On 3/25/22 9:47 PM, olcott wrote:
On 3/25/2022 3:29 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
A SHD computes the mapping from its input to its own accept or reject
state based on whether or not the pure simulation of its simulated
input could reach its own final state in a finite number of simulated
steps.
The following simplifies the syntax for the definition of the Linz
Turing machine Ĥ, it is now a single machine with a single start
state. A copy of Linz H is embedded at Ĥ.qx
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its final
state.
Or, more simply, use Linz's condition: if Ĥ applied to ⟨Ĥ⟩ halts.
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.
Or, more simply, use Linz's condition: if Ĥ applied to ⟨Ĥ⟩ does not >>> halt.
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⟩...
The ... is wrong for any Ĥ that meets Linz's definitions. You can talk >>> about a TM that acts like this, but since it's not what Linz is talking
about any conclusions you draw don't apply the proof.
Even Richard understands that unless embedded_H aborts its simulation
of its input: ⟨Ĥ⟩ ⟨Ĥ⟩ that this simulation never stops.
What Richard get get wrong:
He believes that aborting the simulation of the input causes this
aborted input to reach its final state.
It isn't 'this simulation' that reaches the final state, it is the
CORRECT simulation that reaches the final state. Nobody (but you) cares
about the results of an aborted simulation by H,
olcott <NoOne@NoWhere.com> writes:
A SHD computes the mapping from its input to its own accept or reject
state based on whether or not the pure simulation of its simulated
input could reach its own final state in a finite number of simulated
steps.
The following simplifies the syntax for the definition of the Linz
Turing machine Ĥ, it is now a single machine with a single start
state. A copy of Linz H is embedded at Ĥ.qx
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its final
state.
Or, more simply, use Linz's condition: if Ĥ applied to ⟨Ĥ⟩ halts.
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.
Or, more simply, use Linz's condition: if Ĥ applied to ⟨Ĥ⟩ does not halt.
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⟩...
The ... is wrong for any Ĥ that meets Linz's definitions. You can talk about a TM that acts like this, but since it's not what Linz is talking
about any conclusions you draw don't apply the proof.
On 3/25/22 10:07 PM, olcott wrote:We are not talking about the simulation NITWIT!
On 3/25/2022 9:03 PM, Richard Damon wrote:
On 3/25/22 9:47 PM, olcott wrote:
On 3/25/2022 3:29 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
A SHD computes the mapping from its input to its own accept or reject >>>>>> state based on whether or not the pure simulation of its simulated >>>>>> input could reach its own final state in a finite number of simulated >>>>>> steps.
The following simplifies the syntax for the definition of the Linz >>>>>> Turing machine Ĥ, it is now a single machine with a single start
state. A copy of Linz H is embedded at Ĥ.qx
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its final
state.
Or, more simply, use Linz's condition: if Ĥ applied to ⟨Ĥ⟩ halts. >>>>>
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.
Or, more simply, use Linz's condition: if Ĥ applied to ⟨Ĥ⟩ does not >>>>> halt.
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⟩...
The ... is wrong for any Ĥ that meets Linz's definitions. You can >>>>> talk
about a TM that acts like this, but since it's not what Linz is
talking
about any conclusions you draw don't apply the proof.
Even Richard understands that unless embedded_H aborts its
simulation of its input: ⟨Ĥ⟩ ⟨Ĥ⟩ that this simulation never stops.
What Richard get get wrong:
He believes that aborting the simulation of the input causes this
aborted input to reach its final state.
It isn't 'this simulation' that reaches the final state, it is the
CORRECT simulation that reaches the final state. Nobody (but you)
cares about the results of an aborted simulation by H,
If the simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach its own final >> state then the simulated input (by definition) specifies a non-halting
sequence of configurations.
The CORRECT simulation reaches the final state.
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