Re: DDD correctly simulated by HHH can't possibly reach its own "return" statement --- Liars on comp.theory

On 28/06/2025 16:00, olcott wrote:

On 6/27/2025 7:14 PM, Mike Terry wrote:

On 27/06/2025 20:36, olcott wrote:

I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
>
I have included proof that the people on comp.theory
lied about this at the bottom.
>
typedef void (*ptr)();
int HHH(ptr P);
>
void DDD()
{
   HHH(DDD);
   return;
}
>
int main()
{
   HHH(DDD);
   DDD();
}
>
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
>
On 6/27/2025 12:27 PM, Alan Mackenzie wrote:
 > In comp.theory olcott <polcott333@gmail.com
 > wrote:
 >
 >> I know that DDD .... simulated by HHH cannot
 >> possibly reach its own simulated "return" statement
 >> final halt state because the execution trace
 >> conclusively proves this.
 >
 > Everybody else knows this, too, and nobody has
 > said otherwise. *The conclusion is that the*
 > *simulation by HHH is incorrect*
 >
>
*That last sentence is an intentional falsehood*

>
Well, people here use the term "simulation" in a number of ways, right?

 *There is only one correct way*

That's patently rubbish.
This is one of your major cognitive mistakes - believing that the squiggles which make up words have an absolute meaning independently of the people using them to (hopefully) communicate.

 void DDD()
{
   HHH(DDD);
   return;
}
 The correct simulation of DDD by HHH means that HHH simulates
DDD and then emulates itself simulating DDD a number of times
according to the semantics of C.

"a number of times"???
a)   Does a correct simulation need to simulate until its target computation terminates?
b)   Does a simulation need to also correctly decide halting for its target computation?
[My personal answers would be "no" and "no" - I would consider what others may insist on calling a partial simulation to a correct example of "simulation" (but of course incomplete), provided the steps of the simulation are correctly calculated; and I consider deciding the halt status to be the job of a halt "decider" rather than a simulator [of course a halt decider may /utilise/ simulation as a technique in reaching its decision].  Other ways of looking at this are equally valid and we need to understand each other's positions.]

 Everyone here knows that DDD correctly simulated by HHH
cannot possibly reach its own simulated "return" statement.

That's what Alan said above.  Read his words:  "Everybody else knows this, too, and nobody has said otherwise."  You need to read what people write and keep track of the conversation...

Most everyone on comp.theory lies about this.

Well, not Alan (whom you accused), or me,  or Richard or Mikko or Fred (I expect I could include others but can't be bothered to research the question).
So who lies about it exactly, and you had better provide a (recent) link to them lying!  Or just admit you haven't a clue what people are actually saying to you...

 

Like I've pointed out to you on several occasions.  Maybe you should consider a simple misunderstanding over terminology before assuming bad intentions.
>

 After three years on this same point a simple misunderstanding
has been ruled out.

For a "normal" correspondant perhaps, but you forget your own disabilities - you are incapable of understanding what people say to you, including whether they are agreeing with you (in strictly limited parts) or not.

 

So what does Alan mean by "the simulation by HHH is incorrect", exactly?  (And why do you think it is incorrect?)
>

 He is simply lying. Most of the reviews of my work are
counter-factual dogmatic assertions utterly bereft of
any supporting reasoning.

Lol, you haven't even answered the question of what you think Alan meant by his statement, or why that is a lie.  So how can you be sure he is lying?  All you've talked about above is the bit Alan explicitly said he /agreed/ with.
Mike.