Re: Simulating Termination Analyzer HHH correctly rejects input DDD

On 10/10/24 9:55 PM, olcott wrote:

On 10/9/2024 6:48 PM, Richard Damon wrote:

On 10/9/24 2:46 PM, olcott wrote:

On 10/9/2024 6:46 AM, Richard Damon wrote:

On 10/9/24 7:01 AM, olcott wrote:

On 10/9/2024 1:08 AM, Jeff Barnett wrote:

On 10/8/2024 6:49 AM, Andy Walker wrote:

... after a short break.
>
     Richard -- no-one sane carries on an extended discussion with
someone they [claim to] consider a "stupid liar".  So which are you?
Not sane?  Or stupid enough to try to score points off someone who is
incapable of conceding them?  Or lying when you describe Peter?  You
must surely have better things to do.  Meanwhile, you surely noticed
that Peter is running rings around you.
>
     Peter -- you surely have better things to do.  No-one sensible
is reading the repetitive stuff.  Decades, and myriads of articles, ago
people here tried to help you knock your points into shape, but anything
sensible is swamped by the insults.  Free advice, worth roughly what you
are paying for it:  step back, and summarise [from scratch, not using HHH
and DDD (etc) without explanation] (a) what it is you think you are trying
to prove and (b) what progress you claim to have made.  No more than one
side of paper.  Assume that people who don't actively insult you are, in
fact, trying to help.

>
And this approach has been tried many times. It makes no more progress than the ones you are criticizing. Just assume the regulars are lonesome, very lonesome and USENET keeps everybody off the deserted streets at night.

>
HHH is an emulating termination analyzer that takes the machine
address of DDD as input then emulates the x86 machine language
of DDD until a non-terminating behavior pattern is recognized.

>
But fails, because you provided it with a proven incorrect pattern
>

>
HHH recognizes this pattern when HHH emulates itself emulating DDD
>
void DDD()
{
   HHH(DDD);
   return;
}
>

>
Which isn't a correct analysis (but of course, that is just what you do)
>
Since we know that HHH(DDD) returns 0, it can not be a non- terminating behaivor, but that claim is just a lie.
>

One cannot simply ignore the actual behavior specified by the
finite string x86 machine language of DDD such that
>

>
Right, one can not ignore the fact that HHH(DDD) is determined to return 0.
>

DDD emulated by each corresponding HHH that can possibly
exist never returns

>
More lies. It has been determined that EVERY DDD that calls an HHH(DDD) that returns 0 will halt.
>
The DDDs that don't return are the ones that call an HHH that never returns an answer.
>

>
*Your weasel words are in incorrect paraphrase of this*

>
WHAT PARAPHARSE.
>

>
DDD emulated by each corresponding HHH that can possibly
exist never returns

>
No, that means the behavior of the code of DDD when directly executed. or youy are lying about working on the Halting Problem.
>

 It seems to me that you just said that:
the behavior of DDD emulated by HHH
<is not>
the behavior of DDD emulated by HHH.
 

No, (the behavior of ddd) [that was] emulated by HHH is a different sentence then
the behavior (of DDD) [by] emulation by HHH

Instead you seemed to have said that:
the behavior of DDD emulated by HHH
is the behavior of DDD when directly executed.

(The behavior of DDD) is the behavior of DDD directly executed.
The emulated by HHH, just specifies WHICH DDD was look at.
The sentence, in the context of Computation Theory and the Halting Problem, does not allow the "emulation" to modify behavior (which wasn't in your original sentence to begin with) as which behavior is defined by the technical definiitions of the field, something you have CHOSEN to no learn, thus leading you into self-inflicted stupidity.

 Is this what you mean?
 

You are just proving you are trying to use an equivocation.