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On 4/24/2025 6:07 PM, Richard Damon wrote:Your are forgetting your stipulation, that Halt7.c is part of the input, and thus *THAT* definition of HHH is what you are stuck with, and that ALWAYS aborts at the point it does.On 4/24/25 3:41 PM, olcott wrote:On 4/24/2025 2:12 PM, Fred. Zwarts wrote:>Op 24.apr.2025 om 19:13 schreef olcott:>>
HHH correctly determines through mathematical induction that
DD emulated by HHH (according to the finite string transformations
specified by the x86 language) cannot possibly reach its final
halt state in an infinite number of steps.No, HHH has a bug which makes that it fails to see that there is only a finite recursion,>
*You are technically incompetent on this point*
When the finite string transformation rules of the
x86 language are applied to the input to HHH(DD)
THIS DD CANNOT POSSIBLY REACH ITS FINAL HALT STATE
not even after an infinite number of emulated steps.When the defined finite string trasnsformation rules, thos of the x86 language, are applied to this input, completed with the definitions from Halt7.c as stipulated, we see that DD calls HHH(DD), that it will spend some time emulating DDm then it willCorrectly determine that DD emulated by HHH can never possibly
reach its final halt state even after an infinite number of
steps are emulated. Many C programmers have attested to this.
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
When the smart knowledgeable people here disagree it seems
far too implausible to construe this as any honest mistake.
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