Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--

On 3/8/24 11:08 AM, olcott wrote:

On 3/8/2024 12:58 PM, immibis wrote:

>
So it's impossible to make a Turing machine that writes 12345 onto its tape unless 12345 is a parameter?

 That is a valid point.
It is impossible to get me to talk about that until
we first have full closure that the Linz H correctly
determines the halt status of the Linz Ĥ ⟨Ĥ⟩ when run
in the Olcott master UTM.
 

In other words you are going to ignore the true facts until someone agrees to your lies.
Since it has been shown that for the PROPERLY CONSTRUCTED Linz H^ (built by the semantic definition implemented for Olcott Machines) that Linz H will NOT give the right answser, I guess you are deciding to DIE being still known as the idiot.
(It becomes a bit of a Lie to try to refer to Linz's proof with Olcott Machines since it SPECIFICALLY was talking about Turing Machines)
With the adapted definition of H^ that transforms
H^.q0 (M) <H^> |-> H^.Hq0 (M) (M) <H> to get H's answer to the machine M and thus for the input of the description of itself it does
H^.q0 (H^) <H^> |-> H^.Hq0 (H^) (H^) <H> which will go to the same H state as H (H^) (H^) <H> since it is the same algorithm with the same input, and then Hqy loops forevern and Hqn Halts, we see that H can not get the right answer.
When we look even farther at your description, we see that H, since its first parameter doesn't match it given self description, not even a sub-string of it (since the state number would have gotten remapped), so H doesn't abort it's simulation.
Also H^.H (H^) (H^) <H> doesn't see a match either, as H^ change the description of the running machine to match that of the top level H.
So Nobody in this chain will abort their simulatons, and thus end up in the same case as your old system when H insists on continuing to simulate until it can actually prove its answer (which it never does) and thus nothing ever halts.
Thus, you have an eternity to look at the problem until someone will correctly agree to your incorrect conclusion.
Yes, perhaps with an improperly built H^, H might be able to get the right answer, but that is the problem with "Proof by example", you need to look at an infinite number of examples to be exhaustive.
The claim isn't that H gets NO answers right, just that there is at least one input that it doesn't get right, and it is shown how to build that for any arbitrary machine that might be claimed to be a Halt Decider, thus proving it isn't, and since we can do this for ANY machine, that none are.