Sujet : Re: HHH(DDD) is correct to reject its input as non-halting --- EVIDENCE THAT I AM CORRECT
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory sci.logic comp.ai.philosophyDate : 16. Jun 2025, 00:44:54
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <e56b12c126e72134e8761986f8d2d0d047560a24@i2pn2.org>
References : 1
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On 6/15/25 4:10 PM, olcott wrote:
void DDD()
{
HHH(DDD);
return;
}
When I challenge anyone to show the details of exactly
how DDD correctly simulated by ANY simulating termination
analyzer HHH can possibly reach its own simulated "return"
statement final halt state they ignore this challenge.
And it seems you don't understand that the problem is that while, yes, if HHH does infact do a correct simulation, it will not reach a final state, that fact only applie *IF* HHH does that, and all the other HHHs which differ see different inputs.
You don't seem to understand that you can't use as a premise to a implication a false statement and get anything out of it.
That
THAT FACT THAT NOT ONE PERSON HAS MET THIS CHALLENGE
IN SEVERAL YEARS IS VERY STRONG EVIDENCE THAT I AM CORRECT.
No, you are just showing that you don't understand how logic works.
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
DDD is a simplified version of DD.
So, the above is NOT a complete description of DD, as it doesn't include the code for HHH, so either you admit that it is an invalid input as is, or that you have implicitly inlclude the code of HHH into the "input", and thus there is one and only one HHH that can exist in this problem, the one that DD is calling and has been included in its input.
That also means that HHH has a full definition, and thus HHH(DD) or HHH(DDD) has a defined answer that it gives.
If that answer is 0, the that means that HHH *HAS* aborted its simulaiton, and thus hasn't done a correct simulation of its input, and thus your proposition above doesn't hold as it is based on a false premise. THus the "Halting" property of that input CAN'T be defined by what HHH did, but only by the direct execution or complete and correct simulation of that exact same input (so it calls the HHH(DD) that aborts and returns 0, even though HHH isn't the machine simulating it, as it isn't allowed to know who is looking at it).
This result, as you have previously admitted, is to Halt, and thus your HHH was just wrong.
If you want to try to claim it returns 1, because it simulated to the end, then you are just admitting that your claim is a lie, but it turns out that with your structure, as the only way your HHH was allowed to return 1 was if it simulated to the end, and thus it actually did a correct simulation. But, as you proof has shown, no HHH that actually correctly simulates and never aborts, never reaches that final state, so you claim that HHH(DD) somehow returns 1 is an admittion that you have lied.
The only other poissiblity is that HHH(DD) just doesn't return an answer, and that just makes it not a decider, and thus not a correct halt decider.
Thus, your claim that HHH correctly determines that its input "DD" is non-halting is just proven to be incorrect, and given the number of times that this has been explained, it is impossibe to believe in a truely honest mistake, and thus it is either blatant lies, perhaps based on intentional ignorance, or just pathological lying based on a mental deficiency to understand the material and the meaning of words like "correct" and "truth".
Sorry, you failure to answer nearly uncountable number of errors pointed out just provide that insurmountable proof of your error.