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

On 7/03/24 18:22, olcott wrote:

On 3/7/2024 9:07 AM, immibis wrote:

On 7/03/24 04:21, Richard Damon wrote:

But, if H^.H was constructed by the requirements of being an exact copy of the algoirthm of the compuation H, then H (H^) (H^) and H^H (H^) (H^) need to end up in the equivalent state, so you are just admitting that either they DON'T ACTUALLY GO to the correct states you listed, or that you have LIED about doing it right.

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This is key.
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Supposedly, an Olcott machine is a Turing machine where the initial tape always ends with four spaces, its own machine description, and four more spaces. To make an "embedded copy" of an Olcott machine for the purpose of the Linz proof, simply copying the code/transition table doesn't work. We have to add a section of code/states at the beginning, which deletes the machine description from the tape and then replaces it with the machine description of the original machine. Then we have a "copy" that works for the Linz proof.

 It is still the case that the uncorrupted Linz H ⟨Ĥ⟩ ⟨Ĥ⟩
H can correctly determine whether or not Ĥ ⟨Ĥ⟩ halts.

The Linz proof is for Turing machines. The updated Linz proof for Olcott machines is like this:
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