On 2022-02-25 10:32, olcott wrote:
On 2/25/2022 11:17 AM, André G. Isaak wrote:
On 2022-02-25 09:11, olcott wrote:
On 2/25/2022 9:45 AM, Richard Damon wrote:
On 2/25/22 10:26 AM, olcott wrote:
THIS IS ALWAYS EXACTLY THE SAME
Simulating halt deciders continue to simulate their input until
they determine that this simulated input would never reach its
final state.
But how do they determine that?
The fact that humans can see that in both cases the simulated input
never reaches its final state in any finite number of simulated
steps conclusively proves that it is possible to correctly detect
the infinite loop and the infinitely nested simulation.
What humans can do provides no evidence at all about what algorithms
can do. Humans are not algorithms.
If humans can do thing X then thing X is proven to be possible to do.
People can win olympic pole vaulting competitions. It doesn't follow
from this that a Turing Machine can.
And you, the human, are recognizing something making use of a piece of information with which the TM is *NOT* provided; you are aware of the
fact that the input happens to be a representation of Ĥ, a machine which includes a copy of embedded_H.
embedded_H, on the other hand, is *not* provided with this information.
So how does your embedded_H recognize that the input string includes a
copy of itself?
(and what humans can do with information x, y, and z tells us even
left about what an algorithm can do with only x and y).
If you want to claim it is possible for an algorithm to recognize
infinitely recursive simulation, you need to actually show how that
algorithm works.
The first step of this elaboration requires acknowledgement that:
If humans can do thing X then thing X is proven to be possible to do.
∴ if embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn then embedded_H is correct.
I can't possibly acknowledge anything about embedded_H if you won't
provide the details of how it works such as the one I ask about above.
André
On 2022-03-01 08:50, olcott wrote:
On 2/25/2022 1:00 PM, André G. Isaak wrote:
On 2022-02-25 10:32, olcott wrote:
On 2/25/2022 11:17 AM, André G. Isaak wrote:
On 2022-02-25 09:11, olcott wrote:
On 2/25/2022 9:45 AM, Richard Damon wrote:
On 2/25/22 10:26 AM, olcott wrote:
THIS IS ALWAYS EXACTLY THE SAME
Simulating halt deciders continue to simulate their input until >>>>>>>> they determine that this simulated input would never reach its >>>>>>>> final state.
But how do they determine that?
The fact that humans can see that in both cases the simulated
input never reaches its final state in any finite number of
simulated steps conclusively proves that it is possible to
correctly detect the infinite loop and the infinitely nested
simulation.
What humans can do provides no evidence at all about what
algorithms can do. Humans are not algorithms.
If humans can do thing X then thing X is proven to be possible to do.
People can win olympic pole vaulting competitions. It doesn't follow
from this that a Turing Machine can.
And you, the human, are recognizing something making use of a piece
of information with which the TM is *NOT* provided; you are aware of
the fact that the input happens to be a representation of Ĥ, a
machine which includes a copy of embedded_H.
embedded_H, on the other hand, is *not* provided with this information.
So how does your embedded_H recognize that the input string includes
a copy of itself?
I can't possibly acknowledge anything about embedded_H if you won't(and what humans can do with information x, y, and z tells us even
left about what an algorithm can do with only x and y).
If you want to claim it is possible for an algorithm to recognize
infinitely recursive simulation, you need to actually show how that
algorithm works.
The first step of this elaboration requires acknowledgement that:
If humans can do thing X then thing X is proven to be possible to do.
∴ if embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn then embedded_H is correct. >>>
provide the details of how it works such as the one I ask about above.
André
It is stipulated that the key aspect of simulating halt deciders is
that they examine the behavior patterns of their simulated input.
"Stipulating" this doesn't magically make it possible.
It is universally true that when-so-ever a simulating halt decider
must abort the simulation of its input to prevent the infinite
simulation of this input that thus input specifies an infinite
sequence of configurations.
On the basis of the above it is self evident that the pure simulation
of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never stop running, thus infinite >> behavior is proved. Because infinite behavior is proved then a
transition to Ĥ.qn is proved to be correct.
The above by itself is much further than anyone has ever gotten with
the halting problem.
Because humans can easily see that ⟨Ĥ⟩ ⟨Ĥ⟩ specifies an infinite >> sequence of configurations to simulating halt decider embedded_H it is
reasonable to conclude that a computation could do likewise.
That's not a proof. What a human can do tells us nothing about what is possible for an algorithm. And remember that the human has knowledge
which the Turing Machine does not.
So how exactly does is the Turing Machine Ĥ determine whether its input string is a representation of itself?
Unless it has some way of doing
that, it has no way of recognizing that it is being called recursively.
André
On 2022-03-01 13:59, olcott wrote:
On 3/1/2022 1:59 PM, André G. Isaak wrote:
On 2022-03-01 11:32, olcott wrote:
On 3/1/2022 11:46 AM, André G. Isaak wrote:
On 2022-03-01 08:50, olcott wrote:
On 2/25/2022 1:00 PM, André G. Isaak wrote:"Stipulating" this doesn't magically make it possible.
On 2022-02-25 10:32, olcott wrote:
On 2/25/2022 11:17 AM, André G. Isaak wrote:
On 2022-02-25 09:11, olcott wrote:
On 2/25/2022 9:45 AM, Richard Damon wrote:
On 2/25/22 10:26 AM, olcott wrote:
THIS IS ALWAYS EXACTLY THE SAME
Simulating halt deciders continue to simulate their input >>>>>>>>>>>> until they determine that this simulated input would never >>>>>>>>>>>> reach its final state.
But how do they determine that?
The fact that humans can see that in both cases the simulated >>>>>>>>>> input never reaches its final state in any finite number of >>>>>>>>>> simulated steps conclusively proves that it is possible to >>>>>>>>>> correctly detect the infinite loop and the infinitely nested >>>>>>>>>> simulation.
What humans can do provides no evidence at all about what
algorithms can do. Humans are not algorithms.
If humans can do thing X then thing X is proven to be possible >>>>>>>> to do.
People can win olympic pole vaulting competitions. It doesn't
follow from this that a Turing Machine can.
And you, the human, are recognizing something making use of a
piece of information with which the TM is *NOT* provided; you are >>>>>>> aware of the fact that the input happens to be a representation
of Ĥ, a machine which includes a copy of embedded_H.
embedded_H, on the other hand, is *not* provided with this
information.
So how does your embedded_H recognize that the input string
includes a copy of itself?
(and what humans can do with information x, y, and z tells us >>>>>>>>> even left about what an algorithm can do with only x and y). >>>>>>>>>
If you want to claim it is possible for an algorithm to
recognize infinitely recursive simulation, you need to actually >>>>>>>>> show how that algorithm works.
The first step of this elaboration requires acknowledgement that: >>>>>>>> If humans can do thing X then thing X is proven to be possible >>>>>>>> to do.
∴ if embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn then embedded_H is correct.
I can't possibly acknowledge anything about embedded_H if you
won't provide the details of how it works such as the one I ask
about above.
André
It is stipulated that the key aspect of simulating halt deciders
is that they examine the behavior patterns of their simulated input. >>>>>
In this case it does. You did not pay perfect attention to the exact
words that I used.
It is universally true that when-so-ever a simulating halt decider >>>>>> must abort the simulation of its input to prevent the infinite
simulation of this input that thus input specifies an infinite
sequence of configurations.
On the basis of the above it is self evident that the pure
simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never stop running, thus
infinite behavior is proved. Because infinite behavior is proved
then a transition to Ĥ.qn is proved to be correct.
The above by itself is much further than anyone has ever gotten
with the halting problem.
Because humans can easily see that ⟨Ĥ⟩ ⟨Ĥ⟩ specifies an infinite
sequence of configurations to simulating halt decider embedded_H
it is reasonable to conclude that a computation could do likewise.
That's not a proof. What a human can do tells us nothing about what
is possible for an algorithm. And remember that the human has
knowledge which the Turing Machine does not.
I have a computer program that can do this too, yet it is simply
ruled as not counting on the basis that it does not conform to the
limitations of computable functions.
So how exactly does is the Turing Machine Ĥ determine whether its
input string is a representation of itself?
A RASP machine simply sees that its input is at its same machine
address.
Unless it has some way of doing that, it has no way of recognizing
that it is being called recursively.
André
https://en.wikipedia.org/wiki/Random-access_stored-program_machine
If a RASP machine is strictly more powerful than a Turing machine
then simply abandon the TM model as inherently inferior.
You are horridly confused here.
RASP machines or x86 machines are *not* more powerful than Turing
Machines.
Anything your computer can do is a computation. But it doesn't follow
from that that any arbitrary C function or piece of x86 code is a
computation, because a computation involves a *self-contained*
algorithm.
Thus, unless a C function or piece of x86 code can run *on its own*,
it is not a computation. If it depends on other functions (including
Operating System functions) then the *combination* of that function
and all of the other functions it depends on can be construed as a
computation, but the function on its own cannot be (and your computer
wouldn't be able to run it in absence of those dependencies).
Ah so a C function that has its own machine address directly encoded
into one of its x86 instructions is still a computation. This would be
the machine address of a label ---> __asm lea eax, HERE
Sure, but it's a computation that has that address as part of its
*input*. That means you're not dealing with the same computation as the halting problem.
Again, when dealing with computations, 'input' doesn't just mean things
which are passed to a function as formal parameters; it means *any* data which the function has access to and makes use of.
The problem with all of your solutions to the halting problem isn't
that they aren't computations, but that they are the *wrong*
computation.
So you assume by simply ignoring the self-evident truth of this
statement:
It is universally true that when-so-ever a simulating halt decider
must abort the simulation of its input to prevent the infinite
simulation of this input that thus input specifies an infinite
sequence of configurations.
The input to a computation consists of *all* of the information which
is made available to that computation, not just those bits of
information which are passed as formal parameters to a C function.
Given some C function defined as
My C/x86 H merely needs to know its own machine address, then it can
tell that as soon as P calls H with identical parameters that this is
infinitely nested simulation.
Boolean foo(someType *X) {...}
This function does not necessarily compute a function from X to a
boolean value because a C function can in principle access all sorts
of other data besides its formal parameter X. *Any* data which the
function makes use of must be considered part of its input even if it
isn't declared as a formal parameter.
If a function refers to both some data in memory and the address of
that data, those are two *separate* pieces of information, and when
such a function is construed as a computation, both of those pieces
of information must be treated as part of its input.
The halting problem specifies exactly which information the input to
a halt decider includes, and if a C function makes use of any data
other than that information then the function may be a computation,
but it is *not* the computation that the halting problem is concerned
with.
That seems like an arbitrary constraint.
There's nothing arbitrary about it. When we ask whether W is computable
from X and Y, and a program makes use of X, Y, and Z to calculate W then
it isn't addressing the question being asked; it's addressing whether W
is computable from X, Y, and Z, which is an entirely different question.
As long as there are no inputs that a function cannot decide the halt
status of, then the halting problem is only impossible when arbitrary
constraints are place upon it.
We could for example say that it is impossible to clap your hands
together and then restrict the solution set to people that have no hands.
The halting problem is concerned with whether it is possible to
determine whether a given computation is finite given *only* a
description of a Turing Machine and its input string. There's nothing
in that input which includes information about where in memory this
information might be stored in an x86 implementation.
Still seems like a purely arbitrary constraint.
This is why both C and x86 assembly language are absolutely terrible
models for discussing computational theory. Neither of these things
is designed as a model of computation, and consequently the code for
C or x86 routines don't necessarily make it clear exactly what it is
that is being computed. When declaring a C function you must declare
its formal parameters, but these parameters do *not* necessarily
correspond to the input of the algorithm which that function implements. >>>
This is why you really ought to simply abandon C/x86 and focus on TMs
if you actually want to learn computational theory and what
computations are. Because of the way they work, it is always entirely
clear exactly which computation is involved and what the input is
when working with TMs. Not so with C and/or x86 code.
André
The problem with this approach is that key details must be totally
ignored because TM's are simply not expressive enough to express
algorithm's concisely.
No. The problem is that C/x86 allow you to make use of inputs which are
not *explicitly* declared. But they are still inputs.
André
On 2022-03-01 11:32, olcott wrote:
On 3/1/2022 11:46 AM, André G. Isaak wrote:
On 2022-03-01 08:50, olcott wrote:
On 2/25/2022 1:00 PM, André G. Isaak wrote:
On 2022-02-25 10:32, olcott wrote:
On 2/25/2022 11:17 AM, André G. Isaak wrote:People can win olympic pole vaulting competitions. It doesn't
On 2022-02-25 09:11, olcott wrote:
On 2/25/2022 9:45 AM, Richard Damon wrote:
On 2/25/22 10:26 AM, olcott wrote:
THIS IS ALWAYS EXACTLY THE SAME
Simulating halt deciders continue to simulate their input
until they determine that this simulated input would never >>>>>>>>>> reach its final state.
But how do they determine that?
The fact that humans can see that in both cases the simulated
input never reaches its final state in any finite number of
simulated steps conclusively proves that it is possible to
correctly detect the infinite loop and the infinitely nested
simulation.
What humans can do provides no evidence at all about what
algorithms can do. Humans are not algorithms.
If humans can do thing X then thing X is proven to be possible to do. >>>>>
follow from this that a Turing Machine can.
And you, the human, are recognizing something making use of a piece
of information with which the TM is *NOT* provided; you are aware
of the fact that the input happens to be a representation of Ĥ, a
machine which includes a copy of embedded_H.
embedded_H, on the other hand, is *not* provided with this
information.
So how does your embedded_H recognize that the input string
includes a copy of itself?
(and what humans can do with information x, y, and z tells us
even left about what an algorithm can do with only x and y).
If you want to claim it is possible for an algorithm to recognize >>>>>>> infinitely recursive simulation, you need to actually show how
that algorithm works.
The first step of this elaboration requires acknowledgement that:
If humans can do thing X then thing X is proven to be possible to do. >>>>>> ∴ if embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn then embedded_H is correct.
I can't possibly acknowledge anything about embedded_H if you won't
provide the details of how it works such as the one I ask about above. >>>>>
André
It is stipulated that the key aspect of simulating halt deciders is
that they examine the behavior patterns of their simulated input.
"Stipulating" this doesn't magically make it possible.
In this case it does. You did not pay perfect attention to the exact
words that I used.
It is universally true that when-so-ever a simulating halt decider
must abort the simulation of its input to prevent the infinite
simulation of this input that thus input specifies an infinite
sequence of configurations.
On the basis of the above it is self evident that the pure
simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never stop running, thus
infinite behavior is proved. Because infinite behavior is proved
then a transition to Ĥ.qn is proved to be correct.
The above by itself is much further than anyone has ever gotten with
the halting problem.
Because humans can easily see that ⟨Ĥ⟩ ⟨Ĥ⟩ specifies an infinite >>>> sequence of configurations to simulating halt decider embedded_H it
is reasonable to conclude that a computation could do likewise.
That's not a proof. What a human can do tells us nothing about what
is possible for an algorithm. And remember that the human has
knowledge which the Turing Machine does not.
I have a computer program that can do this too, yet it is simply ruled
as not counting on the basis that it does not conform to the
limitations of computable functions.
So how exactly does is the Turing Machine Ĥ determine whether its
input string is a representation of itself?
A RASP machine simply sees that its input is at its same machine address.
Unless it has some way of doing that, it has no way of recognizing
that it is being called recursively.
André
https://en.wikipedia.org/wiki/Random-access_stored-program_machine
If a RASP machine is strictly more powerful than a Turing machine then
simply abandon the TM model as inherently inferior.
You are horridly confused here.
RASP machines or x86 machines are *not* more powerful than Turing Machines.
Anything your computer can do is a computation. But it doesn't follow
from that that any arbitrary C function or piece of x86 code is a computation, because a computation involves a *self-contained* algorithm.
Thus, unless a C function or piece of x86 code can run *on its own*, it
is not a computation. If it depends on other functions (including
Operating System functions) then the *combination* of that function and
all of the other functions it depends on can be construed as a
computation, but the function on its own cannot be (and your computer wouldn't be able to run it in absence of those dependencies).
On 3/1/22 11:13 PM, olcott wrote:
On 3/1/2022 1:59 PM, André G. Isaak wrote:
On 2022-03-01 11:32, olcott wrote:That is great I have a better understanding of this now.
On 3/1/2022 11:46 AM, André G. Isaak wrote:
On 2022-03-01 08:50, olcott wrote:
On 2/25/2022 1:00 PM, André G. Isaak wrote:"Stipulating" this doesn't magically make it possible.
On 2022-02-25 10:32, olcott wrote:
On 2/25/2022 11:17 AM, André G. Isaak wrote:
On 2022-02-25 09:11, olcott wrote:
On 2/25/2022 9:45 AM, Richard Damon wrote:
On 2/25/22 10:26 AM, olcott wrote:
THIS IS ALWAYS EXACTLY THE SAME
Simulating halt deciders continue to simulate their input >>>>>>>>>>>> until they determine that this simulated input would never >>>>>>>>>>>> reach its final state.
But how do they determine that?
The fact that humans can see that in both cases the simulated >>>>>>>>>> input never reaches its final state in any finite number of >>>>>>>>>> simulated steps conclusively proves that it is possible to >>>>>>>>>> correctly detect the infinite loop and the infinitely nested >>>>>>>>>> simulation.
What humans can do provides no evidence at all about what
algorithms can do. Humans are not algorithms.
If humans can do thing X then thing X is proven to be possible >>>>>>>> to do.
People can win olympic pole vaulting competitions. It doesn't
follow from this that a Turing Machine can.
And you, the human, are recognizing something making use of a
piece of information with which the TM is *NOT* provided; you are >>>>>>> aware of the fact that the input happens to be a representation
of Ĥ, a machine which includes a copy of embedded_H.
embedded_H, on the other hand, is *not* provided with this
information.
So how does your embedded_H recognize that the input string
includes a copy of itself?
(and what humans can do with information x, y, and z tells us >>>>>>>>> even left about what an algorithm can do with only x and y). >>>>>>>>>
If you want to claim it is possible for an algorithm to
recognize infinitely recursive simulation, you need to actually >>>>>>>>> show how that algorithm works.
The first step of this elaboration requires acknowledgement that: >>>>>>>> If humans can do thing X then thing X is proven to be possible >>>>>>>> to do.
∴ if embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn then embedded_H is correct.
I can't possibly acknowledge anything about embedded_H if you
won't provide the details of how it works such as the one I ask
about above.
André
It is stipulated that the key aspect of simulating halt deciders
is that they examine the behavior patterns of their simulated input. >>>>>
In this case it does. You did not pay perfect attention to the exact
words that I used.
It is universally true that when-so-ever a simulating halt decider >>>>>> must abort the simulation of its input to prevent the infinite
simulation of this input that thus input specifies an infinite
sequence of configurations.
On the basis of the above it is self evident that the pure
simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never stop running, thus
infinite behavior is proved. Because infinite behavior is proved
then a transition to Ĥ.qn is proved to be correct.
The above by itself is much further than anyone has ever gotten
with the halting problem.
Because humans can easily see that ⟨Ĥ⟩ ⟨Ĥ⟩ specifies an infinite
sequence of configurations to simulating halt decider embedded_H
it is reasonable to conclude that a computation could do likewise.
That's not a proof. What a human can do tells us nothing about what
is possible for an algorithm. And remember that the human has
knowledge which the Turing Machine does not.
I have a computer program that can do this too, yet it is simply
ruled as not counting on the basis that it does not conform to the
limitations of computable functions.
So how exactly does is the Turing Machine Ĥ determine whether its
input string is a representation of itself?
A RASP machine simply sees that its input is at its same machine
address.
Unless it has some way of doing that, it has no way of recognizing
that it is being called recursively.
André
https://en.wikipedia.org/wiki/Random-access_stored-program_machine
If a RASP machine is strictly more powerful than a Turing machine
then simply abandon the TM model as inherently inferior.
You are horridly confused here.
RASP machines or x86 machines are *not* more powerful than Turing
Machines.
Anything your computer can do is a computation. But it doesn't follow
from that that any arbitrary C function or piece of x86 code is a
computation, because a computation involves a *self-contained*
algorithm.
Thus, unless a C function or piece of x86 code can run *on its own*,
it is not a computation. If it depends on other functions (including
Operating System functions) then the *combination* of that function
and all of the other functions it depends on can be construed as a
computation, but the function on its own cannot be (and your computer
wouldn't be able to run it in absence of those dependencies).
Since H needs nothing outside of itself to determine its own machine
address H is still a computable function, even knowing it own machine
address.
If the behavior of a function changes based on where it is loaded into memory, then its loading address is an input to the function.
A Halt Decider doesn't get that as an input, so if it gets its address,
and uses it to change its behavior, it isn't being the needed computation.
Yes, you can define some computation H(wM, w, addr) that does what you
are trying to define, it isn't a correct Halting Decider, BY DEFINITION.
In computer programming, a pure function is a function that has the
following properties:[1][2]
(1) the function return values are identical for identical arguments
(no variation with local static variables, non-local variables,
mutable reference arguments or input streams), and
(2) the function application has no side effects (no mutation of local
static variables, non-local variables, mutable reference arguments or
input/output streams).
https://en.wikipedia.org/wiki/Pure_function
And being a 'Pure Function' isn't sufficient for a piece of code to be a Computation of its formal parameters.
Even though a C function would know its own machine address, it still
meets the requirements of (1) and (2) because it will always
consistently derive the same accept or reject state on the basis of
the same input and has no side effects.
If H is a computable function and a pure function that seems to be
enough.
Nope, it needs to meet the requirements of a COMPUTATION, which means
that all copies of it must give the same answer for the same input.
You just don't understand the meaning of the word REQUIREMENTS or
DEFINITION do you?
Maybe you should just give up since you don't have time to learn all
this stuff you just forgot to learn and are making very bad guesses abot.
On 2022-03-01 08:50, olcott wrote:
On 2/25/2022 1:00 PM, André G. Isaak wrote:
On 2022-02-25 10:32, olcott wrote:
On 2/25/2022 11:17 AM, André G. Isaak wrote:
On 2022-02-25 09:11, olcott wrote:
On 2/25/2022 9:45 AM, Richard Damon wrote:
On 2/25/22 10:26 AM, olcott wrote:
THIS IS ALWAYS EXACTLY THE SAME
Simulating halt deciders continue to simulate their input until >>>>>>>> they determine that this simulated input would never reach its >>>>>>>> final state.
But how do they determine that?
The fact that humans can see that in both cases the simulated
input never reaches its final state in any finite number of
simulated steps conclusively proves that it is possible to
correctly detect the infinite loop and the infinitely nested
simulation.
What humans can do provides no evidence at all about what
algorithms can do. Humans are not algorithms.
If humans can do thing X then thing X is proven to be possible to do.
People can win olympic pole vaulting competitions. It doesn't follow
from this that a Turing Machine can.
And you, the human, are recognizing something making use of a piece
of information with which the TM is *NOT* provided; you are aware of
the fact that the input happens to be a representation of Ĥ, a
machine which includes a copy of embedded_H.
embedded_H, on the other hand, is *not* provided with this information.
So how does your embedded_H recognize that the input string includes
a copy of itself?
I can't possibly acknowledge anything about embedded_H if you won't(and what humans can do with information x, y, and z tells us even
left about what an algorithm can do with only x and y).
If you want to claim it is possible for an algorithm to recognize
infinitely recursive simulation, you need to actually show how that
algorithm works.
The first step of this elaboration requires acknowledgement that:
If humans can do thing X then thing X is proven to be possible to do.
∴ if embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn then embedded_H is correct. >>>
provide the details of how it works such as the one I ask about above.
André
It is stipulated that the key aspect of simulating halt deciders is
that they examine the behavior patterns of their simulated input.
"Stipulating" this doesn't magically make it possible.
It is universally true that when-so-ever a simulating halt decider
must abort the simulation of its input to prevent the infinite
simulation of this input that thus input specifies an infinite
sequence of configurations.
On the basis of the above it is self evident that the pure simulation
of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never stop running, thus infinite >> behavior is proved. Because infinite behavior is proved then a
transition to Ĥ.qn is proved to be correct.
The above by itself is much further than anyone has ever gotten with
the halting problem.
Because humans can easily see that ⟨Ĥ⟩ ⟨Ĥ⟩ specifies an infinite >> sequence of configurations to simulating halt decider embedded_H it is
reasonable to conclude that a computation could do likewise.
That's not a proof. What a human can do tells us nothing about what is possible for an algorithm. And remember that the human has knowledge
which the Turing Machine does not.
So how exactly does is the Turing Machine Ĥ determine whether its input string is a representation of itself? Unless it has some way of doing
that, it has no way of recognizing that it is being called recursively.
André
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