On 2025-10-28, joes <noreply@example.org> wrote:Actually that is the key computer scienceAm Tue, 21 Oct 2025 18:54:33 -0500 schrieb olcott:Moreover, there cannot be a "this DD" versus "that DD".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:Which it does, therefore DD halts.That DD() halts depends on HHH(DD) rejecting its input.There is no "unless" about it. DD halts.What Sipser did not agree was that even if the direct execution of DIn the case of HHH(DD) even the directly executed DD()
halts if not aborted it is still possible that H correctly determines
that its simulated D would never stop running unless aborted.
does not halt unless HHH(DD) unless HHH aborts the simulation of its
input.
If you're thinking about the problem in such a way that you have to
distinguish different DDs, to all which the DD name refers
simultaneously in the same scenario, you are wrong.
discovery that I made. You have to accept
the first step of my proof before you will
understand the next step:
int D()
{
int Halt_Status = H(D);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
H simulates D
that calls H(D) to simulate D
that calls H(D) to simulate D
that calls H(D) to simulate D
that calls H(D) to simulate D
that calls H(D) to simulate D
until H sees this repeating pattern.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
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