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On 5/14/2025 6:43 PM, Keith Thompson wrote:Claiming and shouting is not proving. It is dead obvious that HHH fails to reach the reachable end of the program, because HHH fails to see the most important part of its input, the part that aborts. That the programmer made HHH blind for that part of the input, does not mean that it is not present.olcott <polcott333@gmail.com> writes:That is ridiculous as you already acknowledged.
[...]HHH does correctly simulate DDD until[...]
HHH correctly determines that its simulated DDD
would never stop running unless aborted.
>
And my understanding is that it is mathematically impossible for HHH
to do what you claim it does,
void DDD()
{
HHH(DDD);
return;
}
It is dead obvious to any expert in C to correctly
determine that DDD cannot possibly stop running
when HHH is a pure simulator.
HHH merely needs to see this exact same DEAD OBVIOUS thing.
>
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