Re: ChatGPT refutes the key rebuttal of my work

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:
>

ChatGPT does correctly apply truth preserving operations to the
premises that it was provided regarding the behavior of DDD and
HHH.
*Try to find a mistake in its reasoning*

No reasoning shown.

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.

It is nonsensical for HHH not to report that DDD terminates.

The explanation is quite good. I will take what you said to mean
that it was over your head or didn't bother to look at it.
You never confirmed that you even know what infinite recursion is.

No, he means your argument is just non-sense, and it is just a
blantant lie that you put forwards because you just don't understand
what you are talking about.,

In other words you coward away from trying to convince ChatGPT that is
is incorrect.

What do you mean. With one statement I got it to admit that the ACTUAL
behavior of DDD was to halt.
 
 

Since you say that it is a YES man it should be easy for you to get it
to admit that it is wrong.

Which I did,
 

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.

You err because you fail to understand how the same C/x86 function
invoked in a different process context can have different behavior.

Do explain how a pure function can change.

--
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.