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

Am Wed, 16 Oct 2024 19:55:18 -0500 schrieb olcott:

On 10/16/2024 7:37 PM, Richard Damon wrote:

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.

You are responding to something that I did not say.

Did you say that HHH does not determine the behaviour of DDD?

HHH correctly emulates N steps of DDD therefore N steps of DDD are
correctly emulated by HHH.

Yes, and the rest are not simulated at all, not even incorrectly.

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Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.