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

On 8/03/24 17:23, olcott wrote:

On 3/8/2024 10:17 AM, immibis wrote:

On 8/03/24 03:16, olcott wrote:

On 3/7/2024 7:37 PM, immibis wrote:

On 8/03/24 00:04, olcott wrote:

On 3/7/2024 4:21 PM, Richard Damon wrote:

On 3/7/24 12:20 PM, olcott wrote:

On 3/7/2024 1:59 PM, Richard Damon wrote:

On 3/6/24 11:11 PM, olcott wrote:

On 3/7/2024 12:37 AM, Richard Damon wrote:

On 3/6/24 10:17 PM, olcott wrote:

>

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn     // Ĥ applied to ⟨Ĥ⟩ does not halt
>
The design of Olcott Machines makes quite easy for Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
to get its abort criteria.
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Which doesn't match the Halting Problem requirements,

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It does match the Halting Problem requirements, when
they are implemented indirectly as "abort criteria".
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Which is a different criteria, so you are just admitting that you are using a strawman desception and thus INTENTIONALLY LYING.
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Somehow you think lies are ok if they help you prove your false statements.

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The Linz second ⊢* enables H to compute any damn
thing as long as this ends up computing halting.
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Note quite, it is whatever the algorithm for H generates.
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That exact same algorithm exists in H^.H, so that WILL get the same answer, and since you logic says it doesn't, that means you are lying that H^ was built by the specification, or as to what H will actually do.

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*Already addressed in my reply to you here*
We finally know exactly how H1(D,D) derives a different result than H(D,D)
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We know it's because H and H1 are different programs, not copies.

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*Even when we remove the infinite loop appended to Ĥ.Hqy*
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
>
An Olcott machine Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> could trivially determine
that it is calling itself in recursive simulation.
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and Olcott machine H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> could trivially determine
that it is NOT calling itself in recursive simulation.
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This is true even when the embedded portion of H is
identical to <H>.
>

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The copy of
   H ⟨Ĥ⟩ ⟨Ĥ⟩ <H>
is not
   Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ>
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it is actually
   Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H>
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this is easily done by deleting <Ĥ> from the tape and inserting <H>
>

 Because my cancer came back too quickly (in less than 24 months)
I can't afford the time to diverge from Linz until I have closure
on Linz. I am on year 4.2 of a 50% five year mortality rate.

Here is the closure on Linz: It does not apply to Olcott machines because it is only a proof about Turing machines. For Olcott machines you need a different proof.