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.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

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?

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.

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.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

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.

--
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.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)