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On 6/19/2025 3:12 AM, Mikko wrote:On 2025-06-18 14:39:02 +0000, olcott said:On 6/16/2025 3:21 AM, Mikko wrote:On 2025-06-15 14:44:47 +0000, olcott said:On 6/15/2025 4:19 AM, Mikko wrote:On 2025-06-13 15:33:45 +0000, olcott said:On 6/13/2025 5:37 AM, Mikko wrote:
Only if that non-decider simulates.>>As you respond to my question without answering it it is obvious>
that you don't see any proofs in your article.
It is a fact that there is no actual input D to any termination
analyzer H that does the opposite of whatever value that H
derives. The key element that all conventional HP proofs depend on
cannot possibly exist.
Nonsense is not a fact.
After studying these things for 22 years I found that every
conventional proof of the halting problem never provides an actual
input that would do the opposite of whatever value that its partial
halt decider (PHD) returns.
How can't it be constructed? You are certainly passing that input to HHH.The core part of those proofs is a constructive specification of thatYet no one ever noticed that the counter-example input cannot even be
test case.
constructed thus the proof itself never actually existed.
Do you think that no programs halt when HHH is prepended? Do you thinkDepending on the style of the proof one can ither prove that theNo this is counter-factual.
counter example exists or that if a halting decider exists then the
caunter example exists, too, and otherwise none is needed.
It has never been possible for *AN ACTUAL INPUT* to do the opposite of
whatever value that it decider decides.
int main()Then explain why you are not passing it to HHH for fuck's sake.
{
DD(); // IS NOT AN ACTUAL INPUT TO THE
} // HHH(DD) THAT THIS DD() CALLS.
In the C programming language it has always been impossible for theNo, C does not forbid recursion.
caller of a function to be an argument to this called function.
The finite string of x86 machine language that is passed as an argumentDD does have the same "string". How else would you pass a program?
to HHH *is not exactly one and the same thing as the directly executed
DD*
Intuition would tell you that the behavior must be the same, yetNo, the definitions tell us the behaviour is the same; your intuition
empirical proof proves they are not the same.
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