On 2024-10-02 18:41:54 +0000, olcott said:
On 10/2/2024 9:36 AM, olcott wrote:
On 10/2/2024 4:42 AM, Alan Mackenzie wrote:
olcott <[email protected]> wrote:
On 10/1/2024 2:15 PM, Alan Mackenzie wrote:
olcott <[email protected]> wrote:
On 10/1/2024 7:39 AM, olcott wrote:
Simulating Termination Analyzer H is Not Fooled by Pathological Input D >>>
[ .... ]
https://chatgpt.com/share/66fbec5c-7b10-8011-9ce6-3c26424cb21c
The above link to a ChatGPT conversation is entirely
complete with the first part being the entire input
provided to ChatGPT. The second part is the output that
ChatGPT deriving from analyzing this input.
You are aware that programs like ChatGPT are know for "hallucinating" >>>>> non-facts? They have even less understanding of the truth than you do. >>>
In fact, they tend to regurgitate whatever "facts" they are fed with.
In other words you can convince it that its analysis
of my work is incorrect. I dare you to try to do that.
I've got better things to do with my time. Real live competent
mathematicians have shown your work to be incorrect.
Try this for yourself.
Real live computer scientists begin with the assumption that I
am incorrect and then try to justified that false assumption.
ChatGPT has not been indoctrinated thus reports on what it
sees.
https://chatgpt.com/share/66fbec5c-7b10-8011-9ce6-3c26424cb21c
Does HHH have to abort its emulation of DDD to prevent the infinite
execution of DDD?
Several software engineers (two with master degrees in computer science)
Agree with the ChatGPT answer to the above question when they were asked
to independently derive this answer. Any C programmer that understands
infinite recursion has agreed.
What a dumb chat
program regurgitates has no relevance to anything.
Unless what this dumb chat program says is irrefutably correct.
To show that it is [INCORRECT] one must find an actual mistake.
No one has ever done that. The most that they did is show that
things did not conform to their provably false assumptions.
int main() { DDD(); } does have a different execution
trace than when it is emulated by the emulator that it
calls: HHH(DDD).
The trace of DDD() is the same. If HHH(DDD) does only a partial
simulation of DDD() or partial tracing then some part of the
trace of DDD() are missing. The trace of main contains an initial
and final part that is not a part of the trace of DDD(). So the
traces are different but there is only one trace of DDD().
--
Mikko
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