Re: A simulating halt decider applied to the The Peter Linz Turing Machine description ⟨Ĥ⟩

On 5/26/2024 7:15 PM, Richard Damon wrote:

On 5/26/24 7:45 PM, olcott wrote:

On 5/26/2024 6:07 PM, Richard Damon wrote:

On 5/26/24 6:47 PM, olcott wrote:

On 5/26/2024 3:20 PM, Richard Damon wrote:

On 5/26/24 3:14 PM, olcott wrote:

When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
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When we see that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H in an
infinite number of steps cannot possibly reach its own simulated
final state of ⟨Ĥ.qn⟩ and halt then we correctly deduce that the
same thing applies when simulating halt decider embedded_H correctly
simulates less than an infinite number of steps of ⟨Ĥ⟩ ⟨Ĥ⟩.
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Nope.
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Since we are talking about Turing Machines, your stipulated POOP definitions go away,

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https://www.liarparadox.org/Linz_Proof.pdf
*Simplified the notation for Ĥ on the top of page three*
and put back in the qy state shown in figure 12.2
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When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
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   Ĥ copies its own Turing machine description: ⟨Ĥ⟩
   then invokes embedded_H that simulates ⟨Ĥ⟩ with ⟨Ĥ⟩ as input.
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It is an easily verified fact that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly reach its own simulated final state of
⟨Ĥ.qn⟩ in any finite sequence of steps.

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Nope, since we are in Turing Machines, the term "Correctly Simulated" means, and can ONLY mean, the resuts of a UTM simulation, which BY DEFINITION is nopt aborted.

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You always seem to make sure respond to a different set of words
than the words I actually said. This could be an honest mistake.
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*I SAID A CORRECT SIMULATION OF A FINITE NUMBER OF STEPS*

 No you didn't, not the last time.
 You said (H^) H^) correctly simulated by embbeded_H cannot ..."
 If embedded_H does a "Correct Simulation", then BY DEFINITION, it never aborts.
 That it doesn't reach a final state in a finte number of steps, and thus, that "Correct Simulation" was non-halting.
 (and your earlier statement tried to assert behavior of THIS H^ based on the behaviof or a DIFFERENT H^ built on a diffferent embedded_H with differet behaivor which is just unsound logic, as the two machines are essentially unrelated as far as this behavior)
 

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*WHEN I EXPLICITLY STATE A FINITE NUMBER OF STEPS THEN YOU ARE*
*FLAT OUT WRONG TO SIMPLY ASSUME AN INFINITE NUMBER OF STEPS*

 Nope, you said it didn't reach a final state in a finite number of steps, i.e the simulation is shown to be non-halting.

*If you need to, reread that many times*
 >>>> It is an easily verified fact that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by
 >>>> embedded_H cannot possibly reach its own simulated final state of
 >>>> ⟨Ĥ.qn⟩ in any finite sequence of steps.
*That it is a finite number of steps PROVES that embedded_H halts*
*That it is a finite number of steps PROVES that embedded_H halts*
*That it is a finite number of steps PROVES that embedded_H halts*
When I needed to make changes to the banks VISA credit
card system I always had to reread the spec fifteen times.
The Discover card spec was much much clearer.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer