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?)
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.
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.
On 3/22/22 10:13 AM, olcott wrote:So basically when the simulated input has had it simulation aborted it continues on and reaches its final state even though it was aborted.
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.
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