XPost: comp.theory, sci.logic, sci.math

On 3/22/2022 5:50 AM, Richard Damon wrote:

On 3/21/22 10:39 PM, olcott wrote:

On 3/21/2022 9:16 PM, Richard Damon wrote:

On 3/21/22 10:05 PM, olcott wrote:

A copy of Linz H is embedded at Ĥ.qx as a simulating halt decider
(SHD).

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its >>>> final state.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach
its final state.

When Ĥ is applied to ⟨Ĥ⟩
   Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then embedded_H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩

Then these steps would keep repeating:
   Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then embedded_H0 simulates ⟨Ĥ1⟩
⟨Ĥ2⟩
   Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H1 simulates ⟨Ĥ2⟩
⟨Ĥ3⟩
   Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H2 simulates ⟨Ĥ3⟩
⟨Ĥ4⟩...

Because we can see that a correct simulation of the input to
embedded_H cannot possibly reach its final state we can see that
this input never halts.

But ONLY if mbedded_H doesn't abort its simulation.

computation that halts … the Turing machine will halt whenever it
enters a final state. (Linz:1990:234)

Never halts means that the simulated input will never reach its final
state whether or not its simulation is aborted.

Never Halts means the TURING MACHINE will never reach a final state.
(Where do you get 'Simulation' out of the definition?)

AS YOU ALREADY KNOW:
An executed Turing machine and a Turing machine description simulated by
a UTM are computationally equivalent.

Pretending to not know these things just so that you are can remain disagreeable is quite a jackass move.

YOu can replace that with a simulation that is ACCURATE, but to be
accurate for non-halting, it must NOT be aborted, i.e., must run forever.

YET ANOTHER JACKASS MOVE:
You know that aborting the simulation of a Turing machine description
that would never reach its final state does not cause this Turing
machine description to reach its final state.

Therefore you know that the input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H specifies a sequence of configuration that never reach their final state under any circumstances.

Therefore you know that the input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H cannot meet the
Linz definition of a

computation that halts … the Turing machine will halt whenever it
enters a final state. (Linz:1990:234) UNDER ANY CIRCUMSTANCES

Therefore you know that when embedded_H rejects its input it is correct.

Therefore you know that when embedded_H rejects its input it and is
correct the Linz proof that asserts that rejecting its input results in
a contradiction is refuted.

The fact that an aborted simulation didn't reach a final state is NOT
proof that the machine in non-halting.

You don't evn understand basic English.

--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

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* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 3/22/2022 5:57 PM, Richard Damon wrote:

On 3/22/22 10:13 AM, olcott wrote:

On 3/22/2022 5:50 AM, Richard Damon wrote:

On 3/21/22 10:39 PM, olcott wrote:

On 3/21/2022 9:16 PM, Richard Damon wrote:

On 3/21/22 10:05 PM, olcott wrote:

A copy of Linz H is embedded at Ĥ.qx as a simulating halt decider >>>>>> (SHD).

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its
final state.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach
its final state.

When Ĥ is applied to ⟨Ĥ⟩
   Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then embedded_H simulates ⟨Ĥ0⟩
⟨Ĥ1⟩

Then these steps would keep repeating:
   Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then embedded_H0 simulates
⟨Ĥ1⟩ ⟨Ĥ2⟩
   Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H1 simulates
⟨Ĥ2⟩ ⟨Ĥ3⟩
   Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H2 simulates
⟨Ĥ3⟩ ⟨Ĥ4⟩...

Because we can see that a correct simulation of the input to
embedded_H cannot possibly reach its final state we can see that
this input never halts.

But ONLY if mbedded_H doesn't abort its simulation.

computation that halts … the Turing machine will halt whenever it
enters a final state. (Linz:1990:234)

Never halts means that the simulated input will never reach its
final state whether or not its simulation is aborted.

Never Halts means the TURING MACHINE will never reach a final state.
(Where do you get 'Simulation' out of the definition?)

AS YOU ALREADY KNOW:
An executed Turing machine and a Turing machine description simulated
by a UTM are computationally equivalent.

Pretending to not know these things just so that you are can remain
disagreeable is quite a jackass move.

YOu can replace that with a simulation that is ACCURATE, but to be
accurate for non-halting, it must NOT be aborted, i.e., must run
forever.

YET ANOTHER JACKASS MOVE:
You know that aborting the simulation of a Turing machine description
that would never reach its final state does not cause this Turing
machine description to reach its final state.

The problem you have is that that the simulation that shows it would
never reach a final state requries that H/embedded_H NEVER abort its simulation.

So basically when the simulated input has had it simulation aborted it continues on and reaches its final state even though it was aborted.

--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)