Re: ChatGPT refutes the key rebuttal of my work --- correct emulation

On 10/16/24 8:25 PM, olcott wrote:

On 10/16/2024 6:44 AM, Richard Damon wrote:

On 10/15/24 10:23 PM, olcott wrote:

On 10/15/2024 9:11 PM, Richard Damon wrote:

On 10/15/24 4:01 PM, olcott wrote:

On 10/15/2024 2:33 PM, joes wrote:

Am Tue, 15 Oct 2024 13:25:36 -0500 schrieb olcott:

On 10/15/2024 10:17 AM, joes wrote:

Am Tue, 15 Oct 2024 08:11:30 -0500 schrieb olcott:

On 10/15/2024 6:35 AM, Richard Damon wrote:

On 10/14/24 10:13 PM, olcott wrote:

On 10/14/2024 6:50 PM, Richard Damon wrote:

On 10/14/24 11:18 AM, olcott wrote:

On 10/14/2024 7:06 AM, joes wrote:

Am Mon, 14 Oct 2024 04:49:22 -0500 schrieb olcott:

On 10/14/2024 4:04 AM, Mikko wrote:

On 2024-10-13 12:53:12 +0000, olcott said:

>

https://chatgpt.com/ share/6709e046-4794-8011-98b7-27066fb49f3e When
you click on the link and try to explain how HHH must be wrong when
it reports that DDD does not terminate because DDD does terminate it
will explain your mistake to you.

I did that, and it admitted that DDD halts, it just tries to justify
why a wrong answer must be right.

It explains in great detail that another different DDD (same machine
code different process context) seems to terminate only because the
recursive emulation that it specifies has been aborted at its second
recursive call.

Yes! It really has different code, by way of the static Root variable.
No wonder it behaves differently.

There are no static root variables. There never has been any "not a pure
function of its inputs" aspect to emulation.

>

Oh, did you take out the check if HHH is the root simulator?
>

>
There is some code that was obsolete several years ago.

>
No, that code is still active. it is the source of the value for the variable Root that is passed around, and is checked in the code to alter the behavior.
>

>
It has no effect on the trace itself.

>
Yes it does.
>

 HHH is correctly emulating (not simulating) the x86 language
finite string of DDD including emulating the finite string of
itself emulating the finite string of DDD up until the point
where the emulated emulated DDD would call HHH(DDD) again.
 

Nope, not to a degree that determine the final behavior of the input.
(What do you consider the difference between "emulation" and "simulation"?)
Since HHH aborts its emulation, that emulation does not define the final behavior of the input. Since the direct execution of that input reaches the return, as does the complete emulation of this exact input (which needs to be by a copy of HHH at a different address, like HHH1, since you aren't allowed to change the code of the input, which includes the code of the HHH that it calls), we show that this DDD will return, and thus the HHH that says otherwise is just wrong.
You are just proving that you don't understand what you are talking about but think by just repeating your lies you can make your statement true, you can't.
This just shows that you are nothing but a stupid pathological liar.
If you had something, you could point out the errors in my comments showing your error, but all you can do is repeat you statements, and make unfounded claims that you can't show any sources for that support your position.
The few that you do try to show, we have shown how you misiterpret.
Sorry, you are just speaking out of your ignrance of having only not-quite-learned of the field by rote memorization without context causing you to not actually understand what you are talking about.