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On 2025-10-22 12:32:48 +0000, olcott said:Crucial to the fact that DD halts.
On 10/22/2025 4:04 AM, Mikko wrote:Irrelevant to the fact that DD halts.On 2025-10-21 23:54:33 +0000, olcott said:>
>On 10/21/2025 4:53 AM, Mikko wrote:>On 2025-10-20 17:03:55 +0000, olcott said:>
>On 10/20/2025 4:18 AM, Mikko wrote:>On 2025-10-19 16:18:38 +0000, olcott said:>
>On 10/19/2025 3:57 AM, Mikko wrote:>On 2025-10-18 11:04:45 +0000, olcott said:>
>On 10/18/2025 4:40 AM, Mikko wrote:>On 2025-10-09 17:27:39 +0000, olcott said:>>>
*Ben already agreed that I met that particular reading*
But Sipser hasn't agreed.
It told me that I could quote his agreement.
That only covers the specific words. He didn't agree to any interpretaion
of the words that deviates from his understanding of them, nor to any
inferences from those words.
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its
input D until H correctly determines that its simulated D
would never stop running unless aborted then
>
H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
What Sipser did not agree was that even if the direct execution of D
halts if not aborted it is still possible that H correctly determines
that its simulated D would never stop running unless aborted.
In the case of HHH(DD) even the directly executed DD()
does not halt unless HHH(DD) unless HHH aborts the
simulation of its input.
There is no "unless" about it. DD halts.
That DD() halts depends on HHH(DD) rejecting its input.
You have posted traces that unambigously show that DD() halts. And anyone
can get from GirHub everything needed to check that DD() halts.
No Turing machine can compute any mapping from
non-inputs. The DD() caller of HHH(DD)
*IS NOT AN ARGUMENT TO DD*
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