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On 5/30/2025 1:57 PM, Richard Heathfield wrote:No, the input to HHH is DD, which calls function. The addresses of these functions are present in DD. So HHH has access to all functions used by DD, including the code that specifies the abort and the halting of the program.On 30/05/2025 19:48, dbush wrote:IT DOES YET YOU INSIST ON LYING ABOUT THIS.On 5/30/2025 2:40 PM, olcott wrote:>On 5/30/2025 1:20 PM, Richard Heathfield wrote:
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Let us suppose that there is such a process; that is to say, that we can invent a machine <D- which, when supplied with the S.D of any computing machine i l will test this S.D and if i l is circular will mark the S.D with the symbol "u" and if it is circle-free will mark it with " s ".
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By "the S.D. of any computing machine" he means the 'standard description' of >>>>any<<<< Turing machine.
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HHH is not that process, and thus HHH has no bearing whatsoever on the Turing proof.
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It is a verified fact that
HHH is not that process, and thus HHH has no bearing whatsoever on the Turing proof.
>
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
The input to HHH(DD) SPECIFIES A NON-HALTING
SEQUENCE OF CONFIGURATIONS.
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