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

On 6/28/2025 12:38 PM, Mike Terry wrote:

On 28/06/2025 17:47, olcott wrote:

On 6/28/2025 11:22 AM, Mike Terry wrote:

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"???
>

>
Do you know what a number is?
In this case I am referring to every non-negative integer.
>

a)   Does a correct simulation need to simulate until its target computation terminates?
>

>
There is no target computation here we only have the
behavior that the input to HHH(DDD) specifies.
>

 ok I see you intend to be completely non-responsive, just as you were in the "diverging simulations" thread.
 I'll just point out for the record that you haven't provided any justification for accusing Alan or anybody else of lying to you.  Indeed, your responses just show that you have no clue what other people are saying to you!
 Mike.
 

*I wasn't clear enough*
Here proof that Alan was dishonest by changing my
quoted words and changing the time date stamp to
make it more difficult to see that he was cheating.
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.
On 6/27/2025 11:34 AM, olcott wrote:
 > I know that DDD correctly simulated by HHH cannot
 > possibly reach its own simulated "return" statement
 > final halt state because the execution trace
 > conclusively proves this.
Since HHH simulates DDD according to the semantics of the C
programming language this simulation is necessarily correct.
Disagreeing with this tries to get away with disagreeing
with the semantics of the C programming language.
How can that be honest?
It is also the case that everyone besides Keith
on comp.theory has disagreed with these words:
DDD correctly simulated by HHH cannot possibly reach
its own simulated "return" statement final halt state.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer