The halting problem theorem and proof thereof [Turing, 1937] (upon
which other currently extant halting problem proofs are derived) is
invalid due to an invalid "impossible program" [Strachey, 1965] that
arises not from a function call-like infinite recursion but from a
category error in the form of an invalid (erroneous) infinite recursion
present in the proof [Wikipedia, 2022].

The categories involved in the category error are the decider and that
which is being decided. Currently extant attempts to conflate the
decider with that which is being decided are infinitely recursive and
thus invalid.

/Flibble

On 2022-05-04 10:46, Mr Flibble wrote:
> The halting problem theorem and proof thereof [Turing, 1937] (upon

The halting *problem* and the halting *theorem* are two different
things. You can't "prove" a problem. The halting theorem states that the
halting problem has no solution. That's what Turing proved.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

On Wed, 4 May 2022 10:53:57 -0600
André G. Isaak <agisaak@gm.invalid> wrote:

> On 2022-05-04 10:46, Mr Flibble wrote:
> > The halting problem theorem and proof thereof [Turing, 1937] (upon
>
> The halting *problem* and the halting *theorem* are two different
> things. You can't "prove" a problem. The halting theorem states that
> the halting problem has no solution. That's what Turing proved.

And if you actually read what I actually wrote: "halting problem
THEOREM".

/Flibble

On 5/4/2022 11:46 AM, Mr Flibble wrote:
> The halting problem theorem and proof thereof [Turing, 1937] (upon
> which other currently extant halting problem proofs are derived) is
> invalid due to an invalid "impossible program" [Strachey, 1965] that
> arises not from a function call-like infinite recursion but from a
> category error in the form of an invalid (erroneous) infinite recursion
> present in the proof [Wikipedia, 2022].
>

Strachey, seemed to have been the first person to simplify this:
https://academic.oup.com/comjnl/article/7/4/313/354243

All of the pseudo-code examples of the HP counter-examples are based on
Strachey's CPL code.

> The categories involved in the category error are the decider and that
> which is being decided. Currently extant attempts to conflate the
> decider with that which is being decided are infinitely recursive and
> thus invalid.
>
> /Flibble
>

--
Copyright 2022 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

On 5/4/2022 11:53 AM, André G. Isaak wrote:
> On 2022-05-04 10:46, Mr Flibble wrote:
>> The halting problem theorem and proof thereof [Turing, 1937] (upon
>
> The halting *problem* and the halting *theorem* are two different
> things. You can't "prove" a problem. The halting theorem states that the
> halting problem has no solution. That's what Turing proved.
>
> André
>
> Yes.

--
Copyright 2022 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

On 5/4/2022 11:56 AM, Mr Flibble wrote:
> On Wed, 4 May 2022 10:53:57 -0600
> André G. Isaak <agisaak@gm.invalid> wrote:
>
>> On 2022-05-04 10:46, Mr Flibble wrote:
>>> The halting problem theorem and proof thereof [Turing, 1937] (upon
>>
>> The halting *problem* and the halting *theorem* are two different
>> things. You can't "prove" a problem. The halting theorem states that
>> the halting problem has no solution. That's what Turing proved.
>
> And if you actually read what I actually wrote: "halting problem
> THEOREM".
>
> /Flibble
>

Seems close enough to me.

--
Copyright 2022 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

On Wed, 4 May 2022 12:53:06 -0500
olcott <polcott2@gmail.com> wrote:

> On 5/4/2022 11:56 AM, Mr Flibble wrote:
> > On Wed, 4 May 2022 10:53:57 -0600
> > André G. Isaak <agisaak@gm.invalid> wrote:
> >
> >> On 2022-05-04 10:46, Mr Flibble wrote:
> >>> The halting problem theorem and proof thereof [Turing, 1937]
> >>> (upon
> >>
> >> The halting *problem* and the halting *theorem* are two different
> >> things. You can't "prove" a problem. The halting theorem states
> >> that the halting problem has no solution. That's what Turing
> >> proved.
> >
> > And if you actually read what I actually wrote: "halting problem
> > THEOREM".
> >
> > /Flibble
> >
>
> Seems close enough to me.
What? I used the word theorem deliberately as I was referring not to
the halting problem itself but Turing's theorem and proof.

/Flibble

On 5/4/22 12:46 PM, Mr Flibble wrote:
> The halting problem theorem and proof thereof [Turing, 1937] (upon
> which other currently extant halting problem proofs are derived) is
> invalid due to an invalid "impossible program" [Strachey, 1965] that
> arises not from a function call-like infinite recursion but from a
> category error in the form of an invalid (erroneous) infinite recursion
> present in the proof [Wikipedia, 2022].
>
> The categories involved in the category error are the decider and that
> which is being decided. Currently extant attempts to conflate the
> decider with that which is being decided are infinitely recursive and
> thus invalid.
>
> /Flibble
>

And what is the error between the decider and the decided.

The Decider is H.

The thing to be decideer is H^ applied to <H^>

Now, if you want to claim that H^ can't use H, then you are saying
either that Turing Macines aren't allowed to use other Turing Machines,
which is crasy, or that H just can't be asked about machines which are
based on it, at which point that is just admitting that the Halting
Problem is in fact impossible, as there exist some VALID programs (which
H^ is) that H just can't be asked to decide on.

Note, H^ is NOT recursive in definition (unless H is). H^ is basically
jast a "call" to H with a little bit of simple logic around it.

The "recursion" (which isn't actually recursion) comes about because we
give H^ and input that just happens to be a representation of itself,
and H^ doesn't actually know that.

If H actually meets the requirements of being a decider (and thus also
of being a Comptation), then we get no infinite "recursion", as the H
inside H^ will, by necessity, return an answer after finite time, and H^
will either Halt or go into a simple infinite loop.

H is thus just proved to give the wrong answer.

If H refuses to give the wrong answer, then it turns out to not be a
decider and just loops forever with an ever growing tape as the level of
simulation just keeps increasing.

Note, we NEVER get back to H^.q0 or any of the states of H^ up to H^.qx,
as the entire execution trace is within its copy of H doing its
simulation to try to decide. There is no actual Recursion.

Yes, in trying to decide how we might program H to try to get the right
answer, we can think about levels of recursion, but those are not
actually in the real trace of the execution of H^. Once we actual make a
decision on how we are going to attempt to make an H, and establish a
fixed algorithm to program into H, we find that we have failed to make a
correct decider, because, as the theorem says, it IS impossible. This
doesn't mean the problem is "invalid" just impossible.

Just like the problem of creating a program to always win at Tic Tac
Toe. The problem is well defined, just impossible due to the nature of
the game.

On Wed, 4 May 2022 19:42:19 -0400
Richard Damon <Richard@Damon-Family.org> wrote:

> On 5/4/22 12:46 PM, Mr Flibble wrote:
> > The halting problem theorem and proof thereof [Turing, 1937] (upon
> > which other currently extant halting problem proofs are derived) is
> > invalid due to an invalid "impossible program" [Strachey, 1965] that
> > arises not from a function call-like infinite recursion but from a
> > category error in the form of an invalid (erroneous) infinite
> > recursion present in the proof [Wikipedia, 2022].
> >
> > The categories involved in the category error are the decider and
> > that which is being decided. Currently extant attempts to conflate
> > the decider with that which is being decided are infinitely
> > recursive and thus invalid.
> >
> > /Flibble
> >
>
> And what is the error between the decider and the decided.
>
> The Decider is H.
>
> The thing to be decideer is H^ applied to <H^>
>
> Now, if you want to claim that H^ can't use H, then you are saying
> either that Turing Macines aren't allowed to use other Turing
> Machines, which is crasy, or that H just can't be asked about
> machines which are based on it, at which point that is just admitting
> that the Halting Problem is in fact impossible, as there exist some
> VALID programs (which H^ is) that H just can't be asked to decide on.
>
> Note, H^ is NOT recursive in definition (unless H is). H^ is
> basically jast a "call" to H with a little bit of simple logic around
> it.
>
> The "recursion" (which isn't actually recursion) comes about because
> we give H^ and input that just happens to be a representation of
> itself, and H^ doesn't actually know that.
>
> If H actually meets the requirements of being a decider (and thus
> also of being a Comptation), then we get no infinite "recursion", as
> the H inside H^ will, by necessity, return an answer after finite
> time, and H^ will either Halt or go into a simple infinite loop.
>
> H is thus just proved to give the wrong answer.
>
> If H refuses to give the wrong answer, then it turns out to not be a
> decider and just loops forever with an ever growing tape as the level
> of simulation just keeps increasing.
>
> Note, we NEVER get back to H^.q0 or any of the states of H^ up to
> H^.qx, as the entire execution trace is within its copy of H doing
> its simulation to try to decide. There is no actual Recursion.
>
>
> Yes, in trying to decide how we might program H to try to get the
> right answer, we can think about levels of recursion, but those are
> not actually in the real trace of the execution of H^. Once we actual
> make a decision on how we are going to attempt to make an H, and
> establish a fixed algorithm to program into H, we find that we have
> failed to make a correct decider, because, as the theorem says, it IS
> impossible. This doesn't mean the problem is "invalid" just
> impossible.
>
> Just like the problem of creating a program to always win at Tic Tac
> Toe. The problem is well defined, just impossible due to the nature
> of the game.

Nope. The infinite recursion is a category error and therefore invalid:
you and the rest of you shower appear to have a blindspot to this fact.

/Flibble

Mr Flibble wrote:
> On Wed, 4 May 2022 19:42:19 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 5/4/22 12:46 PM, Mr Flibble wrote:
>>> The halting problem theorem and proof thereof [Turing, 1937] (upon
>>> which other currently extant halting problem proofs are derived) is
>>> invalid due to an invalid "impossible program" [Strachey, 1965] that
>>> arises not from a function call-like infinite recursion but from a
>>> category error in the form of an invalid (erroneous) infinite
>>> recursion present in the proof [Wikipedia, 2022].
>>>
>>> The categories involved in the category error are the decider and
>>> that which is being decided. Currently extant attempts to conflate
>>> the decider with that which is being decided are infinitely
>>> recursive and thus invalid.
>>>
>>> /Flibble
>>>
>>
>> And what is the error between the decider and the decided.
>>
>> The Decider is H.
>>
>> The thing to be decideer is H^ applied to <H^>
>>
>> Now, if you want to claim that H^ can't use H, then you are saying
>> either that Turing Macines aren't allowed to use other Turing
>> Machines, which is crasy, or that H just can't be asked about
>> machines which are based on it, at which point that is just admitting
>> that the Halting Problem is in fact impossible, as there exist some
>> VALID programs (which H^ is) that H just can't be asked to decide on.
>>
>> Note, H^ is NOT recursive in definition (unless H is). H^ is
>> basically jast a "call" to H with a little bit of simple logic around
>> it.
>>
>> The "recursion" (which isn't actually recursion) comes about because
>> we give H^ and input that just happens to be a representation of
>> itself, and H^ doesn't actually know that.
>>
>> If H actually meets the requirements of being a decider (and thus
>> also of being a Comptation), then we get no infinite "recursion", as
>> the H inside H^ will, by necessity, return an answer after finite
>> time, and H^ will either Halt or go into a simple infinite loop.
>>
>> H is thus just proved to give the wrong answer.
>>
>> If H refuses to give the wrong answer, then it turns out to not be a
>> decider and just loops forever with an ever growing tape as the level
>> of simulation just keeps increasing.
>>
>> Note, we NEVER get back to H^.q0 or any of the states of H^ up to
>> H^.qx, as the entire execution trace is within its copy of H doing
>> its simulation to try to decide. There is no actual Recursion.
>>
>>
>> Yes, in trying to decide how we might program H to try to get the
>> right answer, we can think about levels of recursion, but those are
>> not actually in the real trace of the execution of H^. Once we actual
>> make a decision on how we are going to attempt to make an H, and
>> establish a fixed algorithm to program into H, we find that we have
>> failed to make a correct decider, because, as the theorem says, it IS
>> impossible. This doesn't mean the problem is "invalid" just
>> impossible.
>>
>> Just like the problem of creating a program to always win at Tic Tac
>> Toe. The problem is well defined, just impossible due to the nature
>> of the game.
>
> Nope. The infinite recursion is a category error and therefore invalid:
> you and the rest of you shower appear to have a blindspot to this fact.
>
> /Flibble
>

YOU ARE A IDIOT.

On 5/4/22 7:51 PM, Mr Flibble wrote:
> On Wed, 4 May 2022 19:42:19 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 5/4/22 12:46 PM, Mr Flibble wrote:
>>> The halting problem theorem and proof thereof [Turing, 1937] (upon
>>> which other currently extant halting problem proofs are derived) is
>>> invalid due to an invalid "impossible program" [Strachey, 1965] that
>>> arises not from a function call-like infinite recursion but from a
>>> category error in the form of an invalid (erroneous) infinite
>>> recursion present in the proof [Wikipedia, 2022].
>>>
>>> The categories involved in the category error are the decider and
>>> that which is being decided. Currently extant attempts to conflate
>>> the decider with that which is being decided are infinitely
>>> recursive and thus invalid.
>>>
>>> /Flibble
>>>
>>
>> And what is the error between the decider and the decided.
>>
>> The Decider is H.
>>
>> The thing to be decideer is H^ applied to <H^>
>>
>> Now, if you want to claim that H^ can't use H, then you are saying
>> either that Turing Macines aren't allowed to use other Turing
>> Machines, which is crasy, or that H just can't be asked about
>> machines which are based on it, at which point that is just admitting
>> that the Halting Problem is in fact impossible, as there exist some
>> VALID programs (which H^ is) that H just can't be asked to decide on.
>>
>> Note, H^ is NOT recursive in definition (unless H is). H^ is
>> basically jast a "call" to H with a little bit of simple logic around
>> it.
>>
>> The "recursion" (which isn't actually recursion) comes about because
>> we give H^ and input that just happens to be a representation of
>> itself, and H^ doesn't actually know that.
>>
>> If H actually meets the requirements of being a decider (and thus
>> also of being a Comptation), then we get no infinite "recursion", as
>> the H inside H^ will, by necessity, return an answer after finite
>> time, and H^ will either Halt or go into a simple infinite loop.
>>
>> H is thus just proved to give the wrong answer.
>>
>> If H refuses to give the wrong answer, then it turns out to not be a
>> decider and just loops forever with an ever growing tape as the level
>> of simulation just keeps increasing.
>>
>> Note, we NEVER get back to H^.q0 or any of the states of H^ up to
>> H^.qx, as the entire execution trace is within its copy of H doing
>> its simulation to try to decide. There is no actual Recursion.
>>
>>
>> Yes, in trying to decide how we might program H to try to get the
>> right answer, we can think about levels of recursion, but those are
>> not actually in the real trace of the execution of H^. Once we actual
>> make a decision on how we are going to attempt to make an H, and
>> establish a fixed algorithm to program into H, we find that we have
>> failed to make a correct decider, because, as the theorem says, it IS
>> impossible. This doesn't mean the problem is "invalid" just
>> impossible.
>>
>> Just like the problem of creating a program to always win at Tic Tac
>> Toe. The problem is well defined, just impossible due to the nature
>> of the game.
>
> Nope. The infinite recursion is a category error and therefore invalid:
> you and the rest of you shower appear to have a blindspot to this fact.
>
> /Flibble
>

And what is the infinite recursion in the HALTING PROBLEM?

There is none.

Or in the Theorem, (That there doesn't exist an answer to the Halting
Problem), THERE IS NONE.

In fact, you could say that "infinite recursion" is one of the things
that can be used to PROVE the Theorem, as any answer (by simulation) to
the problem would need to invoke an infinte recursion to answer it. Thus
the ANSWER (which is what is trying to do the infinite recursion) and
not the problem (which has no recursion) is invalid.

On Wed, 4 May 2022 20:02:57 -0400
Richard Damon <Richard@Damon-Family.org> wrote:

> On 5/4/22 7:51 PM, Mr Flibble wrote:
> > On Wed, 4 May 2022 19:42:19 -0400
> > Richard Damon <Richard@Damon-Family.org> wrote:
> >
> >> On 5/4/22 12:46 PM, Mr Flibble wrote:
> >>> The halting problem theorem and proof thereof [Turing, 1937] (upon
> >>> which other currently extant halting problem proofs are derived)
> >>> is invalid due to an invalid "impossible program" [Strachey,
> >>> 1965] that arises not from a function call-like infinite
> >>> recursion but from a category error in the form of an invalid
> >>> (erroneous) infinite recursion present in the proof [Wikipedia,
> >>> 2022].
> >>>
> >>> The categories involved in the category error are the decider and
> >>> that which is being decided. Currently extant attempts to
> >>> conflate the decider with that which is being decided are
> >>> infinitely recursive and thus invalid.
> >>>
> >>> /Flibble
> >>>
> >>
> >> And what is the error between the decider and the decided.
> >>
> >> The Decider is H.
> >>
> >> The thing to be decideer is H^ applied to <H^>
> >>
> >> Now, if you want to claim that H^ can't use H, then you are saying
> >> either that Turing Macines aren't allowed to use other Turing
> >> Machines, which is crasy, or that H just can't be asked about
> >> machines which are based on it, at which point that is just
> >> admitting that the Halting Problem is in fact impossible, as there
> >> exist some VALID programs (which H^ is) that H just can't be asked
> >> to decide on.
> >>
> >> Note, H^ is NOT recursive in definition (unless H is). H^ is
> >> basically jast a "call" to H with a little bit of simple logic
> >> around it.
> >>
> >> The "recursion" (which isn't actually recursion) comes about
> >> because we give H^ and input that just happens to be a
> >> representation of itself, and H^ doesn't actually know that.
> >>
> >> If H actually meets the requirements of being a decider (and thus
> >> also of being a Comptation), then we get no infinite "recursion",
> >> as the H inside H^ will, by necessity, return an answer after
> >> finite time, and H^ will either Halt or go into a simple infinite
> >> loop.
> >>
> >> H is thus just proved to give the wrong answer.
> >>
> >> If H refuses to give the wrong answer, then it turns out to not be
> >> a decider and just loops forever with an ever growing tape as the
> >> level of simulation just keeps increasing.
> >>
> >> Note, we NEVER get back to H^.q0 or any of the states of H^ up to
> >> H^.qx, as the entire execution trace is within its copy of H doing
> >> its simulation to try to decide. There is no actual Recursion.
> >>
> >>
> >> Yes, in trying to decide how we might program H to try to get the
> >> right answer, we can think about levels of recursion, but those are
> >> not actually in the real trace of the execution of H^. Once we
> >> actual make a decision on how we are going to attempt to make an
> >> H, and establish a fixed algorithm to program into H, we find that
> >> we have failed to make a correct decider, because, as the theorem
> >> says, it IS impossible. This doesn't mean the problem is "invalid"
> >> just impossible.
> >>
> >> Just like the problem of creating a program to always win at Tic
> >> Tac Toe. The problem is well defined, just impossible due to the
> >> nature of the game.
> >
> > Nope. The infinite recursion is a category error and therefore
> > invalid: you and the rest of you shower appear to have a blindspot
> > to this fact.
> >
> > /Flibble
> >
>
> And what is the infinite recursion in the HALTING PROBLEM?
>
> There is none.

I know, which is why I explicitly said halting problem THEOREM in this
post.

>
> Or in the Theorem, (That there doesn't exist an answer to the Halting
> Problem), THERE IS NONE.

The invalid infinite recursion is in the THEOREM as stated in
[Wikipedia, 2022].

>
> In fact, you could say that "infinite recursion" is one of the things
> that can be used to PROVE the Theorem, as any answer (by simulation)
> to the problem would need to invoke an infinte recursion to answer
> it. Thus the ANSWER (which is what is trying to do the infinite
> recursion) and not the problem (which has no recursion) is invalid.

However the infinite recursion is a category error in this case so
cannot prove anything.

/Flibble

On 5/4/22 8:05 PM, Mr Flibble wrote:
> On Wed, 4 May 2022 20:02:57 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 5/4/22 7:51 PM, Mr Flibble wrote:
>>> On Wed, 4 May 2022 19:42:19 -0400
>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>
>>>> On 5/4/22 12:46 PM, Mr Flibble wrote:
>>>>> The halting problem theorem and proof thereof [Turing, 1937] (upon
>>>>> which other currently extant halting problem proofs are derived)
>>>>> is invalid due to an invalid "impossible program" [Strachey,
>>>>> 1965] that arises not from a function call-like infinite
>>>>> recursion but from a category error in the form of an invalid
>>>>> (erroneous) infinite recursion present in the proof [Wikipedia,
>>>>> 2022].
>>>>>
>>>>> The categories involved in the category error are the decider and
>>>>> that which is being decided. Currently extant attempts to
>>>>> conflate the decider with that which is being decided are
>>>>> infinitely recursive and thus invalid.
>>>>>
>>>>> /Flibble
>>>>>
>>>>
>>>> And what is the error between the decider and the decided.
>>>>
>>>> The Decider is H.
>>>>
>>>> The thing to be decideer is H^ applied to <H^>
>>>>
>>>> Now, if you want to claim that H^ can't use H, then you are saying
>>>> either that Turing Macines aren't allowed to use other Turing
>>>> Machines, which is crasy, or that H just can't be asked about
>>>> machines which are based on it, at which point that is just
>>>> admitting that the Halting Problem is in fact impossible, as there
>>>> exist some VALID programs (which H^ is) that H just can't be asked
>>>> to decide on.
>>>>
>>>> Note, H^ is NOT recursive in definition (unless H is). H^ is
>>>> basically jast a "call" to H with a little bit of simple logic
>>>> around it.
>>>>
>>>> The "recursion" (which isn't actually recursion) comes about
>>>> because we give H^ and input that just happens to be a
>>>> representation of itself, and H^ doesn't actually know that.
>>>>
>>>> If H actually meets the requirements of being a decider (and thus
>>>> also of being a Comptation), then we get no infinite "recursion",
>>>> as the H inside H^ will, by necessity, return an answer after
>>>> finite time, and H^ will either Halt or go into a simple infinite
>>>> loop.
>>>>
>>>> H is thus just proved to give the wrong answer.
>>>>
>>>> If H refuses to give the wrong answer, then it turns out to not be
>>>> a decider and just loops forever with an ever growing tape as the
>>>> level of simulation just keeps increasing.
>>>>
>>>> Note, we NEVER get back to H^.q0 or any of the states of H^ up to
>>>> H^.qx, as the entire execution trace is within its copy of H doing
>>>> its simulation to try to decide. There is no actual Recursion.
>>>>
>>>>
>>>> Yes, in trying to decide how we might program H to try to get the
>>>> right answer, we can think about levels of recursion, but those are
>>>> not actually in the real trace of the execution of H^. Once we
>>>> actual make a decision on how we are going to attempt to make an
>>>> H, and establish a fixed algorithm to program into H, we find that
>>>> we have failed to make a correct decider, because, as the theorem
>>>> says, it IS impossible. This doesn't mean the problem is "invalid"
>>>> just impossible.
>>>>
>>>> Just like the problem of creating a program to always win at Tic
>>>> Tac Toe. The problem is well defined, just impossible due to the
>>>> nature of the game.
>>>
>>> Nope. The infinite recursion is a category error and therefore
>>> invalid: you and the rest of you shower appear to have a blindspot
>>> to this fact.
>>>
>>> /Flibble
>>>
>>
>> And what is the infinite recursion in the HALTING PROBLEM?
>>
>> There is none.
>
> I know, which is why I explicitly said halting problem THEOREM in this
> post.
>
>>
>> Or in the Theorem, (That there doesn't exist an answer to the Halting
>> Problem), THERE IS NONE.
>
> The invalid infinite recursion is in the THEOREM as stated in
> [Wikipedia, 2022].
>
>>
>> In fact, you could say that "infinite recursion" is one of the things
>> that can be used to PROVE the Theorem, as any answer (by simulation)
>> to the problem would need to invoke an infinte recursion to answer
>> it. Thus the ANSWER (which is what is trying to do the infinite
>> recursion) and not the problem (which has no recursion) is invalid.
>
> However the infinite recursion is a category error in this case so
> cannot prove anything.
>
> /Flibble
>

But what HAS the infinite recursion?

Not the PROBLEM.

Not the Theorem.

Not H^

The only thing that comes close is H, so that means that the H doesn't
exist, and thus the Problem is intact and the Theorem proved.

On Wed, 4 May 2022 20:21:04 -0400
Richard Damon <Richard@Damon-Family.org> wrote:

> On 5/4/22 8:05 PM, Mr Flibble wrote:
> > On Wed, 4 May 2022 20:02:57 -0400
> > Richard Damon <Richard@Damon-Family.org> wrote:
> >
> >> On 5/4/22 7:51 PM, Mr Flibble wrote:
> >>> On Wed, 4 May 2022 19:42:19 -0400
> >>> Richard Damon <Richard@Damon-Family.org> wrote:
> >>>
> >>>> On 5/4/22 12:46 PM, Mr Flibble wrote:
> >>>>> The halting problem theorem and proof thereof [Turing, 1937]
> >>>>> (upon which other currently extant halting problem proofs are
> >>>>> derived) is invalid due to an invalid "impossible program"
> >>>>> [Strachey, 1965] that arises not from a function call-like
> >>>>> infinite recursion but from a category error in the form of an
> >>>>> invalid (erroneous) infinite recursion present in the proof
> >>>>> [Wikipedia, 2022].
> >>>>>
> >>>>> The categories involved in the category error are the decider
> >>>>> and that which is being decided. Currently extant attempts to
> >>>>> conflate the decider with that which is being decided are
> >>>>> infinitely recursive and thus invalid.
> >>>>>
> >>>>> /Flibble
> >>>>>
> >>>>
> >>>> And what is the error between the decider and the decided.
> >>>>
> >>>> The Decider is H.
> >>>>
> >>>> The thing to be decideer is H^ applied to <H^>
> >>>>
> >>>> Now, if you want to claim that H^ can't use H, then you are
> >>>> saying either that Turing Macines aren't allowed to use other
> >>>> Turing Machines, which is crasy, or that H just can't be asked
> >>>> about machines which are based on it, at which point that is just
> >>>> admitting that the Halting Problem is in fact impossible, as
> >>>> there exist some VALID programs (which H^ is) that H just can't
> >>>> be asked to decide on.
> >>>>
> >>>> Note, H^ is NOT recursive in definition (unless H is). H^ is
> >>>> basically jast a "call" to H with a little bit of simple logic
> >>>> around it.
> >>>>
> >>>> The "recursion" (which isn't actually recursion) comes about
> >>>> because we give H^ and input that just happens to be a
> >>>> representation of itself, and H^ doesn't actually know that.
> >>>>
> >>>> If H actually meets the requirements of being a decider (and thus
> >>>> also of being a Comptation), then we get no infinite "recursion",
> >>>> as the H inside H^ will, by necessity, return an answer after
> >>>> finite time, and H^ will either Halt or go into a simple infinite
> >>>> loop.
> >>>>
> >>>> H is thus just proved to give the wrong answer.
> >>>>
> >>>> If H refuses to give the wrong answer, then it turns out to not
> >>>> be a decider and just loops forever with an ever growing tape as
> >>>> the level of simulation just keeps increasing.
> >>>>
> >>>> Note, we NEVER get back to H^.q0 or any of the states of H^ up to
> >>>> H^.qx, as the entire execution trace is within its copy of H
> >>>> doing its simulation to try to decide. There is no actual
> >>>> Recursion.
> >>>>
> >>>>
> >>>> Yes, in trying to decide how we might program H to try to get the
> >>>> right answer, we can think about levels of recursion, but those
> >>>> are not actually in the real trace of the execution of H^. Once
> >>>> we actual make a decision on how we are going to attempt to make
> >>>> an H, and establish a fixed algorithm to program into H, we find
> >>>> that we have failed to make a correct decider, because, as the
> >>>> theorem says, it IS impossible. This doesn't mean the problem is
> >>>> "invalid" just impossible.
> >>>>
> >>>> Just like the problem of creating a program to always win at Tic
> >>>> Tac Toe. The problem is well defined, just impossible due to the
> >>>> nature of the game.
> >>>
> >>> Nope. The infinite recursion is a category error and therefore
> >>> invalid: you and the rest of you shower appear to have a blindspot
> >>> to this fact.
> >>>
> >>> /Flibble
> >>>
> >>
> >> And what is the infinite recursion in the HALTING PROBLEM?
> >>
> >> There is none.
> >
> > I know, which is why I explicitly said halting problem THEOREM in
> > this post.
> >
> >>
> >> Or in the Theorem, (That there doesn't exist an answer to the
> >> Halting Problem), THERE IS NONE.
> >
> > The invalid infinite recursion is in the THEOREM as stated in
> > [Wikipedia, 2022].
> >
> >>
> >> In fact, you could say that "infinite recursion" is one of the
> >> things that can be used to PROVE the Theorem, as any answer (by
> >> simulation) to the problem would need to invoke an infinte
> >> recursion to answer it. Thus the ANSWER (which is what is trying
> >> to do the infinite recursion) and not the problem (which has no
> >> recursion) is invalid.
> >
> > However the infinite recursion is a category error in this case so
> > cannot prove anything.
> >
> > /Flibble
> >
>
> But what HAS the infinite recursion?
>
> Not the PROBLEM.
>
> Not the Theorem.
>
> Not H^
>
> The only thing that comes close is H, so that means that the H
> doesn't exist, and thus the Problem is intact and the Theorem proved.

[Strachey, 1965] simplifies the theorem making the infinite recursion
obvious; see also [Wikipedia, 2022].

/Flibble

On 5/4/22 8:24 PM, Mr Flibble wrote:
> On Wed, 4 May 2022 20:21:04 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 5/4/22 8:05 PM, Mr Flibble wrote:
>>> On Wed, 4 May 2022 20:02:57 -0400
>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>
>>>> On 5/4/22 7:51 PM, Mr Flibble wrote:
>>>>> On Wed, 4 May 2022 19:42:19 -0400
>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>
>>>>>> On 5/4/22 12:46 PM, Mr Flibble wrote:
>>>>>>> The halting problem theorem and proof thereof [Turing, 1937]
>>>>>>> (upon which other currently extant halting problem proofs are
>>>>>>> derived) is invalid due to an invalid "impossible program"
>>>>>>> [Strachey, 1965] that arises not from a function call-like
>>>>>>> infinite recursion but from a category error in the form of an
>>>>>>> invalid (erroneous) infinite recursion present in the proof
>>>>>>> [Wikipedia, 2022].
>>>>>>>
>>>>>>> The categories involved in the category error are the decider
>>>>>>> and that which is being decided. Currently extant attempts to
>>>>>>> conflate the decider with that which is being decided are
>>>>>>> infinitely recursive and thus invalid.
>>>>>>>
>>>>>>> /Flibble
>>>>>>>
>>>>>>
>>>>>> And what is the error between the decider and the decided.
>>>>>>
>>>>>> The Decider is H.
>>>>>>
>>>>>> The thing to be decideer is H^ applied to <H^>
>>>>>>
>>>>>> Now, if you want to claim that H^ can't use H, then you are
>>>>>> saying either that Turing Macines aren't allowed to use other
>>>>>> Turing Machines, which is crasy, or that H just can't be asked
>>>>>> about machines which are based on it, at which point that is just
>>>>>> admitting that the Halting Problem is in fact impossible, as
>>>>>> there exist some VALID programs (which H^ is) that H just can't
>>>>>> be asked to decide on.
>>>>>>
>>>>>> Note, H^ is NOT recursive in definition (unless H is). H^ is
>>>>>> basically jast a "call" to H with a little bit of simple logic
>>>>>> around it.
>>>>>>
>>>>>> The "recursion" (which isn't actually recursion) comes about
>>>>>> because we give H^ and input that just happens to be a
>>>>>> representation of itself, and H^ doesn't actually know that.
>>>>>>
>>>>>> If H actually meets the requirements of being a decider (and thus
>>>>>> also of being a Comptation), then we get no infinite "recursion",
>>>>>> as the H inside H^ will, by necessity, return an answer after
>>>>>> finite time, and H^ will either Halt or go into a simple infinite
>>>>>> loop.
>>>>>>
>>>>>> H is thus just proved to give the wrong answer.
>>>>>>
>>>>>> If H refuses to give the wrong answer, then it turns out to not
>>>>>> be a decider and just loops forever with an ever growing tape as
>>>>>> the level of simulation just keeps increasing.
>>>>>>
>>>>>> Note, we NEVER get back to H^.q0 or any of the states of H^ up to
>>>>>> H^.qx, as the entire execution trace is within its copy of H
>>>>>> doing its simulation to try to decide. There is no actual
>>>>>> Recursion.
>>>>>>
>>>>>>
>>>>>> Yes, in trying to decide how we might program H to try to get the
>>>>>> right answer, we can think about levels of recursion, but those
>>>>>> are not actually in the real trace of the execution of H^. Once
>>>>>> we actual make a decision on how we are going to attempt to make
>>>>>> an H, and establish a fixed algorithm to program into H, we find
>>>>>> that we have failed to make a correct decider, because, as the
>>>>>> theorem says, it IS impossible. This doesn't mean the problem is
>>>>>> "invalid" just impossible.
>>>>>>
>>>>>> Just like the problem of creating a program to always win at Tic
>>>>>> Tac Toe. The problem is well defined, just impossible due to the
>>>>>> nature of the game.
>>>>>
>>>>> Nope. The infinite recursion is a category error and therefore
>>>>> invalid: you and the rest of you shower appear to have a blindspot
>>>>> to this fact.
>>>>>
>>>>> /Flibble
>>>>>
>>>>
>>>> And what is the infinite recursion in the HALTING PROBLEM?
>>>>
>>>> There is none.
>>>
>>> I know, which is why I explicitly said halting problem THEOREM in
>>> this post.
>>>
>>>>
>>>> Or in the Theorem, (That there doesn't exist an answer to the
>>>> Halting Problem), THERE IS NONE.
>>>
>>> The invalid infinite recursion is in the THEOREM as stated in
>>> [Wikipedia, 2022].
>>>
>>>>
>>>> In fact, you could say that "infinite recursion" is one of the
>>>> things that can be used to PROVE the Theorem, as any answer (by
>>>> simulation) to the problem would need to invoke an infinte
>>>> recursion to answer it. Thus the ANSWER (which is what is trying
>>>> to do the infinite recursion) and not the problem (which has no
>>>> recursion) is invalid.
>>>
>>> However the infinite recursion is a category error in this case so
>>> cannot prove anything.
>>>
>>> /Flibble
>>>
>>
>> But what HAS the infinite recursion?
>>
>> Not the PROBLEM.
>>
>> Not the Theorem.
>>
>> Not H^
>>
>> The only thing that comes close is H, so that means that the H
>> doesn't exist, and thus the Problem is intact and the Theorem proved.
>
> [Strachey, 1965] simplifies the theorem making the infinite recursion
> obvious; see also [Wikipedia, 2022].
>
> /Flibble
>

The THEOREM is just that there exist no H that statisfies that Halting
Problem, so that is NOT recursive at all. And has nothing to "Simplify"

You seem to have a category error in distinguishing between the Theorem,
and a proof of it.

So again, WHAT has an invalid infinite recursion.

Not the PROBLEM

Not the THEOREM

Not even the H^ that proves the THEOREM.

The only thing that gets into an infinite recursion is the possible
answer, which by your logic is enough to prove it doesn't exist, and
thus the Theorem is proved.

On Wed, 4 May 2022 20:41:51 -0400
Richard Damon <Richard@Damon-Family.org> wrote:

> On 5/4/22 8:24 PM, Mr Flibble wrote:
> > On Wed, 4 May 2022 20:21:04 -0400
> > Richard Damon <Richard@Damon-Family.org> wrote:
> >
> >> On 5/4/22 8:05 PM, Mr Flibble wrote:
> >>> On Wed, 4 May 2022 20:02:57 -0400
> >>> Richard Damon <Richard@Damon-Family.org> wrote:
> >>>
> >>>> On 5/4/22 7:51 PM, Mr Flibble wrote:
> >>>>> On Wed, 4 May 2022 19:42:19 -0400
> >>>>> Richard Damon <Richard@Damon-Family.org> wrote:
> >>>>>
> >>>>>> On 5/4/22 12:46 PM, Mr Flibble wrote:
> >>>>>>> The halting problem theorem and proof thereof [Turing, 1937]
> >>>>>>> (upon which other currently extant halting problem proofs are
> >>>>>>> derived) is invalid due to an invalid "impossible program"
> >>>>>>> [Strachey, 1965] that arises not from a function call-like
> >>>>>>> infinite recursion but from a category error in the form of an
> >>>>>>> invalid (erroneous) infinite recursion present in the proof
> >>>>>>> [Wikipedia, 2022].
> >>>>>>>
> >>>>>>> The categories involved in the category error are the decider
> >>>>>>> and that which is being decided. Currently extant attempts to
> >>>>>>> conflate the decider with that which is being decided are
> >>>>>>> infinitely recursive and thus invalid.
> >>>>>>>
> >>>>>>> /Flibble
> >>>>>>>
> >>>>>>
> >>>>>> And what is the error between the decider and the decided.
> >>>>>>
> >>>>>> The Decider is H.
> >>>>>>
> >>>>>> The thing to be decideer is H^ applied to <H^>
> >>>>>>
> >>>>>> Now, if you want to claim that H^ can't use H, then you are
> >>>>>> saying either that Turing Macines aren't allowed to use other
> >>>>>> Turing Machines, which is crasy, or that H just can't be asked
> >>>>>> about machines which are based on it, at which point that is
> >>>>>> just admitting that the Halting Problem is in fact impossible,
> >>>>>> as there exist some VALID programs (which H^ is) that H just
> >>>>>> can't be asked to decide on.
> >>>>>>
> >>>>>> Note, H^ is NOT recursive in definition (unless H is). H^ is
> >>>>>> basically jast a "call" to H with a little bit of simple logic
> >>>>>> around it.
> >>>>>>
> >>>>>> The "recursion" (which isn't actually recursion) comes about
> >>>>>> because we give H^ and input that just happens to be a
> >>>>>> representation of itself, and H^ doesn't actually know that.
> >>>>>>
> >>>>>> If H actually meets the requirements of being a decider (and
> >>>>>> thus also of being a Comptation), then we get no infinite
> >>>>>> "recursion", as the H inside H^ will, by necessity, return an
> >>>>>> answer after finite time, and H^ will either Halt or go into a
> >>>>>> simple infinite loop.
> >>>>>>
> >>>>>> H is thus just proved to give the wrong answer.
> >>>>>>
> >>>>>> If H refuses to give the wrong answer, then it turns out to not
> >>>>>> be a decider and just loops forever with an ever growing tape
> >>>>>> as the level of simulation just keeps increasing.
> >>>>>>
> >>>>>> Note, we NEVER get back to H^.q0 or any of the states of H^ up
> >>>>>> to H^.qx, as the entire execution trace is within its copy of H
> >>>>>> doing its simulation to try to decide. There is no actual
> >>>>>> Recursion.
> >>>>>>
> >>>>>>
> >>>>>> Yes, in trying to decide how we might program H to try to get
> >>>>>> the right answer, we can think about levels of recursion, but
> >>>>>> those are not actually in the real trace of the execution of
> >>>>>> H^. Once we actual make a decision on how we are going to
> >>>>>> attempt to make an H, and establish a fixed algorithm to
> >>>>>> program into H, we find that we have failed to make a correct
> >>>>>> decider, because, as the theorem says, it IS impossible. This
> >>>>>> doesn't mean the problem is "invalid" just impossible.
> >>>>>>
> >>>>>> Just like the problem of creating a program to always win at
> >>>>>> Tic Tac Toe. The problem is well defined, just impossible due
> >>>>>> to the nature of the game.
> >>>>>
> >>>>> Nope. The infinite recursion is a category error and therefore
> >>>>> invalid: you and the rest of you shower appear to have a
> >>>>> blindspot to this fact.
> >>>>>
> >>>>> /Flibble
> >>>>>
> >>>>
> >>>> And what is the infinite recursion in the HALTING PROBLEM?
> >>>>
> >>>> There is none.
> >>>
> >>> I know, which is why I explicitly said halting problem THEOREM in
> >>> this post.
> >>>
> >>>>
> >>>> Or in the Theorem, (That there doesn't exist an answer to the
> >>>> Halting Problem), THERE IS NONE.
> >>>
> >>> The invalid infinite recursion is in the THEOREM as stated in
> >>> [Wikipedia, 2022].
> >>>
> >>>>
> >>>> In fact, you could say that "infinite recursion" is one of the
> >>>> things that can be used to PROVE the Theorem, as any answer (by
> >>>> simulation) to the problem would need to invoke an infinte
> >>>> recursion to answer it. Thus the ANSWER (which is what is trying
> >>>> to do the infinite recursion) and not the problem (which has no
> >>>> recursion) is invalid.
> >>>
> >>> However the infinite recursion is a category error in this case so
> >>> cannot prove anything.
> >>>
> >>> /Flibble
> >>>
> >>
> >> But what HAS the infinite recursion?
> >>
> >> Not the PROBLEM.
> >>
> >> Not the Theorem.
> >>
> >> Not H^
> >>
> >> The only thing that comes close is H, so that means that the H
> >> doesn't exist, and thus the Problem is intact and the Theorem
> >> proved.
> >
> > [Strachey, 1965] simplifies the theorem making the infinite
> > recursion obvious; see also [Wikipedia, 2022].
> >
> > /Flibble
> >
>
> The THEOREM is just that there exist no H that statisfies that
> Halting Problem, so that is NOT recursive at all. And has nothing to
> "Simplify"
>
> You seem to have a category error in distinguishing between the
> Theorem, and a proof of it.
>
> So again, WHAT has an invalid infinite recursion.
>
> Not the PROBLEM
>
> Not the THEOREM
>
> Not even the H^ that proves the THEOREM.
>
> The only thing that gets into an infinite recursion is the possible
> answer, which by your logic is enough to prove it doesn't exist, and
> thus the Theorem is proved.

I see that you continue to want to play word games. Oh well. You stated
in an earlier reply that paragraph 2 of [Wikipedia, 2022) was a summary
of the proof and this does indeed contain the erroneous infinite
recursion, erroneous due to a category error. I will make another
post (reprise #3) for your benefit.

/Flibble

On 5/5/2022 11:47 AM, Mr Flibble wrote:
> On Wed, 4 May 2022 20:41:51 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 5/4/22 8:24 PM, Mr Flibble wrote:
>>> On Wed, 4 May 2022 20:21:04 -0400
>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>
>>>> On 5/4/22 8:05 PM, Mr Flibble wrote:
>>>>> On Wed, 4 May 2022 20:02:57 -0400
>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>
>>>>>> On 5/4/22 7:51 PM, Mr Flibble wrote:
>>>>>>> On Wed, 4 May 2022 19:42:19 -0400
>>>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>>>
>>>>>>>> On 5/4/22 12:46 PM, Mr Flibble wrote:
>>>>>>>>> The halting problem theorem and proof thereof [Turing, 1937]
>>>>>>>>> (upon which other currently extant halting problem proofs are
>>>>>>>>> derived) is invalid due to an invalid "impossible program"
>>>>>>>>> [Strachey, 1965] that arises not from a function call-like
>>>>>>>>> infinite recursion but from a category error in the form of an
>>>>>>>>> invalid (erroneous) infinite recursion present in the proof
>>>>>>>>> [Wikipedia, 2022].
>>>>>>>>>
>>>>>>>>> The categories involved in the category error are the decider
>>>>>>>>> and that which is being decided. Currently extant attempts to
>>>>>>>>> conflate the decider with that which is being decided are
>>>>>>>>> infinitely recursive and thus invalid.
>>>>>>>>>
>>>>>>>>> /Flibble
>>>>>>>>>
>>>>>>>>
>>>>>>>> And what is the error between the decider and the decided.
>>>>>>>>
>>>>>>>> The Decider is H.
>>>>>>>>
>>>>>>>> The thing to be decideer is H^ applied to <H^>
>>>>>>>>
>>>>>>>> Now, if you want to claim that H^ can't use H, then you are
>>>>>>>> saying either that Turing Macines aren't allowed to use other
>>>>>>>> Turing Machines, which is crasy, or that H just can't be asked
>>>>>>>> about machines which are based on it, at which point that is
>>>>>>>> just admitting that the Halting Problem is in fact impossible,
>>>>>>>> as there exist some VALID programs (which H^ is) that H just
>>>>>>>> can't be asked to decide on.
>>>>>>>>
>>>>>>>> Note, H^ is NOT recursive in definition (unless H is). H^ is
>>>>>>>> basically jast a "call" to H with a little bit of simple logic
>>>>>>>> around it.
>>>>>>>>
>>>>>>>> The "recursion" (which isn't actually recursion) comes about
>>>>>>>> because we give H^ and input that just happens to be a
>>>>>>>> representation of itself, and H^ doesn't actually know that.
>>>>>>>>
>>>>>>>> If H actually meets the requirements of being a decider (and
>>>>>>>> thus also of being a Comptation), then we get no infinite
>>>>>>>> "recursion", as the H inside H^ will, by necessity, return an
>>>>>>>> answer after finite time, and H^ will either Halt or go into a
>>>>>>>> simple infinite loop.
>>>>>>>>
>>>>>>>> H is thus just proved to give the wrong answer.
>>>>>>>>
>>>>>>>> If H refuses to give the wrong answer, then it turns out to not
>>>>>>>> be a decider and just loops forever with an ever growing tape
>>>>>>>> as the level of simulation just keeps increasing.
>>>>>>>>
>>>>>>>> Note, we NEVER get back to H^.q0 or any of the states of H^ up
>>>>>>>> to H^.qx, as the entire execution trace is within its copy of H
>>>>>>>> doing its simulation to try to decide. There is no actual
>>>>>>>> Recursion.
>>>>>>>>
>>>>>>>>
>>>>>>>> Yes, in trying to decide how we might program H to try to get
>>>>>>>> the right answer, we can think about levels of recursion, but
>>>>>>>> those are not actually in the real trace of the execution of
>>>>>>>> H^. Once we actual make a decision on how we are going to
>>>>>>>> attempt to make an H, and establish a fixed algorithm to
>>>>>>>> program into H, we find that we have failed to make a correct
>>>>>>>> decider, because, as the theorem says, it IS impossible. This
>>>>>>>> doesn't mean the problem is "invalid" just impossible.
>>>>>>>>
>>>>>>>> Just like the problem of creating a program to always win at
>>>>>>>> Tic Tac Toe. The problem is well defined, just impossible due
>>>>>>>> to the nature of the game.
>>>>>>>
>>>>>>> Nope. The infinite recursion is a category error and therefore
>>>>>>> invalid: you and the rest of you shower appear to have a
>>>>>>> blindspot to this fact.
>>>>>>>
>>>>>>> /Flibble
>>>>>>>
>>>>>>
>>>>>> And what is the infinite recursion in the HALTING PROBLEM?
>>>>>>
>>>>>> There is none.
>>>>>
>>>>> I know, which is why I explicitly said halting problem THEOREM in
>>>>> this post.
>>>>>
>>>>>>
>>>>>> Or in the Theorem, (That there doesn't exist an answer to the
>>>>>> Halting Problem), THERE IS NONE.
>>>>>
>>>>> The invalid infinite recursion is in the THEOREM as stated in
>>>>> [Wikipedia, 2022].
>>>>>
>>>>>>
>>>>>> In fact, you could say that "infinite recursion" is one of the
>>>>>> things that can be used to PROVE the Theorem, as any answer (by
>>>>>> simulation) to the problem would need to invoke an infinte
>>>>>> recursion to answer it. Thus the ANSWER (which is what is trying
>>>>>> to do the infinite recursion) and not the problem (which has no
>>>>>> recursion) is invalid.
>>>>>
>>>>> However the infinite recursion is a category error in this case so
>>>>> cannot prove anything.
>>>>>
>>>>> /Flibble
>>>>>
>>>>
>>>> But what HAS the infinite recursion?
>>>>
>>>> Not the PROBLEM.
>>>>
>>>> Not the Theorem.
>>>>
>>>> Not H^
>>>>
>>>> The only thing that comes close is H, so that means that the H
>>>> doesn't exist, and thus the Problem is intact and the Theorem
>>>> proved.
>>>
>>> [Strachey, 1965] simplifies the theorem making the infinite
>>> recursion obvious; see also [Wikipedia, 2022].
>>>
>>> /Flibble
>>>
>>
>> The THEOREM is just that there exist no H that statisfies that
>> Halting Problem, so that is NOT recursive at all. And has nothing to
>> "Simplify"
>>
>> You seem to have a category error in distinguishing between the
>> Theorem, and a proof of it.
>>
>> So again, WHAT has an invalid infinite recursion.
>>
>> Not the PROBLEM
>>
>> Not the THEOREM
>>
>> Not even the H^ that proves the THEOREM.
>>
>> The only thing that gets into an infinite recursion is the possible
>> answer, which by your logic is enough to prove it doesn't exist, and
>> thus the Theorem is proved.
>
> I see that you continue to want to play word games. Oh well. You stated
> in an earlier reply that paragraph 2 of [Wikipedia, 2022) was a summary
> of the proof and this does indeed contain the erroneous infinite
> recursion, erroneous due to a category error. I will make another
> post (reprise #3) for your benefit.
>
> /Flibble
>

On 5/5/22 12:47 PM, Mr Flibble wrote:
> On Wed, 4 May 2022 20:41:51 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 5/4/22 8:24 PM, Mr Flibble wrote:
>>> On Wed, 4 May 2022 20:21:04 -0400
>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>
>>>> On 5/4/22 8:05 PM, Mr Flibble wrote:
>>>>> On Wed, 4 May 2022 20:02:57 -0400
>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>
>>>>>> On 5/4/22 7:51 PM, Mr Flibble wrote:
>>>>>>> On Wed, 4 May 2022 19:42:19 -0400
>>>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>>>
>>>>>>>> On 5/4/22 12:46 PM, Mr Flibble wrote:
>>>>>>>>> The halting problem theorem and proof thereof [Turing, 1937]
>>>>>>>>> (upon which other currently extant halting problem proofs are
>>>>>>>>> derived) is invalid due to an invalid "impossible program"
>>>>>>>>> [Strachey, 1965] that arises not from a function call-like
>>>>>>>>> infinite recursion but from a category error in the form of an
>>>>>>>>> invalid (erroneous) infinite recursion present in the proof
>>>>>>>>> [Wikipedia, 2022].
>>>>>>>>>
>>>>>>>>> The categories involved in the category error are the decider
>>>>>>>>> and that which is being decided. Currently extant attempts to
>>>>>>>>> conflate the decider with that which is being decided are
>>>>>>>>> infinitely recursive and thus invalid.
>>>>>>>>>
>>>>>>>>> /Flibble
>>>>>>>>>
>>>>>>>>
>>>>>>>> And what is the error between the decider and the decided.
>>>>>>>>
>>>>>>>> The Decider is H.
>>>>>>>>
>>>>>>>> The thing to be decideer is H^ applied to <H^>
>>>>>>>>
>>>>>>>> Now, if you want to claim that H^ can't use H, then you are
>>>>>>>> saying either that Turing Macines aren't allowed to use other
>>>>>>>> Turing Machines, which is crasy, or that H just can't be asked
>>>>>>>> about machines which are based on it, at which point that is
>>>>>>>> just admitting that the Halting Problem is in fact impossible,
>>>>>>>> as there exist some VALID programs (which H^ is) that H just
>>>>>>>> can't be asked to decide on.
>>>>>>>>
>>>>>>>> Note, H^ is NOT recursive in definition (unless H is). H^ is
>>>>>>>> basically jast a "call" to H with a little bit of simple logic
>>>>>>>> around it.
>>>>>>>>
>>>>>>>> The "recursion" (which isn't actually recursion) comes about
>>>>>>>> because we give H^ and input that just happens to be a
>>>>>>>> representation of itself, and H^ doesn't actually know that.
>>>>>>>>
>>>>>>>> If H actually meets the requirements of being a decider (and
>>>>>>>> thus also of being a Comptation), then we get no infinite
>>>>>>>> "recursion", as the H inside H^ will, by necessity, return an
>>>>>>>> answer after finite time, and H^ will either Halt or go into a
>>>>>>>> simple infinite loop.
>>>>>>>>
>>>>>>>> H is thus just proved to give the wrong answer.
>>>>>>>>
>>>>>>>> If H refuses to give the wrong answer, then it turns out to not
>>>>>>>> be a decider and just loops forever with an ever growing tape
>>>>>>>> as the level of simulation just keeps increasing.
>>>>>>>>
>>>>>>>> Note, we NEVER get back to H^.q0 or any of the states of H^ up
>>>>>>>> to H^.qx, as the entire execution trace is within its copy of H
>>>>>>>> doing its simulation to try to decide. There is no actual
>>>>>>>> Recursion.
>>>>>>>>
>>>>>>>>
>>>>>>>> Yes, in trying to decide how we might program H to try to get
>>>>>>>> the right answer, we can think about levels of recursion, but
>>>>>>>> those are not actually in the real trace of the execution of
>>>>>>>> H^. Once we actual make a decision on how we are going to
>>>>>>>> attempt to make an H, and establish a fixed algorithm to
>>>>>>>> program into H, we find that we have failed to make a correct
>>>>>>>> decider, because, as the theorem says, it IS impossible. This
>>>>>>>> doesn't mean the problem is "invalid" just impossible.
>>>>>>>>
>>>>>>>> Just like the problem of creating a program to always win at
>>>>>>>> Tic Tac Toe. The problem is well defined, just impossible due
>>>>>>>> to the nature of the game.
>>>>>>>
>>>>>>> Nope. The infinite recursion is a category error and therefore
>>>>>>> invalid: you and the rest of you shower appear to have a
>>>>>>> blindspot to this fact.
>>>>>>>
>>>>>>> /Flibble
>>>>>>>
>>>>>>
>>>>>> And what is the infinite recursion in the HALTING PROBLEM?
>>>>>>
>>>>>> There is none.
>>>>>
>>>>> I know, which is why I explicitly said halting problem THEOREM in
>>>>> this post.
>>>>>
>>>>>>
>>>>>> Or in the Theorem, (That there doesn't exist an answer to the
>>>>>> Halting Problem), THERE IS NONE.
>>>>>
>>>>> The invalid infinite recursion is in the THEOREM as stated in
>>>>> [Wikipedia, 2022].
>>>>>
>>>>>>
>>>>>> In fact, you could say that "infinite recursion" is one of the
>>>>>> things that can be used to PROVE the Theorem, as any answer (by
>>>>>> simulation) to the problem would need to invoke an infinte
>>>>>> recursion to answer it. Thus the ANSWER (which is what is trying
>>>>>> to do the infinite recursion) and not the problem (which has no
>>>>>> recursion) is invalid.
>>>>>
>>>>> However the infinite recursion is a category error in this case so
>>>>> cannot prove anything.
>>>>>
>>>>> /Flibble
>>>>>
>>>>
>>>> But what HAS the infinite recursion?
>>>>
>>>> Not the PROBLEM.
>>>>
>>>> Not the Theorem.
>>>>
>>>> Not H^
>>>>
>>>> The only thing that comes close is H, so that means that the H
>>>> doesn't exist, and thus the Problem is intact and the Theorem
>>>> proved.
>>>
>>> [Strachey, 1965] simplifies the theorem making the infinite
>>> recursion obvious; see also [Wikipedia, 2022].
>>>
>>> /Flibble
>>>
>>
>> The THEOREM is just that there exist no H that statisfies that
>> Halting Problem, so that is NOT recursive at all. And has nothing to
>> "Simplify"
>>
>> You seem to have a category error in distinguishing between the
>> Theorem, and a proof of it.
>>
>> So again, WHAT has an invalid infinite recursion.
>>
>> Not the PROBLEM
>>
>> Not the THEOREM
>>
>> Not even the H^ that proves the THEOREM.
>>
>> The only thing that gets into an infinite recursion is the possible
>> answer, which by your logic is enough to prove it doesn't exist, and
>> thus the Theorem is proved.
>
> I see that you continue to want to play word games. Oh well. You stated
> in an earlier reply that paragraph 2 of [Wikipedia, 2022) was a summary
> of the proof and this does indeed contain the erroneous infinite
> recursion, erroneous due to a category error. I will make another
> post (reprise #3) for your benefit.
>
> /Flibble
>