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

On 10/16/24 8:55 PM, olcott wrote:

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

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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.
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It has no effect on the trace itself.

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Yes it does.
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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.
>
>

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Nope, not to a degree that determine the final behavior of the input.
>

 This is your ADD. You are responding to something that I did not say.
Like I said that I do, try rereading the above paragraph sixteen times.
 I will dumb it down for you so you can get the gist of it.
HHH correctly emulates N steps of DDD therefore N steps of
DDD are correctly emulated by HHH.
 

Right, but just because N steps don't get to the return, doesn't mean that the input doesn't return.
*AND*
Since changing the code for HHH means that you get a different program/funcition DDD as far as behavior is concerned, you can't argue about the results of changing HHH to not abort saying anything about the behavior of the DDD that the original HHH was given, since the non-aborting one had a different input (or the input wasn't suitable for the problem).
Thus, your trying to change the criteria from the actual Behavior of the program/function described by the input to what the emulation by HHH does is invalid, as HHH doesn't do a emulation that qualifies, as it only did a partial emulaiton.
Thus, since the DDD that HHH was actually given will return, HHH can not be correct to say that its doesn't.
The statement that you are actually showing, and using equivocation to try to make look like the needed behavior, is that HHH shows that it did not emulate its input to a final state.
So, all you are doing is ADMITTING that you arguement that HHH is correct is based on equivocation.