On recursion and infinite recursion (reprise)

Mr Flibble

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May 2, 2022, 11:47:35 AMMay 2

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Not all infinitely recursive definitions are invalid however infinitely
recursive definitions that arise out of a category error (as is the
case with the halting problem) are invalid.

The halting problem (as currently defined) is invalid due to the
invalid "impossible program" [Strachey, 1965] that is actually
impossible due to the category error present in its definition and
*not* because of any function call-like recursion; confusion between
these two types of recursion are why Olcott is having difficulty
communicating his ideas with the rest of you shower.

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

olcott

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May 2, 2022, 12:18:40 PMMay 2

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On 5/2/2022 10:47 AM, Mr Flibble wrote:
> Not all infinitely recursive definitions are invalid however infinitely
> recursive definitions that arise out of a category error (as is the
> case with the halting problem) are invalid.
>

It seems to me that all infinitely recursive definitions are invalid and
I am having an excellent dialogue with some Prolog folks about this in
comp.lang.prolog.

> The halting problem (as currently defined) is invalid due to the
> invalid "impossible program" [Strachey, 1965] that is actually
> impossible due to the category error present in its definition and
> *not* because of any function call-like recursion; confusion between
> these two types of recursion are why Olcott is having difficulty
> communicating his ideas with the rest of you shower.
>

I created the x86 operating system taking at least a man-year so that I
could encode the HP counter-example P in C/x86 and then create a halt
decider H in C that examines the behavior of its correct simulation of P.

void P(u32 x)
{
if (H(x, x))
HERE: goto HERE;
return;
}

int main()
{
Output("Input_Halts = ", H((u32)P, (u32)P));
}

_P()
[000009d6](01) 55 push ebp
[000009d7](02) 8bec mov ebp,esp
[000009d9](03) 8b4508 mov eax,[ebp+08]
[000009dc](01) 50 push eax // push P
[000009dd](03) 8b4d08 mov ecx,[ebp+08]
[000009e0](01) 51 push ecx // push P
[000009e1](05) e840feffff call 00000826 // call H
[000009e6](03) 83c408 add esp,+08
[000009e9](02) 85c0 test eax,eax
[000009eb](02) 7402 jz 000009ef
[000009ed](02) ebfe jmp 000009ed
[000009ef](01) 5d pop ebp
[000009f0](01) c3 ret // Final state
Size in bytes:(0027) [000009f0]

machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
...[000009d6][00211368][0021136c] 55 push ebp // enter P
...[000009d7][00211368][0021136c] 8bec mov ebp,esp
...[000009d9][00211368][0021136c] 8b4508 mov eax,[ebp+08]
...[000009dc][00211364][000009d6] 50 push eax // Push P
...[000009dd][00211364][000009d6] 8b4d08 mov ecx,[ebp+08]
...[000009e0][00211360][000009d6] 51 push ecx // Push P
...[000009e1][0021135c][000009e6] e840feffff call 00000826 // Call H
...[000009d6][0025bd90][0025bd94] 55 push ebp // enter P
...[000009d7][0025bd90][0025bd94] 8bec mov ebp,esp
...[000009d9][0025bd90][0025bd94] 8b4508 mov eax,[ebp+08]
...[000009dc][0025bd8c][000009d6] 50 push eax // Push P
...[000009dd][0025bd8c][000009d6] 8b4d08 mov ecx,[ebp+08]
...[000009e0][0025bd88][000009d6] 51 push ecx // Push P
...[000009e1][0025bd84][000009e6] e840feffff call 00000826 // Call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

It is clear that the input to H(P,P) specifies infinitely nested
simulation to H.

> 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
>

I already applied your idea of category error to Gödel's Incompleteness
and Tarski Undefinability. In both of these cases the Gödel H and the
Tarski p are incorrectly placed in the category of truth bearer / logic
sentence. They cannot be correctly evaluated because they are
semantically incorrect.

Now in comp.lang.prolog we are getting Prolog to agree that they are
semantically ill-formed.

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

Ben

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May 2, 2022, 12:39:13 PMMay 2

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olcott <polc...@gmail.com> writes:

> It is clear that the input to H(P,P) specifies infinitely nested
> simulation to H.

What two pointers must be passed to H for H to tell up about the halting
of P(P)? If H can't report on the halting of the computation P(P) it is
not a halt decider, and you have already told use that H(P,P) == false
and that P(P) halts.

You have nowhere to go with H, hence your switching topics to being
wrong about Gödel again.

--
Ben.
"le génie humain a des limites, quand la bêtise humaine n’en a pas"
Alexandre Dumas (fils)

olcott

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May 2, 2022, 4:28:12 PMMay 2

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On 5/2/2022 11:39 AM, Ben wrote:
> olcott <polc...@gmail.com> writes:
>
>> It is clear that the input to H(P,P) specifies infinitely nested
>> simulation to H.
>
> What two pointers must be passed to H for H to tell up about the halting
> of P(P)? If H can't report on the halting of the computation P(P) it is
> not a halt decider, and you have already told use that H(P,P) == false
> and that P(P) halts.

If H can report on the halting of non-input P(P) then it is not a
decider because deciders only compute the mapping from inputs to final
states.

>
> You have nowhere to go with H, hence your switching topics to being
> wrong about Gödel again.
>

That you expect a halt decider to compute the mapping from non-inputs is
a little nuts when you know that deciders can't possibly do this.

It turns out that I can create a whole TM interpreter from scratch
quicker than I can learn the extraneous complexity of the TM Interpreter
http://www.lns.mit.edu/~dsw/turing/turing.html

It will have the same eight-bit quintuples of the TM, The Turing Machine
Interpreter. These will be at the beginning of the file, followed by
BEGIN_TAPE on a line by itself. Followed by the initial tape as
contiguous ASCII characters.

It will output a full execution trace. The only parameter to the command
line system with be filename.tm. This system is easily extended to
16-bits or 32-bits. These larger systems will use 4-8 hex digits for
state / character representation.

Richard Damon

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May 2, 2022, 6:32:16 PMMay 2

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Except that the "impossible program" isn't part of the definition of the
Halting Problem.

By your definition, I suspect much of Mathematics becomes a category error.

If you are willing to say that this is a natural result of your logic
system, fine, but just remember that consequence, which actually says
your logic doesn't refute Gobel, as the Theorm specifically requires
working in a field that supports some minimums that your logic doesn't
handle.

Mr Flibble

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May 2, 2022, 6:38:14 PMMay 2

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On Mon, 2 May 2022 18:32:16 -0400
Richard Damon <Ric...@Damon-Family.org> wrote:

> On 5/2/22 11:47 AM, Mr Flibble wrote:
> > Not all infinitely recursive definitions are invalid however
> > infinitely recursive definitions that arise out of a category error
> > (as is the case with the halting problem) are invalid.
> >
> > The halting problem (as currently defined) is invalid due to the
> > invalid "impossible program" [Strachey, 1965] that is actually
> > impossible due to the category error present in its definition and
> > *not* because of any function call-like recursion; confusion between
> > these two types of recursion are why Olcott is having difficulty
> > communicating his ideas with the rest of you shower.
> >
> > 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
> >
>
> Except that the "impossible program" isn't part of the definition of
> the Halting Problem.

It is according to [Wikipedia, 2022].

/Flibble

Richard Damon

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May 2, 2022, 6:46:01 PMMay 2

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Nope, you comprehend worse that PO.

Note, and Encyclopedic entery, like Wikipedia, is NOT just a definition
but a full article explaining the subject.

Maybe if you look for a FORMAL source, that states what is the ACTUAL
definition, you would learn something.

Mr Flibble

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May 2, 2022, 6:47:15 PMMay 2

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On Mon, 2 May 2022 18:46:00 -0400

If Wikipedia is wrong then correct it and have your corrections
reviewed; until then please shut the fuck up.

/Flibble

Ben

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May 2, 2022, 7:10:49 PMMay 2

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olcott <polc...@gmail.com> writes:

> On 5/2/2022 11:39 AM, Ben wrote:
>> olcott <polc...@gmail.com> writes:
>>
>>> It is clear that the input to H(P,P) specifies infinitely nested
>>> simulation to H.
>> What two pointers must be passed to H for H to tell up about the halting
>> of P(P)? If H can't report on the halting of the computation P(P) it is
>> not a halt decider, and you have already told use that H(P,P) == false
>> and that P(P) halts.
>
> If H can report on the halting of non-input P(P) then it is not a
> decider because deciders only compute the mapping from inputs to final
> states.

TM deciders compute mappings from inputs to final states /according to
some property of the inputs/ -- whether the input represents, for
example, an even number, a prime number or a halting computation.

According to you there is no "input" (in reality a pair of pointers)
that represents the halting computation P(P). Why should anyone care
about this H if it does not decide what we want -- the halting of the
function call represented by the two arguments to H? Whatever H is
actually deciding is not interesting.

Also, I wonder why you wasted so much time justifying the fact that
H(P,P) == false "even though P(P) halts" when H(P,P) is, apparently, not
even supposed to be deciding the halting P(P). Well, we know, of
course. You realised you were in a hole so you started to dig sideways.
You used to know that H(X,Y) had to decide the halting of X(Y). You're
now pretending it never did!

> That you expect a halt decider to compute the mapping from non-inputs
> is a little nuts when you know that deciders can't possibly do this.

Don't be silly. They decide properties of inputs -- parity, halting and
so on. You'd know this if you'd done even the warm-up exercises I set.
How are they coming along? It looks like you have found an excuse to
bail out again:

> It turns out that I can create a whole TM interpreter from scratch
> quicker than I can learn the extraneous complexity of the TM
> Interpreter http://www.lns.mit.edu/~dsw/turing/turing.html

I doubt it. But I suppose you think that's a reasonable excuse. Of
course, some of us remember you saying writing such a thing would take
about a week three years ago. I remember wondering how such a simple
program could take you a week to write.

Of course you don't need an interpreter to write E or specify P, but you
must find some excuse for bailing out.

Richard Damon

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May 2, 2022, 7:16:05 PMMay 2

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It isn't that the article is "Wrong", it is a fairly good Encyclpedic
article. It just is that the first two paragraphs aren't all a
definition, and it doesn't say they are.

Mr Flibble

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May 2, 2022, 7:30:46 PMMay 2

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On Mon, 2 May 2022 19:16:03 -0400

The first two paragraphs define the halting problem as that is what the
currently extant halting problem "proofs" are predicated on (and why
they are invalid).

/Flibble

olcott

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May 2, 2022, 8:35:42 PMMay 2

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I think that the problem is that Richard has disagreeably as his highest
priority, thus doesn't really give a rat's ass for the truth. An

An impossible program C. Strachey
The Computer Journal, Volume 7, Issue 4, January 1965, Page 313,
Published: 01 January 1965
https://academic.oup.com/comjnl/article/7/4/313/354243

It is very common knowledge that the Wikipedia description is true and
this is affirmed in Sipser.

For any program f that might determine if programs halt, a
"pathological" program g, called with some input, can pass its own
source and its input to f and then specifically do the opposite of what
f predicts g will do. https://en.wikipedia.org/wiki/Halting_problem

Now we construct a new Turing machine D with H as a subroutine. This new
TM calls H to determine what M does when the input to M is its own
description ⟨M⟩. Once D has determined this information, it does the
opposite. https://www.liarparadox.org/Sipser_165_167.pdf

Richard Damon

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May 2, 2022, 8:40:16 PMMay 2

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No, lets actually look at what is says, and parse it:

In computability theory, the halting problem is the problem of
determining, from a description of an arbitrary computer program and an
input, whether the program will finish running, or continue to run
forever. Alan Turing proved in 1936 that a general algorithm to solve
the halting problem for all possible program-input pairs cannot exist.

For any program f that might determine if programs halt, a
"pathological" program g, called with some input, can pass its own
source and its input to f and then specifically do the opposite of what

f predicts g will do. No f can exist that handles this case. A key part
of the proof is a mathematical definition of a computer and program,
which is known as a Turing machine; the halting problem is undecidable
over Turing machines. It is one of the first cases of decision problems
proven to be unsolvable. This proof is significant to practical
computing efforts, defining a class of applications which no programming
invention can possibly perform perfectly.

Jack Copeland attributes the introduction of the term halting problem to
the work of Martin Davis in the 1950s.[1]

The FIRST SENTENCE is the definition of the Problem.

The Second Sentence is the Theorem about it that says that no solution
exists.

That ends the first paragraph.

The Second Paragraph, is a continuation of the idea of the Second
Sentence, giving a summary of the proof that no solution exist.

It is a category error to confuse the Statement of the Problem with the
Proof of the Theorem that not answer to the Problem exists.

A Proof is NOT a Problem.

Richard Damon

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May 2, 2022, 8:48:57 PMMay 2

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Thus you have shown you don't even know what a "Definition" is, so it is
impossible for you to reason by the meaning of the words.

You have just proved yourself to be an IDIOT.

Mikko

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May 3, 2022, 5:31:47 AMMay 3

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On 2022-05-02 15:47:32 +0000, Mr Flibble said:

> Not all infinitely recursive definitions are invalid however infinitely
> recursive definitions that arise out of a category error (as is the
> case with the halting problem) are invalid.

An infinite recursion cannot arise out of a category error as the recursion
stops at the category error.

Mikko

Mikko

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May 3, 2022, 5:36:48 AMMay 3

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On 2022-05-02 16:18:36 +0000, olcott said:

> It seems to me that all infinitely recursive definitions are invalid
> and I am having an excellent dialogue with some Prolog folks about this
> in comp.lang.prolog.

One of the rules that define Prolog language is

arguments ::= argument | argument "," arguments

which is infinitely recursive. Is it invalid? Is Prolog invalid because
of this and other infinitely recursive rules?

Mikko

Malcolm McLean

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May 3, 2022, 7:08:12 AMMay 3

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Kind of.
A Prolog program is a physical object, not a mathematical object, so
the recursion has to terminate somewhere.
But it might lead you into strange territory if you tried to define the result
of passing an ininite argument list to some Prolog.

wij wij

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May 3, 2022, 8:12:15 AMMay 3

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richar...@gmail.com 在 2022年5月3日星期二上午8:48:57 [UTC+8] 的信中寫道：

PO is incapable of logic reasoning (PO had shown he cannot even get the truth
table of logical implication/AND right). All he said is delusion including when
words from him happen to be correct to others (no real meaning).

IIRC, PO's revision that H(P,P) has no relation with P(P) is deliberately
fabricated this recent year after PO ran out his reasons to explain why HP is
wrong and he is correct. PO has no trouble to 'lie' to his bible (he can read
it his way), the HP thing is just piece of cake.

olcott

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May 3, 2022, 10:31:21 AMMay 3

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It is an easily verified fact that P(P) and the correct simulation of
the input to H(P,P) specify different sequences of configurations, thus
have different halting behavior. That several people here deny easily
verified facts is a little psychotic on their part.

olcott

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May 3, 2022, 10:33:45 AMMay 3

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Even infinitely recursive math expressions are semantically incorrect in
that they can never be evaluated.

olcott

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May 3, 2022, 10:39:02 AMMay 3

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If would have to be invalid because it can never be resolved.

olcott

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May 3, 2022, 10:42:37 AMMay 3

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The category error is that an expression of language X is construed as a
logic sentence / truth bearer that is true or false. It is because of
the infinitely recursive definition that X is neither of these.

https://en.wikipedia.org/wiki/Sentence_(mathematical_logic)#:~:text=In%20mathematical%20logic%2C%20a%20sentence,must%20be%20true%20or%20false.

Dennis Bush

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May 3, 2022, 10:47:32 AMMay 3

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On Tuesday, May 3, 2022 at 10:31:21 AM UTC-4, olcott wrote:
> On 5/3/2022 7:12 AM, wij wij wrote:

> > Richard Damon 在 2022年5月3日星期二上午8:48:57 [UTC+8] 的信中寫道：

The easily verified fact is that the correct simulation to H(P,P) is performed by Hb(P,P) (which simulates for k more steps than H) which remains in UTM mode while simulating the same input to a final state.

Because H and Hb and both simulating halt deciders and are given the same input, they are deciding on the same sequence of configurations (namely starting with the first instruction of P). Because one answers false and one answers true, one must be wrong.

Since a simulating halt decider that simulates its input to a final state while remaining in UTM mode is necessarily correct, this proves that Hb(P,P) == true is correct and that H(P,P) == false is incorrect, and that H(P,P) does *not* in fact perform a correct simulation of its input because it aborts too soon.

You've been asked several times what input must be given to H to determine if P(P) halts. It turns out that the input (P,P) can be given to Hb to determine exactly that, so the fact that H can't give the same result for the same input just shows that it is wrong and that the halting problem is unsolvable as the existing proofs show.

> That several people here deny easily
> verified facts is a little psychotic on their part.

You're projecting. Again. In fact you're *so* good a projecting that if you opened a movie theater I'll bet the picture quality would be second to none.

olcott

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May 3, 2022, 11:19:33 AMMay 3

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I have no idea what you mean.

> Because H and Hb and both simulating halt deciders and are given the same input, they are deciding on the same sequence of configurations (namely starting with the first instruction of P). Because one answers false and one answers true, one must be wrong.
>

It is ridiculously stupid to assume that an input having pathological
self-reference to its decider would have the same behavior as an input
NOT having pathological to its decider.

> Since a simulating halt decider that simulates its input to a final state while remaining in UTM mode is necessarily correct, this proves that Hb(P,P) == true is correct and that H(P,P) == false is incorrect, and that H(P,P) does *not* in fact perform a correct simulation of its input because it aborts too soon.
>

It is very easy to verify the fact that the simulated input to H(P,P)
would never stop unless aborted. It is pretty psychotic that many of my
reviewers deny easily verified facts.

> You've been asked several times what input must be given to H to determine if P(P) halts. It turns out that the input (P,P) can be given to Hb to determine exactly that, so the fact that H can't give the same result for the same input just shows that it is wrong and that the halting problem is unsolvable as the existing proofs show.
>

It is ridiculously stupid to assume that an input having pathological
self-reference to its decider would have the same behavior as an input
NOT having pathological to its decider.

It is an easily verified fact that H does correctly reject its input and
that deciders only compute the mapping from their inputs.

>> That several people here deny easily
>> verified facts is a little psychotic on their part.
>
> You're projecting. Again. In fact you're *so* good a projecting that if you opened a movie theater I'll bet the picture quality would be second to none.

Anyone that denies easily verified facts has (by definition) a break
from reality.

Dennis Bush

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May 3, 2022, 11:36:07 AMMay 3

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In other words you don't want to admit that this proves you are wrong.

> > Because H and Hb and both simulating halt deciders and are given the same input, they are deciding on the same sequence of configurations (namely starting with the first instruction of P). Because one answers false and one answers true, one must be wrong.
> >
> It is ridiculously stupid to assume that an input having pathological
> self-reference to its decider would have the same behavior as an input
> NOT having pathological to its decider.

Which is another way of saying that H can't give a correct answer for (P,P).

> > Since a simulating halt decider that simulates its input to a final state while remaining in UTM mode is necessarily correct, this proves that Hb(P,P) == true is correct and that H(P,P) == false is incorrect, and that H(P,P) does *not* in fact perform a correct simulation of its input because it aborts too soon.
> >
> It is very easy to verify the fact that the simulated input to H(P,P)
> would never stop unless aborted. It is pretty psychotic that many of my
> reviewers deny easily verified facts.

There is no "unless". The fixed algorithm of H, which will henceforth be referred to as Ha and similarly P will be referred to as Pa, *does* abort. Because of this, Hb(Pa,Pa) explicitly shows that the simulated input to Ha(Pa,Pa) *does* stop. The fact that Pn(Pn) does not halt and that Hn(Pn,Pn) does not halt is irrelevant.

> > You've been asked several times what input must be given to H to determine if P(P) halts. It turns out that the input (P,P) can be given to Hb to determine exactly that, so the fact that H can't give the same result for the same input just shows that it is wrong and that the halting problem is unsolvable as the existing proofs show.
> >
> It is ridiculously stupid to assume that an input having pathological
> self-reference to its decider would have the same behavior as an input
> NOT having pathological to its decider.

Which is another way of saying that Ha can't give a correct answer for (Pa,Pa).

>
> It is an easily verified fact that H does correctly reject its input

Ha does not correctly reject its input as easily verified by Hb.

> and that deciders only compute the mapping from their inputs.

And all halt deciders must compute the same mapping from the same input. Ha(Pa,Pa) and Hb(Pa,Pa) do not perform the same mapping from the same input so one must be wrong.

Since a simulating halt decider that simulates its input to a final state while remaining in UTM mode is necessarily correct, this proves that Hb(Pa,Pa) == true is correct and that Ha(Pa,Pa) == false is incorrect

> >> That several people here deny easily
> >> verified facts is a little psychotic on their part.
> >
> > You're projecting. Again. In fact you're *so* good a projecting that if you opened a movie theater I'll bet the picture quality would be second to none.
> Anyone that denies easily verified facts has (by definition) a break
> from reality.

I can't *wait* to see your movie theater. Such a expert at projection must have a great picture.

olcott

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May 3, 2022, 12:39:49 PMMay 3

to

No I can't understand what you mean.
I think that I see it now, I had forgotten the notation.

An input having a pathological self-reference relationship to its
decider H would necessarily derive a different halt status than an input
not having a pathological self-reference relationship to its decider Hb.

The P having a pathological self-reference relationship to H is not the
same as the Px NOT having a pathological self-reference relationship to
Hb. Because P.H calls itself and Px.Hb does not call itself P is not the
same input as Px.

>
>>> Because H and Hb and both simulating halt deciders and are given the same input, they are deciding on the same sequence of configurations (namely starting with the first instruction of P). Because one answers false and one answers true, one must be wrong.
>>>
>> It is ridiculously stupid to assume that an input having pathological
>> self-reference to its decider would have the same behavior as an input
>> NOT having pathological to its decider.
>
> Which is another way of saying that H can't give a correct answer for (P,P).
>

Different computations must give different answers.
That you don't fully understand all of the nuances of how this applies
to H/P and Hb/Px is OK, it is difficult to understand.

>>> Since a simulating halt decider that simulates its input to a final state while remaining in UTM mode is necessarily correct, this proves that Hb(P,P) == true is correct and that H(P,P) == false is incorrect, and that H(P,P) does *not* in fact perform a correct simulation of its input because it aborts too soon.
>>>
>> It is very easy to verify the fact that the simulated input to H(P,P)
>> would never stop unless aborted. It is pretty psychotic that many of my
>> reviewers deny easily verified facts.
>
> There is no "unless". The fixed algorithm of H, which will henceforth be referred to as Ha and similarly P will be referred to as Pa, *does* abort.

Which is *NOT* halting. A halting input must reach its own final state.

> Because of this, Hb(Pa,Pa) explicitly shows that the simulated input to Ha(Pa,Pa) *does* stop. The fact that Pn(Pn) does not halt and that Hn(Pn,Pn) does not halt is irrelevant.

It it not Hb(Pa,Pa) it is Hb(Px,Px). That P calls H makes it an entirely
different input than Px that does not call Hb.

>>> You've been asked several times what input must be given to H to determine if P(P) halts. It turns out that the input (P,P) can be given to Hb to determine exactly that, so the fact that H can't give the same result for the same input just shows that it is wrong and that the halting problem is unsolvable as the existing proofs show.
>>>
>> It is ridiculously stupid to assume that an input having pathological
>> self-reference to its decider would have the same behavior as an input
>> NOT having pathological to its decider.
>
> Which is another way of saying that Ha can't give a correct answer for (Pa,Pa).
>
>>
>> It is an easily verified fact that H does correctly reject its input
>
> Ha does not correctly reject its input as easily verified by Hb.

That P calls H makes it an entirely different input than Px that does
not call Hb.

>
>> and that deciders only compute the mapping from their inputs.
>
> And all halt deciders must compute the same mapping from the same input. Ha(Pa,Pa) and Hb(Pa,Pa) do not perform the same mapping from the same input so one must be wrong.
>

That P calls H makes it an entirely different input than Px that does
not call Hb.

> Since a simulating halt decider that simulates its input to a final state while remaining in UTM mode is necessarily correct, this proves that Hb(Pa,Pa) == true is correct and that Ha(Pa,Pa) == false is incorrect

That P calls H makes it an entirely different input than Px that does
not call Hb.

>
>>>> That several people here deny easily
>>>> verified facts is a little psychotic on their part.
>>>
>>> You're projecting. Again. In fact you're *so* good a projecting that if you opened a movie theater I'll bet the picture quality would be second to none.
>> Anyone that denies easily verified facts has (by definition) a break
>> from reality.
>
> I can't *wait* to see your movie theater. Such a expert at projection must have a great picture.

Malcolm McLean

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May 3, 2022, 3:17:24 PMMay 3

to

We have very close views on this. Infinite recursion is definitely the
key issue with the HP counter-examples and makes Gödel's G and Tarski's
p semantically incorrect.

Dennis Bush

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May 3, 2022, 5:49:47 PMMay 3

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The P we're talking about is a *specific* P, namely Pa which is built from Ha, and Ha is a *specific* H. So Pa and Px are the *same*.

So just because Pa contains an embedded copy of Ha but not an embedded copy of Hb doesn't means that it's not the same.

Ha(Pa,Pa) and Hb(Pa,Pa) have the *exact* same input.

Just because it appears from a glance that Ha is starting its simulation of Pa "in the middle" doesn't mean that's what's actually happening. That's just how the incorrect simulation is manifesting itself. It's kind of like undefined behavior in a C program.

> >
> >>> Because H and Hb and both simulating halt deciders and are given the same input, they are deciding on the same sequence of configurations (namely starting with the first instruction of P). Because one answers false and one answers true, one must be wrong.
> >>>
> >> It is ridiculously stupid to assume that an input having pathological
> >> self-reference to its decider would have the same behavior as an input
> >> NOT having pathological to its decider.
> >
> > Which is another way of saying that H can't give a correct answer for (P,P).
> >
> Different computations must give different answers.
> That you don't fully understand all of the nuances of how this applies
> to H/P and Hb/Px is OK, it is difficult to understand.

Just because Pa contains an embedded copy of Ha but not an embedded copy of Hb doesn't means that it's not the same.

> >>> Since a simulating halt decider that simulates its input to a final state while remaining in UTM mode is necessarily correct, this proves that Hb(P,P) == true is correct and that H(P,P) == false is incorrect, and that H(P,P) does *not* in fact perform a correct simulation of its input because it aborts too soon.
> >>>
> >> It is very easy to verify the fact that the simulated input to H(P,P)
> >> would never stop unless aborted. It is pretty psychotic that many of my
> >> reviewers deny easily verified facts.
> >
> > There is no "unless". The fixed algorithm of H, which will henceforth be referred to as Ha and similarly P will be referred to as Pa, *does* abort.
> Which is *NOT* halting. A halting input must reach its own final state.
> > Because of this, Hb(Pa,Pa) explicitly shows that the simulated input to Ha(Pa,Pa) *does* stop. The fact that Pn(Pn) does not halt and that Hn(Pn,Pn) does not halt is irrelevant.
> It it not Hb(Pa,Pa) it is Hb(Px,Px). That P calls H makes it an entirely
> different input than Px that does not call Hb.

No it is *exactly* Hb(Pa,Pa). The same encoding passed to Ha is passed to Hb.

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May 3, 2022, 6:11:52 PMMay 3

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It is not at all about physical representations it is about ideas that
directly contradict the verified facts, the square root of a negative
number and parallel lines that meet are both known to be non-existent on
the basis of definitions.

> All these concepts have historically caused great difficulty. Which is why it
> is almost impossible to make progress in mathematics or related disciplines
> without having a deep understanding of what has gone before. Otherwise you
> are doomed to retread old debates.

olcott

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May 3, 2022, 6:15:36 PMMay 3

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Ad Hominem attacks are the first resort of clueless wonders.
You have proven that you are not totally clueless about all of these
things, there are some of these things that you do correctly understand.

Python

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May 3, 2022, 6:21:31 PMMay 3

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Peter Olcott wrote:
> On 5/3/2022 1:59 PM, Mr Flibble wrote:
>> On Tue, 3 May 2022 11:57:39 -0500
>> olcott <polc...@gmail.com> wrote:

...

>>> I don't buy into the whole imaginary numbers game.
>>> We could imagine that 2 + 3 = 17 and call that an imaginary sum.

...

>
> Nice to know, thanks. Thus your rebuttal seems complete it is not an
> infinite anything. Imagining the square root of a negative number or
> that parallel lines meet seems a little nuts to me.

Before jumping to such outrageously uninformed conclusions you may want
to learn how complex numbers are actually defined nowadays.

It is true that, at first, it was used without any proper definition
better than "let's assume we can deal with sqrt(-1) as usual". The
surprising point at that time is it works pretty well.

*Then*, in the XIXth Century, Gallois showed how to define complex
numbers rigorously.

You've never heard of that, Peter, really?

[for the record: C is the set of equivalence classes of polynomials
on R by the relation p ~ q iff p - q = 0 [mod x^2+1], compatibility
of + and * on R and C can be proven easily, R is naturally injected
into C as a set of constant polynomials, i is the equivalence class of
the polynomial x]

Dennis Bush

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May 3, 2022, 6:21:45 PMMay 3

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Hb(Pa,Pa) proves that Ha(Pa,Pa) does NOT perform a correct simulation. Both perform the same mapping so both must answer the same.

Both start simulation at the beginning. At the point that Ha aborts, Hb does not but continues to simulate the same input and sees that it reaches a final state.

There is no infinitely nested simulation in Pa *because* Ha aborts. It *thinks* it sees infinitely nested simulation but it doesn't. While Ha(Pn,Pn) does in fact correctly detect infinitely nested simulation, that's not the case we're interested in.

Hn(Pn,Pn) is infinitely nested simulation, but this has nothing to do with Ha or Pa.

>
> How much longer are you going to continue the verified facts?
> This does make you look quite foolish or dishonest.

So when are you going to open up that movie theater? The picture would just be beautiful from such a master at projection.

Python

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May 3, 2022, 6:23:29 PMMay 3

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You cranks are really unsufferable idiots...

olcott

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May 3, 2022, 6:32:52 PMMay 3

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By looking at the actual execution trace of the simulation of Hb(Pa,Pa)
and Ha(Pa,Pa) it is easy to determine that the simulations are correct
on the basis of the x86 source code of Pa.

I will see if I can create the execution of Hb(Pa,Pa) to derive this
execution trace. We have had the one for Ha(Pa,Pa) for many months now.

It is self evident that Hb will not see itself being called in
infinitely nested simulation and the executuion trace of Ha(Pa,Pa) does
see itself called in infinitely nested simulation. This difference would
very obviously make the results of Ha and Hb differ.

olcott

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May 3, 2022, 6:41:01 PMMay 3

to

On 5/3/2022 5:21 PM, Python wrote:
> Peter Olcott wrote:
>> On 5/3/2022 1:59 PM, Mr Flibble wrote:
>>> On Tue, 3 May 2022 11:57:39 -0500
>>> olcott <polc...@gmail.com> wrote:
> ...
>>>> I don't buy into the whole imaginary numbers game.
>>>> We could imagine that 2 + 3 = 17 and call that an imaginary sum.
> ...
>>
>> Nice to know, thanks. Thus your rebuttal seems complete it is not an
>> infinite anything. Imagining the square root of a negative number or
>> that parallel lines meet seems a little nuts to me.
>
> Before jumping to such outrageously uninformed conclusions you may want
> to learn how complex numbers are actually defined nowadays.
>

They are defined to directly contradict the verified facts.
I really hate anything that directly contradicts the verified facts
because this can result in:
(a) Climate change making humans extinct quite soon,

(b) Nazi "big lie" propaganda about election fraud is making very
significant inroads to transforming Democracy ion the USA to Fascism.

(c) It directly resulted in many covid-19 deaths

> It is true that, at first, it was used without any proper definition
> better than "let's assume we can deal with sqrt(-1) as usual". The
> surprising point at that time is it works pretty well.
>

We can see what happens when we hypothesize (against the facts) that
square roots of negative numbers and parallel lines that meet exist
simply to see where this leads. I am sure that this is the intent.

> *Then*, in the XIXth Century, Gallois showed how to define complex
> numbers rigorously.
>
> You've never heard of that, Peter, really?
>
> [for the record: C is the set of equivalence classes of polynomials
> on R by the relation p ~ q iff p - q = 0 [mod x^2+1], compatibility
> of + and * on R and C can be proven easily, R is naturally injected
> into C as a set of constant polynomials, i is the equivalence class of
> the polynomial x]
>
>

Python

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May 3, 2022, 6:46:05 PMMay 3

to

Peter Olcott wrote:
> On 5/3/2022 5:21 PM, Python wrote:
>> Peter Olcott wrote:
>>> On 5/3/2022 1:59 PM, Mr Flibble wrote:
>>>> On Tue, 3 May 2022 11:57:39 -0500
>>>> olcott <polc...@gmail.com> wrote:
>> ...
>>>>> I don't buy into the whole imaginary numbers game.
>>>>> We could imagine that 2 + 3 = 17 and call that an imaginary sum.
>> ...
>>>
>>> Nice to know, thanks. Thus your rebuttal seems complete it is not an
>>> infinite anything. Imagining the square root of a negative number or
>>> that parallel lines meet seems a little nuts to me.
>>
>> Before jumping to such outrageously uninformed conclusions you may want
>> to learn how complex numbers are actually defined nowadays.
>>
>
> They are defined to directly contradict the verified facts.
> I really hate anything that directly contradicts the verified facts
> because this can result in:
> (a) Climate change making humans extinct quite soon,
>
> (b) Nazi "big lie" propaganda about election fraud is making very
> significant inroads to transforming Democracy ion the USA to Fascism.
>
> (c) It directly resulted in many covid-19 deaths

You should definitely call a doctor and ask for help, your mental
state is utterly ill. What the f* are complex numbers related to
your rants on politics?

>> It is true that, at first, it was used without any proper definition
>> better than "let's assume we can deal with sqrt(-1) as usual". The
>> surprising point at that time is it works pretty well.
>>
>
> We can see what happens when we hypothesize (against the facts) that
> square roots of negative numbers and parallel lines that meet exist
> simply to see where this leads. I am sure that this is the intent.

Not quite you're wrong. The intent was to find *real* roots of degree 3
polynomials even if square root of negative quantities appears at
intermediary steps.

You know *nothing* about history of Science, Peter. Guessing is quite
absurd when it comes to History.

>> *Then*, in the XIXth Century, Gallois showed how to define complex
>> numbers rigorously.
>>
>> You've never heard of that, Peter, really?
>>
>> [for the record: C is the set of equivalence classes of polynomials
>> on R by the relation p ~ q iff p - q = 0 [mod x^2+1], compatibility
>> of + and * on R and C can be proven easily, R is naturally injected
>> into C as a set of constant polynomials, i is the equivalence class of
>> the polynomial x]

No reaction? Well... Not a big surprise, your eyes cancel out any
sensible arguments proving you wrong, as usual.

Die in Hell, idiotic annoying crank. You deserve it.

olcott

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May 3, 2022, 6:49:43 PMMay 3

to

None-the-less as I just said this whole think relies on accepting a
known false premise.

> You know *nothing* about history of Science, Peter. Guessing is quite
> absurd when it comes to History.
>
>>> *Then*, in the XIXth Century, Gallois showed how to define complex
>>> numbers rigorously.
>>>
>>> You've never heard of that, Peter, really?
>>>
>>> [for the record: C is the set of equivalence classes of polynomials
>>> on R by the relation p ~ q iff p - q = 0 [mod x^2+1], compatibility
>>> of + and * on R and C can be proven easily, R is naturally injected
>>> into C as a set of constant polynomials, i is the equivalence class of
>>> the polynomial x]
>
> No reaction? Well... Not a big surprise, your eyes cancel out any
> sensible arguments proving you wrong, as usual.
>
> Die in Hell, idiotic annoying crank. You deserve it.
>
>
>
>

Python

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May 3, 2022, 7:05:24 PMMay 3

to

Let me guess, you are some kind of "information engineer" and consider
yourself also as "one of the greatest logicians Humanity ever had",
right? There is a demented guy of this kind on sci.physics.relativity,
his name is Maciej Wozniak. You guys should definitely mate (NOT).

>> You know *nothing* about history of Science, Peter. Guessing is quite
>> absurd when it comes to History.
>>
>>>> *Then*, in the XIXth Century, Gallois showed how to define complex
>>>> numbers rigorously.
>>>>
>>>> You've never heard of that, Peter, really?
>>>>
>>>> [for the record: C is the set of equivalence classes of polynomials
>>>> on R by the relation p ~ q iff p - q = 0 [mod x^2+1], compatibility
>>>> of + and * on R and C can be proven easily, R is naturally injected
>>>> into C as a set of constant polynomials, i is the equivalence class of
>>>> the polynomial x]
>>
>> No reaction? Well... Not a big surprise, your eyes cancel out any
>> sensible arguments proving you wrong, as usual.
>>
>> Die in Hell, idiotic annoying crank. You deserve it.

(bis)

olcott

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May 3, 2022, 7:48:07 PMMay 3

to

I consider myself to have made significant unique advancements on the
single subject on the philosophical foundation of the notion of logical
truth.

One of my key breakthroughs is redefining the analytic / synthetic
distinction such that analytic means (the same as it did) any expression
of formal or natural language that can be verified as true entirely on
the basis of its meaning. "Dogs are animals"

The somewhat vaguely defined synthetic is renamed as empirical and it is
the same sort of thing as analytic that additionally requires sense data
from the sense organs as an aspect of the truth verification process.
"There is a dog in my living room right now."

This is the most important paper on the subject
Two Dogmas of Empiricism by Willard Van Orman Quine
(Harvard University Press, 1953; second, revised, edition 1961)
https://www.theologie.uzh.ch/dam/jcr:ffffffff-fbd6-1538-0000-000070cf64bc/Quine51.pdf

Quine didn't seem to understand that bachelors are necessarily unmarried.

Meaning Postulates by RUDOLF CARNAP
https://liarparadox.org/Meaning_Postulates_Rudolf_Carnap_1952.pdf
conclusively proved the complete basis of how we know that bachelors are
necessarily unmarried. Quine didn't want to hear this because it
contradicted his paper.

>>> You know *nothing* about history of Science, Peter. Guessing is quite
>>> absurd when it comes to History.
>>>
>>>>> *Then*, in the XIXth Century, Gallois showed how to define complex
>>>>> numbers rigorously.
>>>>>
>>>>> You've never heard of that, Peter, really?
>>>>>
>>>>> [for the record: C is the set of equivalence classes of polynomials
>>>>> on R by the relation p ~ q iff p - q = 0 [mod x^2+1], compatibility
>>>>> of + and * on R and C can be proven easily, R is naturally injected
>>>>> into C as a set of constant polynomials, i is the equivalence class of
>>>>> the polynomial x]
>>>
>>> No reaction? Well... Not a big surprise, your eyes cancel out any
>>> sensible arguments proving you wrong, as usual.
>>>
>>> Die in Hell, idiotic annoying crank. You deserve it.
>
> (bis)
>
>

Richard Damon

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May 3, 2022, 9:38:04 PMMay 3

to

On 5/3/22 2:13 PM, olcott wrote:
> On 5/3/2022 12:27 PM, Mikko wrote:
>> On 2022-05-03 14:42:32 +0000, olcott said:
>>
>>> On 5/3/2022 4:31 AM, Mikko wrote:
>>>> On 2022-05-02 15:47:32 +0000, Mr Flibble said:
>>>>
>>>>> Not all infinitely recursive definitions are invalid however
>>>>> infinitely
>>>>> recursive definitions that arise out of a category error (as is the
>>>>> case with the halting problem) are invalid.
>>>>
>>>> An infinite recursion cannot arise out of a category error as the
>>>> recursion
>>>> stops at the category error.
>>>>
>>>> Mikko
>>>>
>>>
>>> The category error is that an expression of language X is construed
>>> as a logic sentence / truth bearer that is true or false. It is
>>> because of the infinitely recursive definition that X is neither of
>>> these.
>>
>> Only if the recursive expression is used as if it were a truth bearer.
>> Definitions usually don't use expression that way.
>>
>> Mikko
>>
>
> Expressions of language can only be correctly construed as true:
> (a) if they are defined to be true
> (b) have no contradictory elements in (a)
> (c) are derived by applying true preserving operations to (a) or (c)
>

So, you don't believe the Bible verse you like to Quote is actually true?

Then why do you quote it?

There is no "Definiton" that it must be true, or how do you decide WHICH
writings get that status.

And you certainly can't derive it from things actually defined to be true.

This seems to say you have a contradiction in your logic system, so I
guess that mean your logic system isn't correct.

Richard Damon

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May 3, 2022, 9:41:49 PMMay 3

to

On 5/3/22 10:42 AM, olcott wrote:
> On 5/3/2022 4:31 AM, Mikko wrote:
>> On 2022-05-02 15:47:32 +0000, Mr Flibble said:
>>
>>> Not all infinitely recursive definitions are invalid however infinitely
>>> recursive definitions that arise out of a category error (as is the
>>> case with the halting problem) are invalid.
>>
>> An infinite recursion cannot arise out of a category error as the
>> recursion
>> stops at the category error.
>>
>> Mikko
>>
>
> The category error is that an expression of language X is construed as a
> logic sentence / truth bearer that is true or false. It is because of
> the infinitely recursive definition that X is neither of these.
>

> https://en.wikipedia.org/wiki/Sentence_(mathematical_logic)#:~:text=In%20mathematical%20logic%2C%20a%20sentence,must%20be%20true%20or%20false.
>
>

Except that neither G (in F) or the Halting Problem as properly defined
has a recursive definition, so you have no grounds to call it infinitely
recursive.

Yes, G may become recursive in F', but it isn't in F.

Richard Damon

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May 3, 2022, 9:47:13 PMMay 3

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