XPost: comp.theory, sci.logic

Op 17.jul.2025 om 16:44 schreef olcott:

On 7/6/2025 11:02 AM, Alan Mackenzie wrote:

[ Followup-To: set ]

In comp.theory olcott <[email protected]> wrote:

On 7/6/2025 5:16 AM, Alan Mackenzie wrote:

olcott <[email protected]> wrote:

On 7/5/2025 2:07 PM, Alan Mackenzie wrote:

You lie.  You don't have a proof.  Many people in this group have >>>>>> pointed
out lots of errors in various versions of your purported proof,
which you
just ignore.  The section in Professor Linz's book you used to be >>>>>> so fond
of citing will contain plenty of details, if only you would take the >>>>>> trouble to understand it (assuming you're capable of such
understanding).

I have addressed ....

Meaningless pompous word.

.... all of those details that you make sure to ignore so that you can >>>>> baselessly claim that I am wrong.

I vaguely remember rolling my eyes at your hopeless lack of
understanding.  It was like watching a 7 year old trying to do
calculus.
The basic understanding was simply not there.  Years later, it's still >>>> not there.

And yes, you are wrong.  The proofs of the halting theorem which
involve
constructing programs which purported halting deciders cannot decide
correctly are correct.

Yet you cannot point to even one mistake because there are none.

That's what I'm saying.  Those proofs of the halting theorem are free
from mistakes.

More to the point, it is YOU who cannot point to any mistakes in them.
They are valid proofs.  Your work, if it contradicts those proofs (which
isn't at all clear) can thus be dismissed without further consideration.

There cannot possibly be *AN ACTUAL INPUT* that does the
opposite of whatever its decider decides. All of the examples
of this have never been *ACTUAL INPUTS*

That's so sloppily worded, it could mean almost anything.

The standard halting problem proof cannot even be constructed.

It has been constructed, and is valid.  But one would normally talk about >> formulating a proof, rather than constructing one.

[ .... ]

No Turing machine can possibly take another directly executing
Turing machine as in input, thus removing these from the
domain of every halt decider.

And that, too.

*Thus the requirement that HHH report on the behavior*
*of the directly executed DD has always been bogus*

And that makes your hat trick.

Turing machine partial halt deciders compute the mapping
from their actual inputs to the actual behavior that these
inputs specify.

And a fourth.  There's some semblance of truth in there, but it's very >>>> confused.

It is not at all confused. I know exactly what it means.

It's very confused to everybody but you, then.

Sloppy wording is your technique to get people to go down to your level >>>> of discussion.  That involves many posts trying just to tie you down to >>>> specific word meanings, and is very tiresome and unrewarding.  I
decline
to get involved any further.

*Yet as I claimed you found no actual mistake*

I've found plenty of actual mistakes.  I was a software developer by
profession.

Let me tell you the punchline so that you can
see why I said those things.

Despite what I said last post, I will actually go to the trouble of
analysing your sloppy expression.

Because directly executed Turing machines cannot
possibly be inputs to Turing machine deciders this
makes them outside of the domain of these deciders.

It's entirely unclear what a "directly executed Turing machine" is.  Most >> of the time turing machines are theoretical constructs used for proving
theorems.  They can be executed, but rarely are.

It's unclear what you mean by a turing machine being an input to a turing
machine.  Read up about universal turing machines to get a bit of
background.

When a partial halt decider is required to report
on the direct execution of a machine this requirement
is bogus.

See above.  That paragraph is meaningless.

This means that the behavior of DD() is none of the damn
business of HHH, thus does not contradict HHH(DD)==0.
*If you disagree this only proves that you do not understand*

It's fully obscure what DD() and HHH mean, and thus impossible to
affirm or contradict the meaningless "HHH(DD)==0".

HHH(DD) does correctly detect that DD simulated by HHH
according to the semantics pf the C programming language
cannot possibly reach its own "return"statement final
halt state.

See above.  By the way, people concerned with computation theory use
turing machines, which are well-defined, simple, and powerful.  They lack >> the complexity, ambiguity, and unsuitability for theoretical work of real
world programming languages like C.

*If you disagree this only proves that you do not understand*

Any mindless idiot can disagree. Showing an error and proving
that it is an actual mistake requires much more than this.

Indeed.  All you have done is disagree with one of the proofs of the
halting theorem.  You have yet to show an error in it.  That will be
difficult, because there aren't any.

q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ∞,
   if M applied to WM halts, and
q0 WM ⊢* Ĥq0 Wm WM ⊢* Ĥ y1 qn y2,
   if M applied to WM does not halt.

*From the bottom of page 319 has been adapted to this* https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
  if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
  if Ĥ applied to ⟨Ĥ⟩ does not halt.

<*Halting Problem Proof ERROR*>

Requires Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to report on the
direct execution of Ĥ applied to ⟨Ĥ⟩ and thus not
⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by Ĥ.embedded_H.

No Turing Machine decider can ever report on the
behavior of anything that is not an input encoded
as a finite string.

Ĥ is not a finite string input to Ĥ.embedded_H
⟨Ĥ⟩ ⟨Ĥ⟩ are finite string inputs to Ĥ.embedded_H

</*Halting Problem Proof ERROR*>

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
    ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches
    its simulated final halt state of ⟨Ĥ.qn⟩, and

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
    ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly
    reach its simulated final halt state of ⟨Ĥ.qn⟩.

When Ĥ is applied to ⟨Ĥ⟩ and embedded_H is a
simulating partial halt decider
(a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation until
*embedded_H sees the repeating pattern and transitions to Ĥ.qn*

But when this is only finite repeating pattern, it is incorrect to
report non-termination.
Ignoring the conditional branch instructions in the embedded_H, will
cause that it is unable to decide between a finite and an infinite
repeating pattern, between halting and non-halting behaviour.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic

Op 18.jul.2025 om 18:09 schreef olcott:

On 7/18/2025 6:17 AM, Fred. Zwarts wrote:

Op 17.jul.2025 om 16:44 schreef olcott:

On 7/6/2025 11:02 AM, Alan Mackenzie wrote:

[ Followup-To: set ]

In comp.theory olcott <[email protected]> wrote:

On 7/6/2025 5:16 AM, Alan Mackenzie wrote:

olcott <[email protected]> wrote:

On 7/5/2025 2:07 PM, Alan Mackenzie wrote:

You lie.  You don't have a proof.  Many people in this group >>>>>>>> have pointed
out lots of errors in various versions of your purported proof, >>>>>>>> which you
just ignore.  The section in Professor Linz's book you used to >>>>>>>> be so fond
of citing will contain plenty of details, if only you would take >>>>>>>> the
trouble to understand it (assuming you're capable of such
understanding).

I have addressed ....

Meaningless pompous word.

.... all of those details that you make sure to ignore so that
you can
baselessly claim that I am wrong.

I vaguely remember rolling my eyes at your hopeless lack of
understanding.  It was like watching a 7 year old trying to do
calculus.
The basic understanding was simply not there.  Years later, it's
still
not there.

And yes, you are wrong.  The proofs of the halting theorem which
involve
constructing programs which purported halting deciders cannot decide >>>>>> correctly are correct.

Yet you cannot point to even one mistake because there are none.

That's what I'm saying.  Those proofs of the halting theorem are free >>>> from mistakes.

More to the point, it is YOU who cannot point to any mistakes in them. >>>> They are valid proofs.  Your work, if it contradicts those proofs
(which
isn't at all clear) can thus be dismissed without further
consideration.

There cannot possibly be *AN ACTUAL INPUT* that does the
opposite of whatever its decider decides. All of the examples
of this have never been *ACTUAL INPUTS*

That's so sloppily worded, it could mean almost anything.

The standard halting problem proof cannot even be constructed.

It has been constructed, and is valid.  But one would normally talk
about
formulating a proof, rather than constructing one.

[ .... ]

No Turing machine can possibly take another directly executing
Turing machine as in input, thus removing these from the
domain of every halt decider.

And that, too.

*Thus the requirement that HHH report on the behavior*
*of the directly executed DD has always been bogus*

And that makes your hat trick.

Turing machine partial halt deciders compute the mapping
from their actual inputs to the actual behavior that these
inputs specify.

And a fourth.  There's some semblance of truth in there, but it's >>>>>> very
confused.

It is not at all confused. I know exactly what it means.

It's very confused to everybody but you, then.

Sloppy wording is your technique to get people to go down to your
level
of discussion.  That involves many posts trying just to tie you
down to
specific word meanings, and is very tiresome and unrewarding.  I
decline
to get involved any further.

*Yet as I claimed you found no actual mistake*

I've found plenty of actual mistakes.  I was a software developer by
profession.

Let me tell you the punchline so that you can
see why I said those things.

Despite what I said last post, I will actually go to the trouble of
analysing your sloppy expression.

Because directly executed Turing machines cannot
possibly be inputs to Turing machine deciders this
makes them outside of the domain of these deciders.

It's entirely unclear what a "directly executed Turing machine" is.
Most
of the time turing machines are theoretical constructs used for proving >>>> theorems.  They can be executed, but rarely are.

It's unclear what you mean by a turing machine being an input to a
turing
machine.  Read up about universal turing machines to get a bit of
background.

When a partial halt decider is required to report
on the direct execution of a machine this requirement
is bogus.

See above.  That paragraph is meaningless.

This means that the behavior of DD() is none of the damn
business of HHH, thus does not contradict HHH(DD)==0.
*If you disagree this only proves that you do not understand*

It's fully obscure what DD() and HHH mean, and thus impossible to
affirm or contradict the meaningless "HHH(DD)==0".

HHH(DD) does correctly detect that DD simulated by HHH
according to the semantics pf the C programming language
cannot possibly reach its own "return"statement final
halt state.

See above.  By the way, people concerned with computation theory use
turing machines, which are well-defined, simple, and powerful.  They
lack
the complexity, ambiguity, and unsuitability for theoretical work of
real
world programming languages like C.

*If you disagree this only proves that you do not understand*

Any mindless idiot can disagree. Showing an error and proving
that it is an actual mistake requires much more than this.

Indeed.  All you have done is disagree with one of the proofs of the
halting theorem.  You have yet to show an error in it.  That will be >>>> difficult, because there aren't any.

q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ∞,
    if M applied to WM halts, and
q0 WM ⊢* Ĥq0 Wm WM ⊢* Ĥ y1 qn y2,
    if M applied to WM does not halt.

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
   if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
   if Ĥ applied to ⟨Ĥ⟩ does not halt.

<*Halting Problem Proof ERROR*>

Requires Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to report on the
direct execution of Ĥ applied to ⟨Ĥ⟩ and thus not
⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by Ĥ.embedded_H.

No Turing Machine decider can ever report on the
behavior of anything that is not an input encoded
as a finite string.

Ĥ is not a finite string input to Ĥ.embedded_H
⟨Ĥ⟩ ⟨Ĥ⟩ are finite string inputs to Ĥ.embedded_H

</*Halting Problem Proof ERROR*>

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
     ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches
     its simulated final halt state of ⟨Ĥ.qn⟩, and

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
     ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly
     reach its simulated final halt state of ⟨Ĥ.qn⟩.

When Ĥ is applied to ⟨Ĥ⟩ and embedded_H is a
simulating partial halt decider
(a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation until
*embedded_H sees the repeating pattern and transitions to Ĥ.qn*

But when this is only finite repeating pattern,

The infinite simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
cannot possibly reach its own simulated final
halt state of ⟨Ĥ.qn⟩ you fucking moron.

But that infinite simulation exists only in your dreams and is
counter-factual. Ĥ aborts and the same abort code is present in
embedded_H (unless you are cheating), so there is no infinite recursion,
no infinite repeating pattern. The correct simulation will reach the
final halt state. We know that Ĥ fails to do this correct simulation.
And performs a premature abort before it reaches that final halt state. Aborting a simulation before it can reach the final halt state is no
measure for non-halting behaviour, it is only evidence that simulation
is not the right too for this input to determine halting behaviour.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic

On 7/19/25 10:23 AM, olcott wrote:

On 7/19/2025 3:26 AM, Fred. Zwarts wrote:

Op 18.jul.2025 om 18:09 schreef olcott:

On 7/18/2025 6:17 AM, Fred. Zwarts wrote:

Op 17.jul.2025 om 16:44 schreef olcott:

On 7/6/2025 11:02 AM, Alan Mackenzie wrote:

[ Followup-To: set ]

In comp.theory olcott <[email protected]> wrote:

On 7/6/2025 5:16 AM, Alan Mackenzie wrote:

olcott <[email protected]> wrote:

On 7/5/2025 2:07 PM, Alan Mackenzie wrote:

You lie.  You don't have a proof.  Many people in this group >>>>>>>>>> have pointed
out lots of errors in various versions of your purported
proof, which you
just ignore.  The section in Professor Linz's book you used to >>>>>>>>>> be so fond
of citing will contain plenty of details, if only you would >>>>>>>>>> take the
trouble to understand it (assuming you're capable of such
understanding).

I have addressed ....

Meaningless pompous word.

.... all of those details that you make sure to ignore so that >>>>>>>>> you can
baselessly claim that I am wrong.

I vaguely remember rolling my eyes at your hopeless lack of
understanding.  It was like watching a 7 year old trying to do >>>>>>>> calculus.
The basic understanding was simply not there.  Years later, it's >>>>>>>> still
not there.

And yes, you are wrong.  The proofs of the halting theorem which >>>>>>>> involve
constructing programs which purported halting deciders cannot
decide
correctly are correct.

Yet you cannot point to even one mistake because there are none.

That's what I'm saying.  Those proofs of the halting theorem are free >>>>>> from mistakes.

More to the point, it is YOU who cannot point to any mistakes in
them.
They are valid proofs.  Your work, if it contradicts those proofs >>>>>> (which
isn't at all clear) can thus be dismissed without further
consideration.

There cannot possibly be *AN ACTUAL INPUT* that does the
opposite of whatever its decider decides. All of the examples >>>>>>>>> of this have never been *ACTUAL INPUTS*

That's so sloppily worded, it could mean almost anything.

The standard halting problem proof cannot even be constructed.

It has been constructed, and is valid.  But one would normally
talk about
formulating a proof, rather than constructing one.

[ .... ]

No Turing machine can possibly take another directly executing >>>>>>>>> Turing machine as in input, thus removing these from the
domain of every halt decider.

And that, too.

*Thus the requirement that HHH report on the behavior*
*of the directly executed DD has always been bogus*

And that makes your hat trick.

Turing machine partial halt deciders compute the mapping
from their actual inputs to the actual behavior that these
inputs specify.

And a fourth.  There's some semblance of truth in there, but
it's very
confused.

It is not at all confused. I know exactly what it means.

It's very confused to everybody but you, then.

Sloppy wording is your technique to get people to go down to
your level
of discussion.  That involves many posts trying just to tie you >>>>>>>> down to
specific word meanings, and is very tiresome and unrewarding.  I >>>>>>>> decline
to get involved any further.

*Yet as I claimed you found no actual mistake*

I've found plenty of actual mistakes.  I was a software developer by >>>>>> profession.

Let me tell you the punchline so that you can
see why I said those things.

Despite what I said last post, I will actually go to the trouble of >>>>>> analysing your sloppy expression.

Because directly executed Turing machines cannot
possibly be inputs to Turing machine deciders this
makes them outside of the domain of these deciders.

It's entirely unclear what a "directly executed Turing machine"
is. Most
of the time turing machines are theoretical constructs used for
proving
theorems.  They can be executed, but rarely are.

It's unclear what you mean by a turing machine being an input to a >>>>>> turing
machine.  Read up about universal turing machines to get a bit of >>>>>> background.

When a partial halt decider is required to report
on the direct execution of a machine this requirement
is bogus.

See above.  That paragraph is meaningless.

This means that the behavior of DD() is none of the damn
business of HHH, thus does not contradict HHH(DD)==0.
*If you disagree this only proves that you do not understand*

It's fully obscure what DD() and HHH mean, and thus impossible to
affirm or contradict the meaningless "HHH(DD)==0".

HHH(DD) does correctly detect that DD simulated by HHH
according to the semantics pf the C programming language
cannot possibly reach its own "return"statement final
halt state.

See above.  By the way, people concerned with computation theory use >>>>>> turing machines, which are well-defined, simple, and powerful.
They lack
the complexity, ambiguity, and unsuitability for theoretical work
of real
world programming languages like C.

*If you disagree this only proves that you do not understand*

Any mindless idiot can disagree. Showing an error and proving
that it is an actual mistake requires much more than this.

Indeed.  All you have done is disagree with one of the proofs of the >>>>>> halting theorem.  You have yet to show an error in it.  That will be >>>>>> difficult, because there aren't any.

q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ∞,
    if M applied to WM halts, and
q0 WM ⊢* Ĥq0 Wm WM ⊢* Ĥ y1 qn y2,
    if M applied to WM does not halt.

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
   if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
   if Ĥ applied to ⟨Ĥ⟩ does not halt.

<*Halting Problem Proof ERROR*>

Requires Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to report on the
direct execution of Ĥ applied to ⟨Ĥ⟩ and thus not
⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by Ĥ.embedded_H.

No Turing Machine decider can ever report on the
behavior of anything that is not an input encoded
as a finite string.

Ĥ is not a finite string input to Ĥ.embedded_H
⟨Ĥ⟩ ⟨Ĥ⟩ are finite string inputs to Ĥ.embedded_H

</*Halting Problem Proof ERROR*>

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
     ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches
     its simulated final halt state of ⟨Ĥ.qn⟩, and

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
     ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly >>>>>      reach its simulated final halt state of ⟨Ĥ.qn⟩.

When Ĥ is applied to ⟨Ĥ⟩ and embedded_H is a
simulating partial halt decider
(a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ >>>>> (f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation until
*embedded_H sees the repeating pattern and transitions to Ĥ.qn*

But when this is only finite repeating pattern,

The infinite simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
cannot possibly reach its own simulated final
halt state of ⟨Ĥ.qn⟩ you fucking moron.

But that infinite simulation exists only in your dreams and is
counter- factual. Ĥ aborts and the same abort code is present in
embedded_H (unless you are cheating), so there is no infinite
recursion, no infinite repeating pattern. The correct simulation will
reach the final halt state.

When you change my words and use those words as the basis
of your rebuttal you know that you cheat.

The infinite simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
cannot possibly reach its own simulated final
halt state of ⟨Ĥ.qn⟩ you fucking moron.

Which is just a strawman, as the definition of non-halting behavior of
the input, is that when the machine the input describes, in this case
the machibe Ĥ ⟨Ĥ⟩, will never stop running when run, even if carried out to an unbounded number of steps, but since H and thus embedded_H which
is a copy of it DO abort their simulations and go to Qn and thus the
actual machine Ĥ ⟨Ĥ⟩ halts, it must be wrong.

Basing your arguement on Strawmen, even after the error has been pointed
out just establishes that you are just a pathological liar.

 We know that Ĥ fails to do this correct simulation. And performs a
premature abort before it reaches that final halt state. Aborting a
simulation before it can reach the final halt state is no measure for
non-halting behaviour, it is only evidence that simulation is not the
right too for this input to determine halting behaviour.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic

On 7/19/25 11:11 AM, olcott wrote:

On 7/19/2025 3:05 AM, Mikko wrote:

On 2025-07-17 14:44:23 +0000, olcott said:

On 7/6/2025 11:02 AM, Alan Mackenzie wrote:

[ Followup-To: set ]

In comp.theory olcott <[email protected]> wrote:

On 7/6/2025 5:16 AM, Alan Mackenzie wrote:

olcott <[email protected]> wrote:

On 7/5/2025 2:07 PM, Alan Mackenzie wrote:

You lie.  You don't have a proof.  Many people in this group >>>>>>>> have pointed
out lots of errors in various versions of your purported proof, >>>>>>>> which you
just ignore.  The section in Professor Linz's book you used to >>>>>>>> be so fond
of citing will contain plenty of details, if only you would take >>>>>>>> the
trouble to understand it (assuming you're capable of such
understanding).

I have addressed ....

Meaningless pompous word.

.... all of those details that you make sure to ignore so that
you can
baselessly claim that I am wrong.

I vaguely remember rolling my eyes at your hopeless lack of
understanding.  It was like watching a 7 year old trying to do
calculus.
The basic understanding was simply not there.  Years later, it's
still
not there.

And yes, you are wrong.  The proofs of the halting theorem which
involve
constructing programs which purported halting deciders cannot decide >>>>>> correctly are correct.

Yet you cannot point to even one mistake because there are none.

That's what I'm saying.  Those proofs of the halting theorem are free >>>> from mistakes.

More to the point, it is YOU who cannot point to any mistakes in them. >>>> They are valid proofs.  Your work, if it contradicts those proofs
(which
isn't at all clear) can thus be dismissed without further
consideration.

There cannot possibly be *AN ACTUAL INPUT* that does the
opposite of whatever its decider decides. All of the examples
of this have never been *ACTUAL INPUTS*

That's so sloppily worded, it could mean almost anything.

The standard halting problem proof cannot even be constructed.

It has been constructed, and is valid.  But one would normally talk
about
formulating a proof, rather than constructing one.

[ .... ]

No Turing machine can possibly take another directly executing
Turing machine as in input, thus removing these from the
domain of every halt decider.

And that, too.

*Thus the requirement that HHH report on the behavior*
*of the directly executed DD has always been bogus*

And that makes your hat trick.

Turing machine partial halt deciders compute the mapping
from their actual inputs to the actual behavior that these
inputs specify.

And a fourth.  There's some semblance of truth in there, but it's >>>>>> very
confused.

It is not at all confused. I know exactly what it means.

It's very confused to everybody but you, then.

Sloppy wording is your technique to get people to go down to your
level
of discussion.  That involves many posts trying just to tie you
down to
specific word meanings, and is very tiresome and unrewarding.  I
decline
to get involved any further.

*Yet as I claimed you found no actual mistake*

I've found plenty of actual mistakes.  I was a software developer by
profession.

Let me tell you the punchline so that you can
see why I said those things.

Despite what I said last post, I will actually go to the trouble of
analysing your sloppy expression.

Because directly executed Turing machines cannot
possibly be inputs to Turing machine deciders this
makes them outside of the domain of these deciders.

It's entirely unclear what a "directly executed Turing machine" is.
Most
of the time turing machines are theoretical constructs used for proving >>>> theorems.  They can be executed, but rarely are.

It's unclear what you mean by a turing machine being an input to a
turing
machine.  Read up about universal turing machines to get a bit of
background.

When a partial halt decider is required to report
on the direct execution of a machine this requirement
is bogus.

See above.  That paragraph is meaningless.

This means that the behavior of DD() is none of the damn
business of HHH, thus does not contradict HHH(DD)==0.
*If you disagree this only proves that you do not understand*

It's fully obscure what DD() and HHH mean, and thus impossible to
affirm or contradict the meaningless "HHH(DD)==0".

HHH(DD) does correctly detect that DD simulated by HHH
according to the semantics pf the C programming language
cannot possibly reach its own "return"statement final
halt state.

See above.  By the way, people concerned with computation theory use
turing machines, which are well-defined, simple, and powerful.  They
lack
the complexity, ambiguity, and unsuitability for theoretical work of
real
world programming languages like C.

*If you disagree this only proves that you do not understand*

Any mindless idiot can disagree. Showing an error and proving
that it is an actual mistake requires much more than this.

Indeed.  All you have done is disagree with one of the proofs of the
halting theorem.  You have yet to show an error in it.  That will be >>>> difficult, because there aren't any.

q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ∞,
    if M applied to WM halts, and
q0 WM ⊢* Ĥq0 Wm WM ⊢* Ĥ y1 qn y2,
    if M applied to WM does not halt.

This means nothing as long as symbols are undefined.

They are defined on the link provided on the next line.

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
   if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
   if Ĥ applied to ⟨Ĥ⟩ does not halt.

<*Halting Problem Proof ERROR*>

Requires Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to report on the
direct execution of Ĥ applied to ⟨Ĥ⟩ and thus not
⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by Ĥ.embedded_H.

It is not clear what is the subject of the sentence. It makes a big
difference who or what requires. Some requirements are essential to
the topic but others may be irrelevant or even bogus.

This notation refers to Turing Machine: Ĥ
This notation refers to TM description: ⟨Ĥ⟩

The complete context is provided in the original proof https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

*Here is the corrected proof*
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
    ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches
    its simulated final halt state of ⟨Ĥ.qn⟩, and

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
    ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly
    reach its simulated final halt state of ⟨Ĥ.qn⟩.

Ĥ.embedded_H correctly transitions to Ĥ.qn.

Nope, the problem is you altered the definition.

The criterial is *NOT* if the simulation by the decider (or embedded_H)
reaches the final state, but it the program described by the input would
reach the final state when run.

If the decider simulates the input, and reaches a final state, it would
be correct to say that it halts.

If the decider chooses to stop its simulate, the fact that this
simulation did not reach a final state yet is NOT proof that it will not
reach a final state.

The decider needs to prove that the UTM simulation of THIS input, which
for Ĥ includes the code for THIS embedded_H which is the same as the
code for H. Since the ACTUAL H will have this abort and return 0 if that
is what you say is correct, such a proof can not be done, and thus H
never proved it was correct, and thus can be in error.

The fact that you continue to LIE about the definition of the problem,
even after the error has been explained, just shows that you are just a pathological liar, that either is morally deficient and does care abort
being wrong, or mentally deficient so can not learn the elementary
basics of the field.

Either of these could be considered making you a danger to society.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic

On 7/19/25 3:59 PM, olcott wrote:

On 7/19/2025 12:23 PM, Richard Damon wrote:

On 7/19/25 10:23 AM, olcott wrote:>>

When you change my words and use those words as the basis
of your rebuttal you know that you cheat.

The infinite simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
cannot possibly reach its own simulated final
halt state of ⟨Ĥ.qn⟩ you fucking moron.

Which is just a strawman, as the definition of non-halting behavior of
the input, is that when the machine the input describes,

If you only learn by rote memorization it may seem that way.
When you have a depth of understanding then you see that
what you said is merely an incorrect paraphrase of this:

That your are quoting from never-learned ignorance doesn't releave you
of the sin of lying, since it has been explained to you what the true definition is,

Sorry, you are just proving that you think lies are valid logic, and
thus have earned your place in that lake of fire.

Turing machine (at least partial) halt deciders only compute
the mapping from their finite string inputs to the actual
behavior that this input finite string actually specifies.

And that behavior is DEFINED by the behavior of the direct exectution of
the program it represents (and it must represent a program, including
ALL the code it uses).

It can also equivalent be seen as the UTM simulation of that exact same
input, which also will include the code of the same HHH as the input
given to HHH.

Conventional notation of a Turing Machine: Ĥ
Conventional notation of a TM description: ⟨Ĥ⟩

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
  if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
  if Ĥ applied to ⟨Ĥ⟩ does not halt.

*Is corrected to this*

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
    ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches
    its simulated final halt state of ⟨Ĥ.qn⟩, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
    ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly
    reach its simulated final halt state of ⟨Ĥ.qn⟩.

WRONG, READ THE SPECS AGAIN.

THEY SAY NOTHING ABOUT SIMULATION

YOU ARE JUST PROVING YOU ARE A STUPID PATHOLOGICAL LIAR.

the criteria is if Ĥ applied to ⟨Ĥ⟩ will halt when run.

That you can't uinderstand that just is an admission of your stupid
ignorance.

*Original proof*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Just rote repeating your lies just proves you are a pathological liar.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic

Op 19.jul.2025 om 16:23 schreef olcott:

On 7/19/2025 3:26 AM, Fred. Zwarts wrote:

Op 18.jul.2025 om 18:09 schreef olcott:

On 7/18/2025 6:17 AM, Fred. Zwarts wrote:

Op 17.jul.2025 om 16:44 schreef olcott:

On 7/6/2025 11:02 AM, Alan Mackenzie wrote:

[ Followup-To: set ]

In comp.theory olcott <[email protected]> wrote:

On 7/6/2025 5:16 AM, Alan Mackenzie wrote:

olcott <[email protected]> wrote:

On 7/5/2025 2:07 PM, Alan Mackenzie wrote:

You lie.  You don't have a proof.  Many people in this group >>>>>>>>>> have pointed
out lots of errors in various versions of your purported
proof, which you
just ignore.  The section in Professor Linz's book you used to >>>>>>>>>> be so fond
of citing will contain plenty of details, if only you would >>>>>>>>>> take the
trouble to understand it (assuming you're capable of such
understanding).

I have addressed ....

Meaningless pompous word.

.... all of those details that you make sure to ignore so that >>>>>>>>> you can
baselessly claim that I am wrong.

I vaguely remember rolling my eyes at your hopeless lack of
understanding.  It was like watching a 7 year old trying to do >>>>>>>> calculus.
The basic understanding was simply not there.  Years later, it's >>>>>>>> still
not there.

And yes, you are wrong.  The proofs of the halting theorem which >>>>>>>> involve
constructing programs which purported halting deciders cannot
decide
correctly are correct.

Yet you cannot point to even one mistake because there are none.

That's what I'm saying.  Those proofs of the halting theorem are free >>>>>> from mistakes.

More to the point, it is YOU who cannot point to any mistakes in
them.
They are valid proofs.  Your work, if it contradicts those proofs >>>>>> (which
isn't at all clear) can thus be dismissed without further
consideration.

There cannot possibly be *AN ACTUAL INPUT* that does the
opposite of whatever its decider decides. All of the examples >>>>>>>>> of this have never been *ACTUAL INPUTS*

That's so sloppily worded, it could mean almost anything.

The standard halting problem proof cannot even be constructed.

It has been constructed, and is valid.  But one would normally
talk about
formulating a proof, rather than constructing one.

[ .... ]

No Turing machine can possibly take another directly executing >>>>>>>>> Turing machine as in input, thus removing these from the
domain of every halt decider.

And that, too.

*Thus the requirement that HHH report on the behavior*
*of the directly executed DD has always been bogus*

And that makes your hat trick.

Turing machine partial halt deciders compute the mapping
from their actual inputs to the actual behavior that these
inputs specify.

And a fourth.  There's some semblance of truth in there, but
it's very
confused.

It is not at all confused. I know exactly what it means.

It's very confused to everybody but you, then.

Sloppy wording is your technique to get people to go down to
your level
of discussion.  That involves many posts trying just to tie you >>>>>>>> down to
specific word meanings, and is very tiresome and unrewarding.  I >>>>>>>> decline
to get involved any further.

*Yet as I claimed you found no actual mistake*

I've found plenty of actual mistakes.  I was a software developer by >>>>>> profession.

Let me tell you the punchline so that you can
see why I said those things.

Despite what I said last post, I will actually go to the trouble of >>>>>> analysing your sloppy expression.

Because directly executed Turing machines cannot
possibly be inputs to Turing machine deciders this
makes them outside of the domain of these deciders.

It's entirely unclear what a "directly executed Turing machine"
is. Most
of the time turing machines are theoretical constructs used for
proving
theorems.  They can be executed, but rarely are.

It's unclear what you mean by a turing machine being an input to a >>>>>> turing
machine.  Read up about universal turing machines to get a bit of >>>>>> background.

When a partial halt decider is required to report
on the direct execution of a machine this requirement
is bogus.

See above.  That paragraph is meaningless.

This means that the behavior of DD() is none of the damn
business of HHH, thus does not contradict HHH(DD)==0.
*If you disagree this only proves that you do not understand*

It's fully obscure what DD() and HHH mean, and thus impossible to
affirm or contradict the meaningless "HHH(DD)==0".

HHH(DD) does correctly detect that DD simulated by HHH
according to the semantics pf the C programming language
cannot possibly reach its own "return"statement final
halt state.

See above.  By the way, people concerned with computation theory use >>>>>> turing machines, which are well-defined, simple, and powerful.
They lack
the complexity, ambiguity, and unsuitability for theoretical work
of real
world programming languages like C.

*If you disagree this only proves that you do not understand*

Any mindless idiot can disagree. Showing an error and proving
that it is an actual mistake requires much more than this.

Indeed.  All you have done is disagree with one of the proofs of the >>>>>> halting theorem.  You have yet to show an error in it.  That will be >>>>>> difficult, because there aren't any.

q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ∞,
    if M applied to WM halts, and
q0 WM ⊢* Ĥq0 Wm WM ⊢* Ĥ y1 qn y2,
    if M applied to WM does not halt.

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
   if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
   if Ĥ applied to ⟨Ĥ⟩ does not halt.

<*Halting Problem Proof ERROR*>

Requires Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to report on the
direct execution of Ĥ applied to ⟨Ĥ⟩ and thus not
⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by Ĥ.embedded_H.

No Turing Machine decider can ever report on the
behavior of anything that is not an input encoded
as a finite string.

Ĥ is not a finite string input to Ĥ.embedded_H
⟨Ĥ⟩ ⟨Ĥ⟩ are finite string inputs to Ĥ.embedded_H

</*Halting Problem Proof ERROR*>

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
     ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches
     its simulated final halt state of ⟨Ĥ.qn⟩, and

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
     ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly >>>>>      reach its simulated final halt state of ⟨Ĥ.qn⟩.

When Ĥ is applied to ⟨Ĥ⟩ and embedded_H is a
simulating partial halt decider
(a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ >>>>> (f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation until
*embedded_H sees the repeating pattern and transitions to Ĥ.qn*

But when this is only finite repeating pattern,

The infinite simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
cannot possibly reach its own simulated final
halt state of ⟨Ĥ.qn⟩ you fucking moron.

But that infinite simulation exists only in your dreams and is
counter- factual. Ĥ aborts and the same abort code is present in
embedded_H (unless you are cheating), so there is no infinite
recursion, no infinite repeating pattern. The correct simulation will
reach the final halt state.

When you change my words and use those words as the basis
of your rebuttal you know that you cheat.

The infinite simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
cannot possibly reach its own simulated final
halt state of ⟨Ĥ.qn⟩ you fucking moron.

No rebuttal, only repeated claims without evidence. Ĥ and embedded_H
have the same code, including the abort code. Your claim of an infinite simulation is a counter-factual dream.
And ad hominem attacks are evidence for a lack of arguments.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic

Op 20.jul.2025 om 17:16 schreef olcott:

On 7/20/2025 2:54 AM, Fred. Zwarts wrote:

Op 19.jul.2025 om 16:23 schreef olcott:

On 7/19/2025 3:26 AM, Fred. Zwarts wrote:

Op 18.jul.2025 om 18:09 schreef olcott:

On 7/18/2025 6:17 AM, Fred. Zwarts wrote:

Op 17.jul.2025 om 16:44 schreef olcott:

On 7/6/2025 11:02 AM, Alan Mackenzie wrote:

[ Followup-To: set ]

In comp.theory olcott <[email protected]> wrote:

On 7/6/2025 5:16 AM, Alan Mackenzie wrote:

olcott <[email protected]> wrote:

On 7/5/2025 2:07 PM, Alan Mackenzie wrote:

You lie.  You don't have a proof.  Many people in this group >>>>>>>>>>>> have pointed
out lots of errors in various versions of your purported >>>>>>>>>>>> proof, which you
just ignore.  The section in Professor Linz's book you used >>>>>>>>>>>> to be so fond
of citing will contain plenty of details, if only you would >>>>>>>>>>>> take the
trouble to understand it (assuming you're capable of such >>>>>>>>>>>> understanding).

I have addressed ....

Meaningless pompous word.

.... all of those details that you make sure to ignore so >>>>>>>>>>> that you can
baselessly claim that I am wrong.

I vaguely remember rolling my eyes at your hopeless lack of >>>>>>>>>> understanding.  It was like watching a 7 year old trying to do >>>>>>>>>> calculus.
The basic understanding was simply not there.  Years later, >>>>>>>>>> it's still
not there.

And yes, you are wrong.  The proofs of the halting theorem >>>>>>>>>> which involve
constructing programs which purported halting deciders cannot >>>>>>>>>> decide
correctly are correct.

Yet you cannot point to even one mistake because there are none. >>>>>>>>

That's what I'm saying.  Those proofs of the halting theorem are >>>>>>>> free
from mistakes.

More to the point, it is YOU who cannot point to any mistakes in >>>>>>>> them.
They are valid proofs.  Your work, if it contradicts those
proofs (which
isn't at all clear) can thus be dismissed without further
consideration.

There cannot possibly be *AN ACTUAL INPUT* that does the >>>>>>>>>>> opposite of whatever its decider decides. All of the examples >>>>>>>>>>> of this have never been *ACTUAL INPUTS*

That's so sloppily worded, it could mean almost anything.

The standard halting problem proof cannot even be constructed. >>>>>>>>

It has been constructed, and is valid.  But one would normally >>>>>>>> talk about
formulating a proof, rather than constructing one.

[ .... ]

No Turing machine can possibly take another directly executing >>>>>>>>>>> Turing machine as in input, thus removing these from the >>>>>>>>>>> domain of every halt decider.

And that, too.

*Thus the requirement that HHH report on the behavior*
*of the directly executed DD has always been bogus*

And that makes your hat trick.

Turing machine partial halt deciders compute the mapping >>>>>>>>>>> from their actual inputs to the actual behavior that these >>>>>>>>>>> inputs specify.

And a fourth.  There's some semblance of truth in there, but >>>>>>>>>> it's very
confused.

It is not at all confused. I know exactly what it means.

It's very confused to everybody but you, then.

Sloppy wording is your technique to get people to go down to >>>>>>>>>> your level
of discussion.  That involves many posts trying just to tie >>>>>>>>>> you down to
specific word meanings, and is very tiresome and unrewarding. >>>>>>>>>> I decline
to get involved any further.

*Yet as I claimed you found no actual mistake*

I've found plenty of actual mistakes.  I was a software
developer by
profession.

Let me tell you the punchline so that you can
see why I said those things.

Despite what I said last post, I will actually go to the trouble of >>>>>>>> analysing your sloppy expression.

Because directly executed Turing machines cannot
possibly be inputs to Turing machine deciders this
makes them outside of the domain of these deciders.

It's entirely unclear what a "directly executed Turing machine" >>>>>>>> is. Most
of the time turing machines are theoretical constructs used for >>>>>>>> proving
theorems.  They can be executed, but rarely are.

It's unclear what you mean by a turing machine being an input to >>>>>>>> a turing
machine.  Read up about universal turing machines to get a bit of >>>>>>>> background.

When a partial halt decider is required to report
on the direct execution of a machine this requirement
is bogus.

See above.  That paragraph is meaningless.

This means that the behavior of DD() is none of the damn
business of HHH, thus does not contradict HHH(DD)==0.
*If you disagree this only proves that you do not understand* >>>>>>>>

It's fully obscure what DD() and HHH mean, and thus impossible to >>>>>>>> affirm or contradict the meaningless "HHH(DD)==0".

HHH(DD) does correctly detect that DD simulated by HHH
according to the semantics pf the C programming language
cannot possibly reach its own "return"statement final
halt state.

See above.  By the way, people concerned with computation theory >>>>>>>> use
turing machines, which are well-defined, simple, and powerful. >>>>>>>> They lack
the complexity, ambiguity, and unsuitability for theoretical
work of real
world programming languages like C.

*If you disagree this only proves that you do not understand* >>>>>>>>
Any mindless idiot can disagree. Showing an error and proving >>>>>>>>> that it is an actual mistake requires much more than this.

Indeed.  All you have done is disagree with one of the proofs of >>>>>>>> the
halting theorem.  You have yet to show an error in it.  That >>>>>>>> will be
difficult, because there aren't any.

q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ∞,
    if M applied to WM halts, and
q0 WM ⊢* Ĥq0 Wm WM ⊢* Ĥ y1 qn y2,
    if M applied to WM does not halt.

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞, >>>>>>>    if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
   if Ĥ applied to ⟨Ĥ⟩ does not halt.

<*Halting Problem Proof ERROR*>

Requires Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to report on the
direct execution of Ĥ applied to ⟨Ĥ⟩ and thus not
⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by Ĥ.embedded_H.

No Turing Machine decider can ever report on the
behavior of anything that is not an input encoded
as a finite string.

Ĥ is not a finite string input to Ĥ.embedded_H
⟨Ĥ⟩ ⟨Ĥ⟩ are finite string inputs to Ĥ.embedded_H

</*Halting Problem Proof ERROR*>

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
     ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches
     its simulated final halt state of ⟨Ĥ.qn⟩, and

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
     ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly >>>>>>>      reach its simulated final halt state of ⟨Ĥ.qn⟩.

When Ĥ is applied to ⟨Ĥ⟩ and embedded_H is a
simulating partial halt decider
(a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ >>>>>>> (f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation until
*embedded_H sees the repeating pattern and transitions to Ĥ.qn*

But when this is only finite repeating pattern,

The infinite simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
cannot possibly reach its own simulated final
halt state of ⟨Ĥ.qn⟩ you fucking moron.

But that infinite simulation exists only in your dreams and is
counter- factual. Ĥ aborts and the same abort code is present in
embedded_H (unless you are cheating), so there is no infinite
recursion, no infinite repeating pattern. The correct simulation
will reach the final halt state.

When you change my words and use those words as the basis
of your rebuttal you know that you cheat.

The infinite simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
cannot possibly reach its own simulated final
halt state of ⟨Ĥ.qn⟩ you fucking moron.

No rebuttal, only repeated claims without evidence. Ĥ and embedded_H
have the same code, including the abort code. Your claim of an
infinite simulation is a counter-factual dream.
And ad hominem attacks are evidence for a lack of arguments.

As usual, no rebuttal, but only a repetition of incorrect claims without
any evidence:

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
    ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches
    its simulated final halt state of ⟨Ĥ.qn⟩, and

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
    ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly
    reach its simulated final halt state of ⟨Ĥ.qn⟩.

When Ĥ is applied to ⟨Ĥ⟩ and embedded_H is a
simulating partial halt decider
(a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation until
embedded_H sees the repeating pattern and transitions to Ĥ.qn.

This has already been proven to be incomplete.
in (g), we need to replace 'repeating pattern' with 'finite repeating
pattern', because not all repetitions are infinite and a finite
repetition is not a measure for non-halting. So we can continue:

(h) Ĥ has the same detection and abort code as embedded_H.
(i) A correct simulation would reach the final halt state after a few repetitions.
(j) When embedded_h fails to reach this final halt state, it guesses incorrectly that no such final halt state exists.
(k) Embedded_ reports the invalid result that Ĥ does not halt, based on
an aborted finite repetition and a wild guess.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic

Op 21.jul.2025 om 16:04 schreef olcott:

On 7/21/2025 4:28 AM, Mikko wrote:

On 2025-07-20 15:02:30 +0000, olcott said:

On 7/20/2025 3:39 AM, Mikko wrote:

On 2025-07-19 15:11:21 +0000, olcott said:

On 7/19/2025 3:05 AM, Mikko wrote:

On 2025-07-17 14:44:23 +0000, olcott said:

On 7/6/2025 11:02 AM, Alan Mackenzie wrote:

[ Followup-To: set ]

In comp.theory olcott <[email protected]> wrote:

On 7/6/2025 5:16 AM, Alan Mackenzie wrote:

olcott <[email protected]> wrote:

On 7/5/2025 2:07 PM, Alan Mackenzie wrote:

You lie.  You don't have a proof.  Many people in this group >>>>>>>>>>>> have pointed
out lots of errors in various versions of your purported >>>>>>>>>>>> proof, which you
just ignore.  The section in Professor Linz's book you used >>>>>>>>>>>> to be so fond
of citing will contain plenty of details, if only you would >>>>>>>>>>>> take the
trouble to understand it (assuming you're capable of such >>>>>>>>>>>> understanding).

I have addressed ....

Meaningless pompous word.

.... all of those details that you make sure to ignore so >>>>>>>>>>> that you can
baselessly claim that I am wrong.

I vaguely remember rolling my eyes at your hopeless lack of >>>>>>>>>> understanding.  It was like watching a 7 year old trying to do >>>>>>>>>> calculus.
The basic understanding was simply not there.  Years later, >>>>>>>>>> it's still
not there.

And yes, you are wrong.  The proofs of the halting theorem >>>>>>>>>> which involve
constructing programs which purported halting deciders cannot >>>>>>>>>> decide
correctly are correct.

Yet you cannot point to even one mistake because there are none. >>>>>>>>

That's what I'm saying.  Those proofs of the halting theorem are >>>>>>>> free
from mistakes.

More to the point, it is YOU who cannot point to any mistakes in >>>>>>>> them.
They are valid proofs.  Your work, if it contradicts those
proofs (which
isn't at all clear) can thus be dismissed without further
consideration.

There cannot possibly be *AN ACTUAL INPUT* that does the >>>>>>>>>>> opposite of whatever its decider decides. All of the examples >>>>>>>>>>> of this have never been *ACTUAL INPUTS*

That's so sloppily worded, it could mean almost anything.

The standard halting problem proof cannot even be constructed. >>>>>>>>

It has been constructed, and is valid.  But one would normally >>>>>>>> talk about
formulating a proof, rather than constructing one.

[ .... ]

No Turing machine can possibly take another directly executing >>>>>>>>>>> Turing machine as in input, thus removing these from the >>>>>>>>>>> domain of every halt decider.

And that, too.

*Thus the requirement that HHH report on the behavior*
*of the directly executed DD has always been bogus*

And that makes your hat trick.

Turing machine partial halt deciders compute the mapping >>>>>>>>>>> from their actual inputs to the actual behavior that these >>>>>>>>>>> inputs specify.

And a fourth.  There's some semblance of truth in there, but >>>>>>>>>> it's very
confused.

It is not at all confused. I know exactly what it means.

It's very confused to everybody but you, then.

Sloppy wording is your technique to get people to go down to >>>>>>>>>> your level
of discussion.  That involves many posts trying just to tie >>>>>>>>>> you down to
specific word meanings, and is very tiresome and unrewarding. >>>>>>>>>> I decline
to get involved any further.

*Yet as I claimed you found no actual mistake*

I've found plenty of actual mistakes.  I was a software
developer by
profession.

Let me tell you the punchline so that you can
see why I said those things.

Despite what I said last post, I will actually go to the trouble of >>>>>>>> analysing your sloppy expression.

Because directly executed Turing machines cannot
possibly be inputs to Turing machine deciders this
makes them outside of the domain of these deciders.

It's entirely unclear what a "directly executed Turing machine" >>>>>>>> is.  Most
of the time turing machines are theoretical constructs used for >>>>>>>> proving
theorems.  They can be executed, but rarely are.

It's unclear what you mean by a turing machine being an input to >>>>>>>> a turing
machine.  Read up about universal turing machines to get a bit of >>>>>>>> background.

When a partial halt decider is required to report
on the direct execution of a machine this requirement
is bogus.

See above.  That paragraph is meaningless.

This means that the behavior of DD() is none of the damn
business of HHH, thus does not contradict HHH(DD)==0.
*If you disagree this only proves that you do not understand* >>>>>>>>

It's fully obscure what DD() and HHH mean, and thus impossible to >>>>>>>> affirm or contradict the meaningless "HHH(DD)==0".

HHH(DD) does correctly detect that DD simulated by HHH
according to the semantics pf the C programming language
cannot possibly reach its own "return"statement final
halt state.

See above.  By the way, people concerned with computation theory >>>>>>>> use
turing machines, which are well-defined, simple, and powerful. >>>>>>>> They lack
the complexity, ambiguity, and unsuitability for theoretical
work of real
world programming languages like C.

*If you disagree this only proves that you do not understand* >>>>>>>>
Any mindless idiot can disagree. Showing an error and proving >>>>>>>>> that it is an actual mistake requires much more than this.

Indeed.  All you have done is disagree with one of the proofs of >>>>>>>> the
halting theorem.  You have yet to show an error in it.  That >>>>>>>> will be
difficult, because there aren't any.

q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ∞,
    if M applied to WM halts, and
q0 WM ⊢* Ĥq0 Wm WM ⊢* Ĥ y1 qn y2,
    if M applied to WM does not halt.

This means nothing as long as symbols are undefined.

They are defined on the link provided on the next line.

That was not said in the quoted message, and some of the symbols
are not defined there.

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞, >>>>>>>    if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
   if Ĥ applied to ⟨Ĥ⟩ does not halt.

<*Halting Problem Proof ERROR*>

Requires Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to report on the
direct execution of Ĥ applied to ⟨Ĥ⟩ and thus not
⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by Ĥ.embedded_H.

It is not clear what is the subject of the sentence. It makes a big >>>>>> difference who or what requires. Some requirements are essential to >>>>>> the topic but others may be irrelevant or even bogus.

<ChatGPT>
Misrepresentation of Input:
The standard proof assumes a decider
H(M,x) that determines whether machine
M halts on input x.

But this formulation is flawed, because:
Turing machines can only process finite
encodings (e.g. ⟨M⟩), not executable entities
like M.

So the valid formulation must be
H(⟨M⟩,x), where ⟨M⟩ is a string.
</ChatGPT>

This notation refers to Turing Machine: Ĥ
This notation refers to TM description: ⟨Ĥ⟩

The complete context is provided in the original proof
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

*Here is the corrected proof*
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
     ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches
     its simulated final halt state of ⟨Ĥ.qn⟩, and

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
     ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly >>>>>      reach its simulated final halt state of ⟨Ĥ.qn⟩.

Ĥ.embedded_H correctly transitions to Ĥ.qn.

The question about the missing subject remains unanswered. Apparently
you don't know what you were talking about.

The key subject to me is this:
How does Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ transition to Ĥ.qn correctly?

(a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation until
embedded_H sees the repeating pattern and transitions to Ĥ.qn.

None of which is relevant to the point what is not a garamamtical
sentence cannot be true.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
   if Ĥ applied to ⟨Ĥ⟩ halts, and

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
   if Ĥ applied to ⟨Ĥ⟩ does not halt.

*The above incorrect proof is corrected below*

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
   ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches
   its simulated final halt state of ⟨Ĥ.qn⟩, and

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
   ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly
   reach its simulated final halt state of ⟨Ĥ.qn⟩.

It is unable to prove either halting or non-halting, showing that it is
not the right tool for this input. Other deciders are able to get the
correct result for exctly the same input.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic

Op 21.jul.2025 om 15:36 schreef olcott:

On 7/21/2025 3:10 AM, Fred. Zwarts wrote:

Op 20.jul.2025 om 17:16 schreef olcott:

On 7/20/2025 2:54 AM, Fred. Zwarts wrote:

Op 19.jul.2025 om 16:23 schreef olcott:

On 7/19/2025 3:26 AM, Fred. Zwarts wrote:

Op 18.jul.2025 om 18:09 schreef olcott:

On 7/18/2025 6:17 AM, Fred. Zwarts wrote:

Op 17.jul.2025 om 16:44 schreef olcott:

On 7/6/2025 11:02 AM, Alan Mackenzie wrote:

[ Followup-To: set ]

In comp.theory olcott <[email protected]> wrote:

On 7/6/2025 5:16 AM, Alan Mackenzie wrote:

olcott <[email protected]> wrote:

On 7/5/2025 2:07 PM, Alan Mackenzie wrote:

You lie.  You don't have a proof.  Many people in this >>>>>>>>>>>>>> group have pointed
out lots of errors in various versions of your purported >>>>>>>>>>>>>> proof, which you
just ignore.  The section in Professor Linz's book you >>>>>>>>>>>>>> used to be so fond
of citing will contain plenty of details, if only you >>>>>>>>>>>>>> would take the
trouble to understand it (assuming you're capable of such >>>>>>>>>>>>>> understanding).

I have addressed ....

Meaningless pompous word.

.... all of those details that you make sure to ignore so >>>>>>>>>>>>> that you can
baselessly claim that I am wrong.

I vaguely remember rolling my eyes at your hopeless lack of >>>>>>>>>>>> understanding.  It was like watching a 7 year old trying to >>>>>>>>>>>> do calculus.
The basic understanding was simply not there.  Years later, >>>>>>>>>>>> it's still
not there.

And yes, you are wrong.  The proofs of the halting theorem >>>>>>>>>>>> which involve
constructing programs which purported halting deciders >>>>>>>>>>>> cannot decide
correctly are correct.

Yet you cannot point to even one mistake because there are none. >>>>>>>>>>

That's what I'm saying.  Those proofs of the halting theorem >>>>>>>>>> are free
from mistakes.

More to the point, it is YOU who cannot point to any mistakes >>>>>>>>>> in them.
They are valid proofs.  Your work, if it contradicts those >>>>>>>>>> proofs (which
isn't at all clear) can thus be dismissed without further
consideration.

There cannot possibly be *AN ACTUAL INPUT* that does the >>>>>>>>>>>>> opposite of whatever its decider decides. All of the examples >>>>>>>>>>>>> of this have never been *ACTUAL INPUTS*

That's so sloppily worded, it could mean almost anything. >>>>>>>>>>

The standard halting problem proof cannot even be constructed. >>>>>>>>>>

It has been constructed, and is valid.  But one would normally >>>>>>>>>> talk about
formulating a proof, rather than constructing one.

[ .... ]

No Turing machine can possibly take another directly executing >>>>>>>>>>>>> Turing machine as in input, thus removing these from the >>>>>>>>>>>>> domain of every halt decider.

And that, too.

*Thus the requirement that HHH report on the behavior* >>>>>>>>>>>>> *of the directly executed DD has always been bogus*

And that makes your hat trick.

Turing machine partial halt deciders compute the mapping >>>>>>>>>>>>> from their actual inputs to the actual behavior that these >>>>>>>>>>>>> inputs specify.

And a fourth.  There's some semblance of truth in there, but >>>>>>>>>>>> it's very
confused.

It is not at all confused. I know exactly what it means.

It's very confused to everybody but you, then.

Sloppy wording is your technique to get people to go down to >>>>>>>>>>>> your level
of discussion.  That involves many posts trying just to tie >>>>>>>>>>>> you down to
specific word meanings, and is very tiresome and
unrewarding. I decline
to get involved any further.

*Yet as I claimed you found no actual mistake*

I've found plenty of actual mistakes.  I was a software
developer by
profession.

Let me tell you the punchline so that you can
see why I said those things.

Despite what I said last post, I will actually go to the
trouble of
analysing your sloppy expression.

Because directly executed Turing machines cannot
possibly be inputs to Turing machine deciders this
makes them outside of the domain of these deciders.

It's entirely unclear what a "directly executed Turing
machine" is. Most
of the time turing machines are theoretical constructs used >>>>>>>>>> for proving
theorems.  They can be executed, but rarely are.

It's unclear what you mean by a turing machine being an input >>>>>>>>>> to a turing
machine.  Read up about universal turing machines to get a bit of >>>>>>>>>> background.

When a partial halt decider is required to report
on the direct execution of a machine this requirement
is bogus.

See above.  That paragraph is meaningless.

This means that the behavior of DD() is none of the damn >>>>>>>>>>> business of HHH, thus does not contradict HHH(DD)==0.
*If you disagree this only proves that you do not understand* >>>>>>>>>>

It's fully obscure what DD() and HHH mean, and thus impossible to >>>>>>>>>> affirm or contradict the meaningless "HHH(DD)==0".

HHH(DD) does correctly detect that DD simulated by HHH
according to the semantics pf the C programming language >>>>>>>>>>> cannot possibly reach its own "return"statement final
halt state.

See above.  By the way, people concerned with computation >>>>>>>>>> theory use
turing machines, which are well-defined, simple, and powerful. >>>>>>>>>> They lack
the complexity, ambiguity, and unsuitability for theoretical >>>>>>>>>> work of real
world programming languages like C.

*If you disagree this only proves that you do not understand* >>>>>>>>>>
Any mindless idiot can disagree. Showing an error and proving >>>>>>>>>>> that it is an actual mistake requires much more than this. >>>>>>>>>>

Indeed.  All you have done is disagree with one of the proofs >>>>>>>>>> of the
halting theorem.  You have yet to show an error in it.  That >>>>>>>>>> will be
difficult, because there aren't any.

q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ∞,
    if M applied to WM halts, and
q0 WM ⊢* Ĥq0 Wm WM ⊢* Ĥ y1 qn y2,
    if M applied to WM does not halt.

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞, >>>>>>>>>    if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn >>>>>>>>>    if Ĥ applied to ⟨Ĥ⟩ does not halt.

<*Halting Problem Proof ERROR*>

Requires Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to report on the
direct execution of Ĥ applied to ⟨Ĥ⟩ and thus not
⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by Ĥ.embedded_H.

No Turing Machine decider can ever report on the
behavior of anything that is not an input encoded
as a finite string.

Ĥ is not a finite string input to Ĥ.embedded_H
⟨Ĥ⟩ ⟨Ĥ⟩ are finite string inputs to Ĥ.embedded_H

</*Halting Problem Proof ERROR*>

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞ >>>>>>>>>      ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches >>>>>>>>>      its simulated final halt state of ⟨Ĥ.qn⟩, and

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn >>>>>>>>>      ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly >>>>>>>>>      reach its simulated final halt state of ⟨Ĥ.qn⟩. >>>>>>>>>
When Ĥ is applied to ⟨Ĥ⟩ and embedded_H is a
simulating partial halt decider
(a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ >>>>>>>>> (f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation until
*embedded_H sees the repeating pattern and transitions to Ĥ.qn* >>>>>>>>

But when this is only finite repeating pattern,

The infinite simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
cannot possibly reach its own simulated final
halt state of ⟨Ĥ.qn⟩ you fucking moron.

But that infinite simulation exists only in your dreams and is
counter- factual. Ĥ aborts and the same abort code is present in
embedded_H (unless you are cheating), so there is no infinite
recursion, no infinite repeating pattern. The correct simulation
will reach the final halt state.

When you change my words and use those words as the basis
of your rebuttal you know that you cheat.

The infinite simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
cannot possibly reach its own simulated final
halt state of ⟨Ĥ.qn⟩ you fucking moron.

No rebuttal, only repeated claims without evidence. Ĥ and embedded_H
have the same code, including the abort code. Your claim of an
infinite simulation is a counter-factual dream.
And ad hominem attacks are evidence for a lack of arguments.

As usual, no rebuttal, but only a repetition of incorrect claims
without any evidence:

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
     ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches
     its simulated final halt state of ⟨Ĥ.qn⟩, and

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
     ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly
     reach its simulated final halt state of ⟨Ĥ.qn⟩.

When Ĥ is applied to ⟨Ĥ⟩ and embedded_H is a
simulating partial halt decider
(a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation until
embedded_H sees the repeating pattern and transitions to Ĥ.qn.

This has already been proven to be incomplete.
in (g), we need to replace 'repeating pattern' with 'finite repeating
pattern', because not all repetitions are infinite and a finite
repetition is not a measure for non-halting. So we can continue:

void Infinite_Loop()
{
  HERE: goto HERE;
  return;
}

HHH(Infinite_Loop) is the same sort of finite repeating pattern.

No, Finite_Recursion looks like the recursion specified in the input for
HHH:

void Finite_Recursion (int N) {
if (N > 0) Finite_Recursion (N - 1);
printf ("Olcott thinks this is never printed.\n");
}

You fails to see the difference between infinite recursion and finite recursion. Probably because you are still dreaming of an HHH that does
not abort, even when the HHH is specified in the input to abort and
return. That your HHH fails to reach that point is no measure for the
behaviour specified in the input.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)