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On 8/13/2024 2:29 AM, Fred. Zwarts wrote:No evidence, except four unspecified 'experts' against hundreds of other experts.Op 12.aug.2024 om 14:42 schreef olcott:*Every attempt at rebutting this has been*On 8/11/2024 2:54 PM, Fred. Zwarts wrote:Many more experts with master degrees tell you that it does halt.Op 11.aug.2024 om 13:45 schreef olcott:>>>
void DDD()
{
HHH(DDD);
return;
}
>
None-the-less it is clear that of the above specified infinite
set DDD correctly emulated by each element of that set never
reaches its own "return" instruction halt state.
Since no DDD is correctly simulated by HHH, we are talking about the properties of an empty set.
But, indeed, the simulation of DDD by HHH fails to reach the halt state. It aborts one cycle before the simulated HHH would reach its 'return' instruction, after which DDD would reach its halt state.
>>Every C programmer understands that a simulation fails if it does not reach the end of a halting program.
My words must be understandable by ordinary C programmers
and computer scientists. The latter tend to conclude that
my work is incorrect as soon as they know the subject matter
before actually seeing what I said.
>
Four expert C programmers (two with masters degrees in
computer science) agree that DDD correctly simulated by
HHH does not halt.
>
Show evidence instead of authority.
(a) Denying verified facts
(b) Strawman-deception of changing what I said and rebutting that
(c) Pure ad hominem insults with zero reasoning
Mike is the only one here that seems to have enough technicalYou did not. You showed a simulation that fails to reach the end of a halting program. So, the simulation is incomplete and therefore incorrect.
skill to understand the verified facts. Everyone else denies them
entirely on the basis of their own ignorance.
Overview of proof that the input to HHH(DDD) specifies non-halting behavior
We prove that the simulation is correct.
Then we prove that this simulation cannot possiblyNo proof has been provided.
reach its final halt state / ever stop running without being aborted.
The semantics of the x86 language conclusive proves this is true.
Thus when we measure the behavior specified by this finiteNo it only shows a failed attempt to create a simulating decider.
string by DDD correctly simulated/emulated by HHH it specifies
non-halting behavior.
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