Re: HHH maps its input to the behavior specified by it --- never reaches its halt state

On 8/7/24 10:53 PM, olcott wrote:

On 8/7/2024 8:22 PM, Richard Damon wrote:

On 8/7/24 9:12 PM, olcott wrote:

On 8/7/2024 8:03 PM, Richard Damon wrote:

On 8/7/24 2:14 PM, olcott wrote:

On 8/7/2024 1:02 PM, joes wrote:

Am Wed, 07 Aug 2024 08:54:41 -0500 schrieb olcott:

On 8/7/2024 2:29 AM, Mikko wrote:

On 2024-08-05 13:49:44 +0000, olcott said:

>

I know what it means. But the inflected form "emulated" does not mean
what you apparently think it means. You seem to think that "DDD
emulated by HHH" means whatever HHH thinks DDD means but it does not.
DDD means what it means whether HHH emulates it or not.
>

In other words when DDD is defined to have a pathological relationship
to HHH we can just close our eyes and ignore it and pretend that it
doesn't exist?

It doesn't change anything about DDD. HHH was supposed to decide anything
and can't fulfill that promise. That doesn't mean that DDD is somehow
faulty, it's just a counterexample.
>

>
void DDD()
{
   HHH(DDD);
   return;
}
>
*HHH is required to report on the behavior of DDD*
Anyone that does not understand that HHH meets this criteria
has insufficient understanding.

>
But it doesn't, as a correct simulation of a DDD that calls an HHH that returns will stop running,

>
I really think that you must be a liar here because
you have known this for years:
>
On 8/2/2024 11:32 PM, Jeff Barnett wrote:
 > ...In some formulations, there are specific states
 >    defined as "halting states" and the machine only
 >    halts if either the start state is a halt state...
>
 > ...these and many other definitions all have
 >    equivalent computing prowess...
>
Anyone that knows C knows that DDD correctly simulated
by any HHH cannot possibly reach its "return" {halt state}.
>

>
But the problem is that you HHH ODESN'T correctly emulate the DDD it is given, because it aborts its emulation.
>

 Each HHH of every HHH that can possibly exist definitely
emulates zero to infinity instructions correctly. In
none of these cases does the emulated DDD ever reach
its "return" instruction halt state.

Nope, only the ONE HHH that never aborts emulates its input completely correctly, and that one fails to answer.
Every other one proves your next sstatement is a Projection of your own LIES that partial emulation is a complete correct emulation.

 *There are no double-talk weasel words around this*
*There are no double-talk weasel words around this*
*There are no double-talk weasel words around this*]\

Except for every word you say.

 Everyone can tell that Richard is trying to get way
with disagreeing with a tautology.

No, every one sees YOU are LYING with weasel words that try to give double-meaning
You failure to answer ANY of the question just proves your total ignorance of the tpo[pic.

 The strawman deception won't work any more Richard!
When you switch from DDD correctly simulated by HHH
everyone can tell.
 

Ecept that you never had a HHH that has completely correctly emulated DDD that you have produced, so you whole arguement is built on your lies. How can I switch from something that never was.
And, since the STARTING POINT it the behavior of DDD directly run, your simulation is itself the strawman, and he is smarter than you.
So, you just keep repeating your disproven LIES, PROVING YOUR STUPIDITY
Every HHH that simulates for a finite number of steps, and then aborts and returns (which is the class that you claim to be correct) only PARTIALLY simulate the input, and thus only get PARTIAL details of the behavior of the PROGRAM, a thing you don't seemt o understand about, which makes your logic so broken.
Since it is the behavior of the PROGRAM that is what the decider needs to answer about, since the question is about if the PROGRAM given as the input will halt, and the behavior of the progrma is defined by the semantics of the x86 language which says that the PROGRAM doesn't stop unless it reaches a terminal instruction, means that HHH stopping its emulation (which just makes its emulation incorrect) doesn't stop the behavior of the program, that will continue until the HHH that it called, which behaves IDENTICALLY to the HHH that was emulating it, also stops its emulation and returns to DDD which then reaches its terminal instruction and halts.
Thus, the CORRECT answer about the halting of the program represented by the input to HHH, if HHH answers, is that it halts, even though the correct emulation of a finite number of steps didn't get to there. But it is a simple fact that the correct emulation of a finite number of steps doesn't correctly show the behavior of a larger finite number of steps, and thus isn't a totally correct emulation.
You logic is just based on double-talk and weasel words where you need ot use words with shades of meaning and use them in multiple contradictory meanings in one statement.
This just shows how poor your actual position is, as you can't actually point to one actual respected source for any part of your logic, at best you get people with minor names whi make similar errors.
Sorry, you are just proving your ignorance and stupidity.
Just repeating the same sentences just proves it.