On Tuesday, 22 March 2022 at 10:05:09 UTC+8, 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.
computation that halts … the Turing machine will halt whenever it enters >> a final state. (Linz:1990:234)
Whether or not it is even possible for embedded_H to recognize this
infinitely nested simulation does not matter (for refuting Linz).
As long as embedded_H transitions to Ĥ.qn it refutes the Linz conclusion
that rejecting its input forms a necessary contradiction.
(bottom half of last page) https://www.liarparadox.org/Linz_Proof.pdf
After we have mutual agreement on the above point we can move on to how
embedded_H would use finite string comparison to detect that its input
specifies infinitely nested simulation.
Halting problem undecidability and infinitely nested simulation (V4)
https://www.researchgate.net/publication/359349179_Halting_problem_undecidability_and_infinitely_nested_simulation_V4
--
Copyright 2021 Pete Olcott
Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer
The H of the Halting Problem is not required to detect any infinite loop. H is
only required to report whether the given algorithm instance P halts or not.
Firstly, proving the conventional liar-paradox-like instance is wrong to prove
the Halting Problem is decidable. At most, such proof proves HP is still unsolved. That HP remains unsolved is the most what such an approach can get.
Secondly, you modified Linz's proof to your POOP. Nothing you say is therefore
valid to the true/false status of Linz's proof.
The discussion is at most about your POOP.
olcott <NoOne@NoWhere.com> writes:
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.
But for your "PO-machines":
"Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn
corresponds to
H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
and
"The copy of H at Ĥ.qx correctly decides that its input never halts.
H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input halts"
so this has nothing to do with Linz. He is talking about Turing
machines.
olcott <NoOne@NoWhere.com> writes:
On 3/21/2022 10:22 PM, Ben Bacarisse wro0te:
olcott <NoOne@NoWhere.com> writes:
A copy of Linz H is embedded at Ĥ.qx as a simulating halt decider (SHD). >>>>But for your "PO-machines":
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.
"Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn
corresponds to
H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
and
"The copy of H at Ĥ.qx correctly decides that its input never halts. >>> H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input halts" >>> so this has nothing to do with Linz. He is talking about Turing
machines.
The Linz conclusion only pertains to the behavior the copy of H
embedded within Ĥ applied to ⟨Ĥ⟩ ⟨Ĥ⟩.
Everything Linz says, everything, is predicated on what a Turing machine
is. Unlike Turing machines, your machines are magic -- identical state transition functions can entail different configuration sequences for
the same input. Nothing you say has any relevance to Linz's Turing
machines until you categorically repudiate this nonsense.
I don't have time to diverge from this point:
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its final
state.
And, since the action of your magic machines are not constrained by the
state transition function and the input, I can assert that a copy of
our Ĥ does this:
On 3/22/22 10:16 AM, olcott wrote:
On 3/22/2022 6:00 AM, Richard Damon wrote:
On 3/21/22 11:59 PM, olcott wrote:
On 3/21/2022 10:22 PM, Ben Bacarisse wro0te:
olcott <NoOne@NoWhere.com> writes:
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.
But for your "PO-machines":
"Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn
corresponds to
H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
and
"The copy of H at Ĥ.qx correctly decides that its input never
halts.
H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input halts"
so this has nothing to do with Linz. He is talking about Turing
machines.
The Linz conclusion only pertains to the behavior the copy of H
embedded within Ĥ applied to ⟨Ĥ⟩ ⟨Ĥ⟩. I don't have time to diverge
from this point:
No, it applies to ALL copies of H, because they ALL behave the same.
Because the outermost simulation sees that its abort criteria is met
before any of the inner simulations have seen this criteria the
outermost simulation aborts all the rest which cuts off their behavior.
But we don't care about simulations, we care about actual machine exectuitons.
On 3/22/22 10:46 AM, olcott wrote:And it doesn't halt if embedded_H DOES abort its simulation, therefore
On 3/22/2022 9:32 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/21/2022 10:22 PM, Ben Bacarisse wro0te:
olcott <NoOne@NoWhere.com> writes:
A copy of Linz H is embedded at Ĥ.qx as a simulating halt decider >>>>>> (SHD).But for your "PO-machines":
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.
"Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn
corresponds to
H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
and
"The copy of H at Ĥ.qx correctly decides that its input never >>>>> halts.
H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input halts"
so this has nothing to do with Linz. He is talking about Turing
machines.
The Linz conclusion only pertains to the behavior the copy of H
embedded within Ĥ applied to ⟨Ĥ⟩ ⟨Ĥ⟩.
Everything Linz says, everything, is predicated on what a Turing machine >>> is. Unlike Turing machines, your machines are magic -- identical state >>> transition functions can entail different configuration sequences for
the same input. Nothing you say has any relevance to Linz's Turing
machines until you categorically repudiate this nonsense.
That your only rebuttal to what I say now is dredging up what I said
many months ago proves that you are being dishonest.
It is the case that the pure simulation of the input to the simulating
halt decider Linz H embedded at Ĥ.qx would never reach the final state
of this simulated input.
You ignore this and change the subject only because you know that it
is correct and have denigration rather than honest critique as your
only goal.
I don't have time to diverge from this point:
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its final
state.
And, since the action of your magic machines are not constrained by the
state transition function and the input, I can assert that a copy of
our Ĥ does this:
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 above shows that the simulated input to embedded_H will never
reach its own final state whether or not its simulation is aborted
thus never meets the Linz definition of halting:
computation that halts … the Turing machine will halt whenever it
enters a final state. (Linz:1990:234)
Halting problem undecidability and infinitely nested simulation (V4)
https://www.researchgate.net/publication/359349179_Halting_problem_undecidability_and_infinitely_nested_simulation_V4
The above only doesn't halt if embedded_H never aborts its simulation,
On 3/22/22 7:13 PM, olcott wrote:
On 3/22/2022 6:00 PM, Richard Damon wrote:
On 3/22/22 10:46 AM, olcott wrote:And it doesn't halt if embedded_H DOES abort its simulation, therefore
On 3/22/2022 9:32 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/21/2022 10:22 PM, Ben Bacarisse wro0te:
olcott <NoOne@NoWhere.com> writes:
A copy of Linz H is embedded at Ĥ.qx as a simulating haltBut for your "PO-machines":
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.
"Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn
corresponds to
H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
and
"The copy of H at Ĥ.qx correctly decides that its input never >>>>>>> halts.
H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input halts"
so this has nothing to do with Linz. He is talking about Turing >>>>>>> machines.
The Linz conclusion only pertains to the behavior the copy of H
embedded within Ĥ applied to ⟨Ĥ⟩ ⟨Ĥ⟩.
Everything Linz says, everything, is predicated on what a Turing
machine
is. Unlike Turing machines, your machines are magic -- identical
state
transition functions can entail different configuration sequences for >>>>> the same input. Nothing you say has any relevance to Linz's Turing >>>>> machines until you categorically repudiate this nonsense.
That your only rebuttal to what I say now is dredging up what I said
many months ago proves that you are being dishonest.
It is the case that the pure simulation of the input to the
simulating halt decider Linz H embedded at Ĥ.qx would never reach
the final state of this simulated input.
You ignore this and change the subject only because you know that it
is correct and have denigration rather than honest critique as your
only goal.
I don't have time to diverge from this point:
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its final
state.
And, since the action of your magic machines are not constrained by
the
state transition function and the input, I can assert that a copy of >>>>> our Ĥ does this:
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 above shows that the simulated input to embedded_H will never
reach its own final state whether or not its simulation is aborted
thus never meets the Linz definition of halting:
computation that halts … the Turing machine will halt whenever it
enters a final state. (Linz:1990:234)
Halting problem undecidability and infinitely nested simulation (V4)
https://www.researchgate.net/publication/359349179_Halting_problem_undecidability_and_infinitely_nested_simulation_V4
The above only doesn't halt if embedded_H never aborts its simulation,
it doesn't halt under any condition.
Except that the actual machine DOES Halt, as does the UTM simulation.
olcott <NoOne@NoWhere.com> writes:
On 3/22/2022 9:32 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/21/2022 10:22 PM, Ben Bacarisse wro0te:
olcott <NoOne@NoWhere.com> writes:
A copy of Linz H is embedded at Ĥ.qx as a simulating halt decider (SHD).But for your "PO-machines":
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.
"Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn
corresponds to
H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
and
"The copy of H at Ĥ.qx correctly decides that its input never halts.
H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input halts"
so this has nothing to do with Linz. He is talking about Turing
machines.
The Linz conclusion only pertains to the behavior the copy of H
embedded within Ĥ applied to ⟨Ĥ⟩ ⟨Ĥ⟩.
Everything Linz says, everything, is predicated on what a Turing machine >>> is. Unlike Turing machines, your machines are magic -- identical state
transition functions can entail different configuration sequences for
the same input. Nothing you say has any relevance to Linz's Turing
machines until you categorically repudiate this nonsense.
That your only rebuttal to what I say now is dredging up what I said
many months ago proves that you are being dishonest.
You said this:
"The copy of H at Ĥ.qx correctly decides that its input never halts.
H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input halts"
four days ago and you haven't retracted it. Until you do, when you
write Ĥ your readers must assume that you are referring to something
about which this quote applies.
What's more, for your remarks to have any bearing on Linz's Ĥ you must
not only repudiate what you said, you must accept the converse,
i.e. that if
Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* Ĥ.qn
then
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qn
So, do you retract what you said and accept this fact about Linz's H and
Ĥ?
olcott <NoOne@NoWhere.com> writes:
On 3/22/2022 9:39 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
Yet you cannot point out a single error in my halting problemThe mistakes have been pointed out so many times that it's reasonable to >>> assume you can't see them or are simply ignoring them. The latest
refutation.
monster error is that if (as you claim) this is true of your Ĥ:
"Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn"
but
"H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
then Ĥ is not even a Truing machine, let alone the specific Ĥ in Linz's >>> proof.
If the halt deciding criteria compares the finite strings of Turing
machine descriptions...
But Ĥ is not a Turing machine and ⟨Ĥ⟩ is not a Turing machine
description. You are not talking about Turing machines. Turing
machines are not magic --
identical state transition functions entail
identical configuration sequences for the identical inputs.
as its halt deciding basis then it will find that H and the copy of H
embedded at Ĥ.qx are not the identical (embedded_H is a longer finite
string) thus providing the basis for H to see that embedded_H will
transition to Ĥ.qn and halt.
H and Ĥ have to be the Turing machines. When you write these symbols,
you are referring to entities with magic properties. For actual Turing machines, If Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn then H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qn.
On 3/23/22 12:00 AM, olcott wrote:
On 3/22/2022 10:37 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/22/2022 9:32 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/21/2022 10:22 PM, Ben Bacarisse wro0te:
olcott <NoOne@NoWhere.com> writes:
A copy of Linz H is embedded at Ĥ.qx as a simulating haltBut for your "PO-machines":
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.
"Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn
corresponds to
H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
and
"The copy of H at Ĥ.qx correctly decides that its input >>>>>>> never halts.
H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input halts"
so this has nothing to do with Linz. He is talking about Turing >>>>>>> machines.
The Linz conclusion only pertains to the behavior the copy of H
embedded within Ĥ applied to ⟨Ĥ⟩ ⟨Ĥ⟩.
Everything Linz says, everything, is predicated on what a Turing
machine
is. Unlike Turing machines, your machines are magic -- identical
state
transition functions can entail different configuration sequences for >>>>> the same input. Nothing you say has any relevance to Linz's Turing >>>>> machines until you categorically repudiate this nonsense.
That your only rebuttal to what I say now is dredging up what I said
many months ago proves that you are being dishonest.
You said this:
"The copy of H at Ĥ.qx correctly decides that its input never halts. >>> H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input halts" >>>
If ⟨H⟩ and ⟨Ĥ⟩ were identical finite strings then they must derive the
same result. They are not identical final strings.
four days ago and you haven't retracted it. Until you do, when you
write Ĥ your readers must assume that you are referring to something
about which this quote applies.
What's more, for your remarks to have any bearing on Linz's Ĥ you must
not only repudiate what you said, you must accept the converse,
i.e. that if
Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* Ĥ.qn
then
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qn
So, do you retract what you said and accept this fact about Linz's H and >>> Ĥ?
You you continue to say that you believe that a decider must report on
its own behavior when you already know damn well that a decider only
computes the mapping from its inputs to its own final state.
A Decider must report on its own behavior (or the behavior of a copy of
it) if that is what the input asks for.
On 3/22/22 11:51 PM, olcott wrote:
On 3/22/2022 10:36 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/22/2022 9:39 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
Yet you cannot point out a single error in my halting problemThe mistakes have been pointed out so many times that it's
refutation.
reasonable to
assume you can't see them or are simply ignoring them. The latest
monster error is that if (as you claim) this is true of your Ĥ:
"Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn"
but
"H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
then Ĥ is not even a Truing machine, let alone the specific Ĥ in
Linz's
proof.
If the halt deciding criteria compares the finite strings of Turing
machine descriptions...
But Ĥ is not a Turing machine and ⟨Ĥ⟩ is not a Turing machine
Technically Linz refers to a whole class of computations. You can
assume that this class is empty on the basis of assuming that the Linz
proof is correct.
description. You are not talking about Turing machines. Turing
machines are not magic --
identical state transition functions entail
identical configuration sequences for the identical inputs.
Yes that is correct.
as its halt deciding basis then it will find that H and the copy of H
embedded at Ĥ.qx are not the identical (embedded_H is a longer finite >>>> string) thus providing the basis for H to see that embedded_H will
transition to Ĥ.qn and halt.
H and Ĥ have to be the Turing machines. When you write these symbols,
If you start with the foundational assumption that Linz is correct you
can simply assume that my rebuttal is incorrect without even looking
at it. This is not how correct reasoning works.
you are referring to entities with magic properties. For actual Turing >>> machines, If Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn then H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qn.
Not when Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn on the basis that Ĥ is invoking an
identical machine description with identical inputs twice in sequence.
Except that this is not a CORRECT descriptor of an infinite behVIOR.
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