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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory,sci.logic
Subject: Re: My reviewers think that halt deciders must report on the behavior
 of their caller
Date: Tue, 8 Jul 2025 10:10:37 +0200
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Op 07.jul.2025 om 15:23 schreef olcott:
> On 7/7/2025 2:36 AM, Fred. Zwarts wrote:
>> Op 07.jul.2025 om 05:12 schreef olcott:
>>> On 7/6/2025 9:09 PM, Richard Damon wrote:
>>>> On 7/6/25 4:06 PM, olcott wrote:
>>>>> On 7/6/2025 12:00 PM, Richard Damon wrote:
>>>>>> On 7/6/25 11:19 AM, olcott wrote:
>>>>>>>
>>>>>>> void DDD()
>>>>>>> {
>>>>>>>    HHH(DDD);
>>>>>>>    return;
>>>>>>> }
>>>>>>>
>>>>>>> *EVERY BOT FIGURES THIS OUT ON ITS OWN*
>>>>>>
>>>>>> No, it just isn't smart enough to detect that you lied in your 
>>>>>> premise.
>>>>>>
>>>>>>> There is no way that DDD simulated by HHH (according
>>>>>>> to the semantics of the C programming language)
>>>>>>> can possibly reach its own "return" statement final
>>>>>>> halt state.
>>>>>>
>>>>>> And there is no way for HHH to correctly simulate its input and 
>>>>>> return an answer
>>>>>>
>>>>>
>>>>> You insistence that a non-terminating input be simulated
>>>>> until non-existent completion is especially nuts because
>>>>> you have been told about this dozens of times.
>>>>>
>>>>> What the F is wrong with you?
>>>>>
>>>>
>>>> It seems you don't understand those words.
>>>>
>>>> I don't say that the decider needs to simulate the input to 
>>>> completion, but that it needs to be able to actually PROVE that if 
>>>> this exact input WAS given to a correct simultor (which won't be 
>>>> itself, since it isn't doing the complete simulation) will run for 
>>>> an unbounded number of steps.
>>>>
>>>
>>> No decider is ever allowed to report on anything
>>> besides the actual behavior that its input actually
>>> specifies.
>>>
>> And HHH does not do that. The input specifies a halting program, 
>> because it includes the abort code. 
> 
> 
> void DDD()
> {
>    HHH(DDD);
>    return;
> }
> 
> _DDD()
> [00002192] 55             push ebp
> [00002193] 8bec           mov ebp,esp
> [00002195] 6892210000     push 00002192  // push DDD
> [0000219a] e833f4ffff     call 000015d2  // call HHH
> [0000219f] 83c404         add esp,+04
> [000021a2] 5d             pop ebp
> [000021a3] c3             ret
> Size in bytes:(0018) [000021a3]
> 
> That does have an effect on DDD emulated by HHH according
> to the semantics of the x86 language stopping running.
> It has no effect on this DDD every reaching its final halt
> state. I have corrected your error on this too many times
> you don't seem to want an honest dialogue.

As usual repeated claims without evidence.
It is a pity that you ignore or do not understand the corrections that 
have been pointed out to you. Not understanding something is not stupid, 
but resistance against learning from errors is.

We have pointed out already many times that this cannot be the whole 
input. At 0000219a there is a call to 000015d2 , but the code at 
000015d2 is not specified. We have to take it from another source, your 
Halt7.c. There we see that HHH is a function that returns with a value 0 
with this input. So, a correct simulation will then continue at 0000219f 
with this value and even a beginner sees that the simulation will then 
reach a natural end.
This is the full specification of the input.
This is also proven when exactly the same input is used in direct 
execution, or by world-class simulators, even by HHH1.
So, the following still holds:

> 
>> But HHH gives up before it reaches that part of the specification and 
>> the final halt state.
If HHH does not see the full specification, it does not change the 
specification. It only illustrates the failure of the simulation to 
complete.
There is nothing wrong with aborting a simulation, because it is known 
that simulation is not always the correct tool to determine 
halting/non-halting behaviour. In such cases other tools are needed to 
determine halting/non-halting behaviour.