Re: How do simulating termination analyzers work?

On 6/24/2025 3:53 AM, Mikko wrote:

On 2025-06-23 16:51:23 +0000, olcott said:
 

On 6/23/2025 2:12 AM, Mikko wrote:

On 2025-06-22 19:16:24 +0000, olcott said:
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On 6/22/2025 3:59 AM, Mikko wrote:

On 2025-06-21 17:34:55 +0000, olcott said:
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On 6/21/2025 4:52 AM, Mikko wrote:

On 2025-06-20 13:59:02 +0000, olcott said:
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On 6/20/2025 4:20 AM, Fred. Zwarts wrote:

Op 19.jun.2025 om 17:17 schreef olcott:

On 6/19/2025 4:21 AM, Fred. Zwarts wrote:

Op 18.jun.2025 om 15:46 schreef olcott:

On 6/18/2025 5:12 AM, Fred. Zwarts wrote:

Op 18.jun.2025 om 03:54 schreef olcott:

On 6/17/2025 8:19 PM, Richard Damon wrote:

On 6/17/25 4:34 PM, olcott wrote:

void Infinite_Recursion()
{
   Infinite_Recursion();
   return;
}
>
void Infinite_Loop()
{
   HERE: goto HERE;
   return;
}
>
void DDD()
{
   HHH(DDD);
   return;
}
>
When it is understood that HHH does simulate itself
simulating DDD then any first year CS student knows
that when each of the above are correctly simulated
by HHH that none of them ever stop running unless aborted.

>
WHich means that the code for HHH is part of the input, and thus there is just ONE HHH in existance at this time.
>
Since that code aborts its simulation to return the answer that you claim, you are just lying that it did a correct simulation (which in this context means complete)
>

>
*none of them ever stop running unless aborted*

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All of them do abort and their simulation does not need an abort.
>

>
*It is not given that any of them abort*
>

>
At least it is true for all aborting ones, such as the one you presented in Halt7.c.

>
My claim is that each of the above functions correctly
simulated by any termination analyzer HHH that can possibly
exist will never stop running unless aborted by HHH.
Can you affirm or correctly refute this?

>

Yes, I confirmed many times that we can confirm this vacuous claim, because no such HHH exists. All of them fail to do a correct simulation up to the point where they can see whether the input specifies a halting program.

>
if DDD correctly simulated by any simulating termination
analyzer HHH never aborts its simulation of DDD then

>
that HHH is not interesting.

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*then the HP proofs are proved to be wrong*

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No, they are not. You have not solved the halting problem and that
(in addition to all proofs) supports the claim that halting problem
is unsolvable.

>
ChatGPT corrected my words and agreed that I have
correctly refuted the generic HP proof technique
where an input has been defined to only do the
opposite of whatever value that its decider decides.
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76

>
Doesn't matter. Only proofs matter. So far you have not proven anything
and it is unlikely you could prove anything even after asking ChatGPT
for help.
>

The ChatGPT that evaluated and affirmed my analysis
of HHH(DDD) one year ago could only handle 4000 tokens
thus could not understand HHH(DD).
>
ChatGPT with GPT-4-turbo — can handle up to 128,000 tokens
of context in a single conversation, immediately understood
HHH(DD) within the context of the conversation of HHH(DDD).

>
ChatGPT does not understand. Whether you do is still not determined.
>
Anyway,

In order to show that a proof is wrong you need to show an error
in the proof. Even then the conclusion is proven unless you can
show an error in every proof of that conclusion.

>

>
That you do not understand that any set of expressions of
language that show another expression of language is
necessarily true is its proof is your ignorance not mine.

 Only a proof is a proof.
 

There are things called proofs that have a certain
form and there is the broader concept of proof that
does not require this certain form.
A proof is any set of expressions of language that
correctly concludes that another expression of
language is definitely true.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer