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

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
>
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⟩.
>

>
Nope.
>
Since we are talking about Turing Machines, your stipulated POOP definitions go away,

>
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
>
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
   Ĥ copies its own Turing machine description: ⟨Ĥ⟩
   then invokes embedded_H that simulates ⟨Ĥ⟩ with ⟨Ĥ⟩ as input.
>
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.

>
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.

 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.
 *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)

 *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.

 The above is STEP TWO  of my four step proof. STEP ONE can be
discarded if it is easy enough for you and others to accept
this STEP TWO. I already covered most of STEP THREE.
 

And thus your proof is shown to be just wrong.
Rmember, when we are talking about Turing Machines (and Compuation theory in general) a "Correct Simulation" means a simulation that is never aborted, and thus WILL correctly recreate the behavior of the machine being simulated.
That means the correct simulation of a non-halting machine description, IS non-halting.
That is why using "correct simulation" as your method of halt deciding doesn't work.
You need to use partial simulation that actually PROVES what the Correct Simulation (that isn't actually done) would do.
I.e. your wording of "Correct Simulation by H" automattically means H must fail for non-halting inputs, as if H DOES do a correct simulation (as the criteria states) it must be non-halting.
You need your H to try to analyze the Correct Simulation by a UTM (which isn't H) would never halt, but that gets you right back to where you started.