On 3/31/22 9:40 PM, olcott wrote:
On 3/31/2022 8:37 PM, Richard Damon wrote:
On 3/31/22 9:18 PM, olcott wrote:
On 3/31/2022 7:09 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/31/2022 3:15 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
The directly executed Ĥ applied to ⟨Ĥ⟩ is the first invocation ofThis is the old "it only halts because" ruse...
infinite recursion that only terminates normally because of its >>>>>>>> one-way dependency relationship on embedded_H aborting the second >>>>>>>> invocation of this otherwise infinite recursion.
DIFFERENT SEQUENCES OF CONFIGURATIONS WILL HAVE DIFFERENT BEHAVIOR: >>>>>>>> This makes the sequence of configurations of the simulation of >>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩
outside of Ĥ different than the the sequence of configurations >>>>>>>> of the
simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ inside of Ĥ. Different sequences of >>>>>>>> configurations will have different behavior.
But this is magic PO-machines again. I thought you had decided >>>>>>> that was
a non-starter?
DIFFERENT SEQUENCES OF CONFIGURATIONS WILL HAVE DIFFERENT BEHAVIOR. >>>>>>
That there is conditional branch on one path and no conditional
branch
on the other path makes the behavior vary between paths.
Ĥ applied to ⟨Ĥ⟩ depends on the decision made by embedded_H.
Yes, you are clear that
Ĥ.q0 ⟨Ĥ⟩ ⊦* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* Ĥ.qn
but please complete the following line for me:
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.q?
Not until after you agree that Ĥ ⟨Ĥ⟩ ⊦* Ĥ.qn is correct on the basis
that the correctly simulated input to embedded_H would never reach
its own final state.
Since that isn't True, you are asking for a lot.
It is true on the basis that I provided and every other basis is
incorrect.
Nope, the DEFINITION is correct, because it is the DEFINITION.
On 4/1/22 9:20 AM, olcott wrote:
On 4/1/2022 7:48 AM, Richard Damon wrote:
On 3/31/22 11:06 PM, olcott wrote:
On 3/31/2022 8:57 PM, Richard Damon wrote:
On 3/31/22 9:40 PM, olcott wrote:
On 3/31/2022 8:37 PM, Richard Damon wrote:
On 3/31/22 9:18 PM, olcott wrote:
On 3/31/2022 7:09 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/31/2022 3:15 PM, Ben Bacarisse wrote:Yes, you are clear that
olcott <NoOne@NoWhere.com> writes:
The directly executed Ĥ applied to ⟨Ĥ⟩ is the first >>>>>>>>>>>> invocation ofThis is the old "it only halts because" ruse...
infinite recursion that only terminates normally because of its >>>>>>>>>>>> one-way dependency relationship on embedded_H aborting the >>>>>>>>>>>> second
invocation of this otherwise infinite recursion.
DIFFERENT SEQUENCES OF CONFIGURATIONS WILL HAVE DIFFERENT >>>>>>>>>>>> BEHAVIOR:
This makes the sequence of configurations of the simulation >>>>>>>>>>>> of ⟨Ĥ⟩ ⟨Ĥ⟩
outside of Ĥ different than the the sequence of
configurations of the
simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ inside of Ĥ. Different sequences of
configurations will have different behavior.
But this is magic PO-machines again. I thought you had >>>>>>>>>>> decided that was
a non-starter?
DIFFERENT SEQUENCES OF CONFIGURATIONS WILL HAVE DIFFERENT
BEHAVIOR.
That there is conditional branch on one path and no
conditional branch
on the other path makes the behavior vary between paths.
Ĥ applied to ⟨Ĥ⟩ depends on the decision made by embedded_H. >>>>>>>>>
Ĥ.q0 ⟨Ĥ⟩ ⊦* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* Ĥ.qn
but please complete the following line for me:
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.q?
Not until after you agree that Ĥ ⟨Ĥ⟩ ⊦* Ĥ.qn is correct on the
basis that the correctly simulated input to embedded_H would
never reach its own final state.
Since that isn't True, you are asking for a lot.
It is true on the basis that I provided and every other basis is
incorrect.
Nope, the DEFINITION is correct, because it is the DEFINITION.
The way that you are construing that definition contradicts its
computer science component parts:
CORRECT DEFINITION OF HALT DECIDING CRITERIA
(1) All deciders compute the mapping of their input finite strings
to an accept or reject state.
Right, so H needs to map <H^> <H^> to Qy or Qn.
YES
(2) The direct execution of a Turing machine is computationally
equivalent to the UTM simulation of its Turing machine description.
Right.
YES
(3) Halt deciders compute the mapping of their input finite strings
to an accept or reject state on the basis of the actual behavior
specified by their input.
Right.
YES
(4) The actual behavior specified by the input is measured by the
behavior of a UTM simulation of this input at the same point in the
execution trace as the simulating halt decider. (defined in (2) above)
Nonsense statement since the UTM simulation of a string is ALWAYS the
same regardless of 'point in the execution trace'
Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
UTM ⟨Ĥ⟩ ⟨Ĥ⟩ derives a different result than >
Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
Ĥ ⟨Ĥ⟩
You don't seem to understand what you are saying, and are thus just
making a Strawman.
NOTHING said to repleace H with a UTM.
That would only be a correct operation if H actually WAS the
computational equivalent of a UTM, and we have previously proven that if
H is that, it fails to answer.
On 3/31/22 11:06 PM, olcott wrote:
On 3/31/2022 8:57 PM, Richard Damon wrote:
On 3/31/22 9:40 PM, olcott wrote:
On 3/31/2022 8:37 PM, Richard Damon wrote:
On 3/31/22 9:18 PM, olcott wrote:
On 3/31/2022 7:09 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/31/2022 3:15 PM, Ben Bacarisse wrote:Yes, you are clear that
olcott <NoOne@NoWhere.com> writes:
The directly executed Ĥ applied to ⟨Ĥ⟩ is the first invocation ofThis is the old "it only halts because" ruse...
infinite recursion that only terminates normally because of its >>>>>>>>>> one-way dependency relationship on embedded_H aborting the second >>>>>>>>>> invocation of this otherwise infinite recursion.
DIFFERENT SEQUENCES OF CONFIGURATIONS WILL HAVE DIFFERENT
BEHAVIOR:
This makes the sequence of configurations of the simulation of >>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩
outside of Ĥ different than the the sequence of configurations >>>>>>>>>> of the
simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ inside of Ĥ. Different sequences of >>>>>>>>>> configurations will have different behavior.
But this is magic PO-machines again. I thought you had decided >>>>>>>>> that was
a non-starter?
DIFFERENT SEQUENCES OF CONFIGURATIONS WILL HAVE DIFFERENT BEHAVIOR. >>>>>>>>
That there is conditional branch on one path and no conditional >>>>>>>> branch
on the other path makes the behavior vary between paths.
Ĥ applied to ⟨Ĥ⟩ depends on the decision made by embedded_H. >>>>>>>
Ĥ.q0 ⟨Ĥ⟩ ⊦* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* Ĥ.qn
but please complete the following line for me:
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.q?
Not until after you agree that Ĥ ⟨Ĥ⟩ ⊦* Ĥ.qn is correct on the >>>>>> basis that the correctly simulated input to embedded_H would never >>>>>> reach its own final state.
Since that isn't True, you are asking for a lot.
It is true on the basis that I provided and every other basis is
incorrect.
Nope, the DEFINITION is correct, because it is the DEFINITION.
The way that you are construing that definition contradicts its
computer science component parts:
CORRECT DEFINITION OF HALT DECIDING CRITERIA
(1) All deciders compute the mapping of their input finite strings to
an accept or reject state.
Right, so H needs to map <H^> <H^> to Qy or Qn.
(2) The direct execution of a Turing machine is computationally
equivalent to the UTM simulation of its Turing machine description.
Right.
(3) Halt deciders compute the mapping of their input finite strings to
an accept or reject state on the basis of the actual behavior
specified by their input.
Right.
(4) The actual behavior specified by the input is measured by the
behavior of a UTM simulation of this input at the same point in the
execution trace as the simulating halt decider. (defined in (2) above)
Nonsense statement since the UTM simulation of a string is ALWAYS the
same regardless of 'point in the execution trace'
(5) Linz: computation that halts … the Turing machine will halt
whenever it enters a final state. (Linz:1990:234)
Right, the TURING Machine, that is H^ applied to <H^> is the measure
(6) A correct simulation of a Turing machine description that would
never reach its final state is computationally equivalent to the
direct execution of this same Turing machine never reaching its final
state and thus specifies a non-halting sequence of configurations.
Right, if the direct execution would never halt, then the UTM simulation would never halt.
You seemed to have forgetten to prove something.
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