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 of >>>> infinite recursion that only terminates normally because of itsThis is the old "it only halts because" ruse...
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?
olcott <NoOne@NoWhere.com> writes:
On 3/31/2022 11:09 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
IT IS ONLY THIS SINGLE POINT THAT CAUSES MY PROOF TO BE REJECTED:Here's that question you would not answer without equivocating, even
Linz and everyone here believes that deciders must base their decision >>>> on non-finite string non-inputs Ĥ applied to ⟨Ĥ⟩ over-ruling the >>>> actual behavior specified by the actual finite string actual input.
after my asking it more than 12 times in a row. André also asked many, >>> many times and got no answer.
What string must be passed to H so that H can tell us whether or not Ĥ
applied to ⟨Ĥ⟩ halts? Do you reject even the idea that a halt decider >>> could tell us whether a particular TM does or does not halt when given
some particular input? Isn't that what the theorem is about? (The
answer is, of course, yes.)
DIFFERENT SEQUENCES OF CONFIGURATIONS WILL HAVE DIFFERENT BEHAVIOR.
The behavior of ⟨Ĥ⟩ ⟨Ĥ⟩ simulated outside of Ĥ must be computationally
equivalent to the direct execution of Ĥ applied to ⟨Ĥ⟩ yet not the
same as ⟨Ĥ⟩ ⟨Ĥ⟩ simulated inside of Ĥ.
Failure to answer number one (or 13 if you count my previous attempts). Here's the question in case you missed it:
What string must be passed to H so that H can tell us whether or not Ĥ applied to ⟨Ĥ⟩ halts?
olcott <NoOne@NoWhere.com> writes:
On 3/31/2022 3:27 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/31/2022 11:09 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
IT IS ONLY THIS SINGLE POINT THAT CAUSES MY PROOF TO BE REJECTED:What string must be passed to H so that H can tell us whether or not Ĥ >>>>> applied to ⟨Ĥ⟩ halts? Do you reject even the idea that a halt decider
Linz and everyone here believes that deciders must base their decision >>>>>> on non-finite string non-inputs Ĥ applied to ⟨Ĥ⟩ over-ruling the >>>>>> actual behavior specified by the actual finite string actual input. >>>>> Here's that question you would not answer without equivocating, even >>>>> after my asking it more than 12 times in a row. André also asked many, >>>>> many times and got no answer.
could tell us whether a particular TM does or does not halt when given >>>>> some particular input? Isn't that what the theorem is about? (The
answer is, of course, yes.)
DIFFERENT SEQUENCES OF CONFIGURATIONS WILL HAVE DIFFERENT BEHAVIOR.
The behavior of ⟨Ĥ⟩ ⟨Ĥ⟩ simulated outside of Ĥ must be computationally
equivalent to the direct execution of Ĥ applied to ⟨Ĥ⟩ yet not the >>>> same as ⟨Ĥ⟩ ⟨Ĥ⟩ simulated inside of Ĥ.
Failure to answer number one (or 13 if you count my previous attempts).
Here's the question in case you missed it:
What string must be passed to H so that H can tell us whether or not Ĥ
applied to ⟨Ĥ⟩ halts?
I worked out many of the details of this, and can see why you believe
it is an important point, I will not begin to discuss this 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 in any finite number of simulated steps.
Failure to answer number two (well, 14). It's a simple question and
it's central what the halting problem is, but it's also central to why
you are wrong, which is why you know you must avoid answering it.
On 4/1/22 8:24 AM, olcott wrote:
On 4/1/2022 5:10 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/31/2022 3:27 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/31/2022 11:09 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
IT IS ONLY THIS SINGLE POINT THAT CAUSES MY PROOF TO BE REJECTED: >>>>>>>>many times and got no answer.
Linz and everyone here believes that deciders must base their
decision
on non-finite string non-inputs Ĥ applied to ⟨Ĥ⟩ over-ruling the >>>>>>>> actual behavior specified by the actual finite string actual input. >>>>>>> Here's that question you would not answer without equivocating, even >>>>>>> after my asking it more than 12 times in a row. André also asked >>>>>>> many,
What string must be passed to H so that H can tell us whether or >>>>>>> not Ĥ
applied to ⟨Ĥ⟩ halts? Do you reject even the idea that a halt >>>>>>> decider
could tell us whether a particular TM does or does not halt when >>>>>>> given
some particular input? Isn't that what the theorem is about? (The >>>>>>> answer is, of course, yes.)
DIFFERENT SEQUENCES OF CONFIGURATIONS WILL HAVE DIFFERENT BEHAVIOR. >>>>>>
The behavior of ⟨Ĥ⟩ ⟨Ĥ⟩ simulated outside of Ĥ must be
computationally
equivalent to the direct execution of Ĥ applied to ⟨Ĥ⟩ yet not the >>>>>> same as ⟨Ĥ⟩ ⟨Ĥ⟩ simulated inside of Ĥ.
Failure to answer number one (or 13 if you count my previous
attempts).
Here's the question in case you missed it:
What string must be passed to H so that H can tell us whether or not Ĥ >>>>> applied to ⟨Ĥ⟩ halts?
I worked out many of the details of this, and can see why you believe
it is an important point, I will not begin to discuss this 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 in any finite number of simulated steps.
Failure to answer number two (well, 14). It's a simple question and
it's central what the halting problem is, but it's also central to why
you are wrong, which is why you know you must avoid answering it.
CORRECT DEFINITION OF HALT DECIDING CRITERIA
(1) All deciders compute the mapping of their input finite strings to
an accept or reject state.
(2) The direct execution of a Turing machine is computationally
equivalent to the UTM simulation of its Turing machine description.
(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.
(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)
Meaningless statement, as the UTM simulation of an input will ALWAYS be
the same, regardless of 'point in the execution trace'
(7) Mark for below
(5) Linz: computation that halts … the Turing machine will halt
whenever it enters a final state. (Linz:1990:234)
(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.
The above CORRECT DEFINITION OF HALT DECIDING CRITERIA
Along with these simplified notation conventions
H ⟨p⟩ ⟨i⟩ ⊢* H.qy iff UTM simulated ⟨p⟩ ⟨i⟩ reaches its final state
H ⟨p⟩ ⟨i⟩ ⊢* H.qn iff UTM simulated ⟨p⟩ ⟨i⟩ would never reach its
final state
Simplified Ĥ directly calls H --- infinite loop has been removed.
Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
Proves that Ĥ ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn is correct.
No, it doesn't. By (7) H applied to <H^> <H^> -> H.Qn is correct only if
UTM simulation of <H^< <H^> would never reach its final state.
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