thread:
>>>>> It is a verified fact that the finite string of machine code of DDD
>>>>> emulated by HHH according to the semantics of the x86 language has
>>>>> different behavior than DDD emulated by HHH1 according to the
>>>>> semantics of the x86 language.
Non sequitur.

>>>> Where did you "verify" that LIE. You claim fails the simple test:
>>>> What is the first instruction actually correctly emulated by the
>>>> rules of the x86 language by HHH and HHH1 that had a different
>>>> result.
>>>>
>>> When DDD emulated by HHH calls HHH(DDD) this call NEVER returns.
>>> When DDD emulated by HHH1 calls HHH(DDD) this call returns.
>> 
>> When DDD is correctly emulated the call HHH(DDD) returns.
>> 
> When are you going to understand that disagreeing with the semantics of
> the x86 language IS NOT ALLOWED?
Disagree with what semantics exactly? The call to HHH *must* return.

-- 
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.