On 5/3/2022 10:36 AM, Dennis Bush wrote:
> On Tuesday, May 3, 2022 at 11:19:33 AM UTC-4, olcott wrote:
>> On 5/3/2022 9:47 AM, Dennis Bush wrote:
>>> 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：
>>>>>> On 5/2/22 8:35 PM, olcott wrote:
>>>>>>> On 5/2/2022 5:47 PM, Mr Flibble wrote:
>>>>>>>> On Mon, 2 May 2022 18:46:00 -0400
>>>>>>>> Richard Damon wrote:
>>>>>>>>
>>>>>>>>> On 5/2/22 6:38 PM, Mr Flibble wrote:
>>>>>>>>>> On Mon, 2 May 2022 18:32:16 -0400
>>>>>>>>>> Richard Damon 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
>>>>>>>>>
>>>>>>>>> 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.
>>>>>>>>
>>>>>>>> If Wikipedia is wrong then correct it and have your corrections
>>>>>>>> reviewed; until then please shut the fuck up.
>>>>>>>>
>>>>>>>> /Flibble
>>>>>>>>
>>>>>>>
>>>>>>> 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
>>>>>>>
>>>>>>>
>>>>>> 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.
>>>>>
>>>>> 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.
>>>> 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.
>>>
>>> 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.
>>>
>> I have no idea what you mean.
>
> In other words you don't want to admit that this proves you are wrong.
>

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.

On Tuesday, 3 May 2022 at 15:33:45 UTC+1, olcott wrote:
> On 5/3/2022 6:08 AM, Malcolm McLean wrote:
> > On Tuesday, 3 May 2022 at 10:36:48 UTC+1, Mikko wrote:
> >> 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?
> >>
> > 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.
> Even infinitely recursive math expressions are semantically incorrect in
> that they can never be evaluated.
>
What's e ^ (PI * i) ?

e is Euler's number.
PI is the ratio of the diameter of a circle to its circumference
i is the square root of -1.

On 5/3/2022 11:41 AM, Malcolm McLean wrote:
> On Tuesday, 3 May 2022 at 15:33:45 UTC+1, olcott wrote:
>> On 5/3/2022 6:08 AM, Malcolm McLean wrote:
>>> On Tuesday, 3 May 2022 at 10:36:48 UTC+1, Mikko wrote:
>>>> 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?
>>>>
>>> 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.

>> Even infinitely recursive math expressions are semantically incorrect in
>> that they can never be evaluated.
>>
> What's e ^ (PI * i) ?
>
> e is Euler's number.
> PI is the ratio of the diameter of a circle to its circumference
> i is the square root of -1.
>

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

That is not an infinitely recursive math expression.
It is a math expression that can be evaluated on the basis of the
numerical constants specified by e, PI and i. That it cannot be resolved
to a finite string of digits does not make it invalid.

--
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 2022-05-03 14:38:57 +0000, olcott said:

> On 5/3/2022 4:36 AM, Mikko wrote:
>> 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
>>
>
> If would have to be invalid because it can never be resolved.

What would be invalid? Prolog? Definition of Prolog?
Why "would be" and not "is"?

Mikko

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

On 5/3/2022 12:17 PM, Mikko wrote:
> On 2022-05-03 14:38:57 +0000, olcott said:
>
>> On 5/3/2022 4:36 AM, Mikko wrote:
>>> 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
>>>
>>
>> If would have to be invalid because it can never be resolved.
>
> What would be invalid? Prolog? Definition of Prolog?
> Why "would be" and not "is"?
>
> Mikko
>

Expressions that cannot be resolved in Prolog that fail the
unify_with_occurs_check test proves that these expressions are
semantically incorrect.

It is generally the case that every expression of any natural of formal
language that cannot be derived by applying truth preserving operations
(such as Prolog rules) to expressions known to be true (such as Prolog
facts) cannot possibly be correctly construed as true.

Dogs are animals (purely analytic)
There is a small dog in my living room right now (Empirical).

This is true for the entire body of analytic knowledge which only
excludes expressions of language that rely on sense data from the sense
organs to verify truth.

The proof that this is correct is that no counter-examples exist.
When G is considered true and unprovable there is some way the "true" is
derived, it is not merely a wild guess.

Just like Prolog databases True is limited to a specific formal system,
one formal system is the entire body of analytic knowledge: EBAK. This
is an entirely different formal system than PA.

unprovable in PA and true in EBAC is not the same thing as true and
unprovable. unprovable in PA means not true in PA, and true in EBAC
means provable in EBAC.

--
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 5/3/2022 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)

--
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 5/3/2022 12:17 PM, Mikko wrote:
> On 2022-05-03 14:38:57 +0000, olcott said:
>
>> On 5/3/2022 4:36 AM, Mikko wrote:
>>> 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
>>>
>>
>> If would have to be invalid because it can never be resolved.
>
> What would be invalid? Prolog? Definition of Prolog?
> Why "would be" and not "is"?
>
> Mikko
>

Failing a unify_with_occurs_check which would otherwise derive this:

[trace] ?- LP = \+(LP).
LP = (\+LP).

[trace] ?- LP.
% ... 1,000,000 ............ 10,000,000 years later
% % >> 42 << (last release gives the question)
[trace] ?-

--
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 Tue, 3 May 2022 12:31:44 +0300
Mikko <mikko.levanto@iki.fi> 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.

Which is kind of my point: the category error is what makes the
infinite recursion invalid thus rendering the halting problem
definition itself invalid and any proofs predicated on it.

/Flibble

On 5/3/2022 10:57 AM, olcott wrote:
> On 5/3/2022 11:41 AM, Malcolm McLean wrote:
>> On Tuesday, 3 May 2022 at 15:33:45 UTC+1, olcott wrote:
>>> On 5/3/2022 6:08 AM, Malcolm McLean wrote:
>>>> On Tuesday, 3 May 2022 at 10:36:48 UTC+1, Mikko wrote:
>>>>> 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?
>>>>>
>>>> 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.
>
>>> Even infinitely recursive math expressions are semantically incorrect in
>>> that they can never be evaluated.
>>>
>> What's e ^ (PI * i) ?
>>
>> e is Euler's number.
>> PI is the ratio of the diameter of a circle to its circumference
>> i is the square root of -1.
>>
>
> I don't buy into the whole imaginary numbers game.
> We could imagine that 2 + 3 = 17 and call that an imaginary sum.
>
> That is not an infinitely recursive math expression.
> It is a math expression that can be evaluated on the basis of the
> numerical constants specified by e, PI and i. That it cannot be resolved
> to a finite string of digits does not make it invalid.
I first saw that expression in high school. It was claimed to be a key
part of what was claimed to be the world's "most intriguing" formula.
Too bad you missed math in high school and the rest of your life. [Try
Google maybe you'll find it and can copy and paste some nonsense about
it. SQUAWK Polly want a cracker?]
--
Jeff Barnett

On Tue, 3 May 2022 11:57:39 -0500
olcott <polcott2@gmail.com> wrote:

> On 5/3/2022 11:41 AM, Malcolm McLean wrote:
> > On Tuesday, 3 May 2022 at 15:33:45 UTC+1, olcott wrote:
> >> On 5/3/2022 6:08 AM, Malcolm McLean wrote:
> >>> On Tuesday, 3 May 2022 at 10:36:48 UTC+1, Mikko wrote:
> >>>> 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?
> >>>>
> >>> 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.
>
> >> Even infinitely recursive math expressions are semantically
> >> incorrect in that they can never be evaluated.
> >>
> > What's e ^ (PI * i) ?
> >
> > e is Euler's number.
> > PI is the ratio of the diameter of a circle to its circumference
> > i is the square root of -1.
> >
>
> I don't buy into the whole imaginary numbers game.
> We could imagine that 2 + 3 = 17 and call that an imaginary sum.
>
> That is not an infinitely recursive math expression.
> It is a math expression that can be evaluated on the basis of the
> numerical constants specified by e, PI and i. That it cannot be
> resolved to a finite string of digits does not make it invalid.

It evaluates to -1, so is a finite string of digits. Google Euler's
Identity.

/Flibble

On 2022-05-03 12:53, Jeff Barnett wrote:

> I first saw that expression in high school. It was claimed to be a key
> part of what was claimed to be the world's "most intriguing" formula.
> Too bad you missed math in high school and the rest of your life. [Try
> Google maybe you'll find it and can copy and paste some nonsense about
> it. SQUAWK Polly want a cracker?]

An ASCII formula might be hard to google. Euler's Identity is probably a
better search term.

https://en.wikipedia.org/wiki/Euler%27s_identity

André

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

On 5/3/2022 1:59 PM, Mr Flibble wrote:
> On Tue, 3 May 2022 11:57:39 -0500
> olcott <polcott2@gmail.com> wrote:
>
>> On 5/3/2022 11:41 AM, Malcolm McLean wrote:
>>> On Tuesday, 3 May 2022 at 15:33:45 UTC+1, olcott wrote:
>>>> On 5/3/2022 6:08 AM, Malcolm McLean wrote:
>>>>> On Tuesday, 3 May 2022 at 10:36:48 UTC+1, Mikko wrote:
>>>>>> 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?
>>>>>>
>>>>> 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.
>>
>>>> Even infinitely recursive math expressions are semantically
>>>> incorrect in that they can never be evaluated.
>>>>
>>> What's e ^ (PI * i) ?
>>>
>>> e is Euler's number.
>>> PI is the ratio of the diameter of a circle to its circumference
>>> i is the square root of -1.
>>>
>>
>> I don't buy into the whole imaginary numbers game.
>> We could imagine that 2 + 3 = 17 and call that an imaginary sum.
>>
>> That is not an infinitely recursive math expression.
>> It is a math expression that can be evaluated on the basis of the
>> numerical constants specified by e, PI and i. That it cannot be
>> resolved to a finite string of digits does not make it invalid.
>
> It evaluates to -1, so is a finite string of digits. Google Euler's
> Identity.
>
> /Flibble
>

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.

We might as well imagine that a cat is an office building and the ask:
What should we feed this imaginary cat.

--
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 Mon, 2 May 2022 20:40:13 -0400
Richard Damon <Richard@Damon-Family.org> wrote:

> On 5/2/22 7:30 PM, Mr Flibble wrote:
> > On Mon, 2 May 2022 19:16:03 -0400
> > Richard Damon <Richard@Damon-Family.org> wrote:
> >
> >> On 5/2/22 6:47 PM, Mr Flibble wrote:
> >>> On Mon, 2 May 2022 18:46:00 -0400
> >>> Richard Damon <Richard@Damon-Family.org> wrote:
> >>>
> >>>> On 5/2/22 6:38 PM, Mr Flibble wrote:
> >>>>> On Mon, 2 May 2022 18:32:16 -0400
> >>>>> Richard Damon <Richard@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
> >>>>>
> >>>>
> >>>> 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.
> >>>
> >>> If Wikipedia is wrong then correct it and have your corrections
> >>> reviewed; until then please shut the fuck up.
> >>>
> >>> /Flibble
> >>>
> >>
> >> 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.
> >
> > 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
> >
>
> 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.

Wrong; the wording in the third paragraph suggests the prior
paragraphs refer to the halting problem itself, i.e. its definition.

Stop playing word games. The halting problem as defined in [Wikipedia,
2022] is erroneous as it contains a category error in the form of an
erroneous infinite recursion. The fact that currently extant halting
problem proofs are predicated on this erroneous infinite recursion
tells us that the second paragraph *is* part of the halting problem
definition and thus are invalid.

/Flibble

On 5/3/2022 2:06 PM, Mr Flibble wrote:
> On Mon, 2 May 2022 20:40:13 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 5/2/22 7:30 PM, Mr Flibble wrote:
>>> On Mon, 2 May 2022 19:16:03 -0400
>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>
>>>> On 5/2/22 6:47 PM, Mr Flibble wrote:
>>>>> On Mon, 2 May 2022 18:46:00 -0400
>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>
>>>>>> On 5/2/22 6:38 PM, Mr Flibble wrote:
>>>>>>> On Mon, 2 May 2022 18:32:16 -0400
>>>>>>> Richard Damon <Richard@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
>>>>>>>
>>>>>>
>>>>>> 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.
>>>>>
>>>>> If Wikipedia is wrong then correct it and have your corrections
>>>>> reviewed; until then please shut the fuck up.
>>>>>
>>>>> /Flibble
>>>>>
>>>>
>>>> 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.
>>>
>>> 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
>>>
>>
>> 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.
>
> Wrong; the wording in the third paragraph suggests the prior
> paragraphs refer to the halting problem itself, i.e. its definition.
>
> Stop playing word games. The halting problem as defined in [Wikipedia,
> 2022] is erroneous as it contains a category error in the form of an
> erroneous infinite recursion. The fact that currently extant halting
> problem proofs are predicated on this erroneous infinite recursion
> tells us that the second paragraph *is* part of the halting problem
> definition and thus are invalid.
>
> /Flibble
>

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.

--
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 Tuesday, May 3, 2022 at 12:39:49 PM UTC-4, olcott wrote:
> On 5/3/2022 10:36 AM, Dennis Bush wrote:
> > On Tuesday, May 3, 2022 at 11:19:33 AM UTC-4, olcott wrote:
> >> On 5/3/2022 9:47 AM, Dennis Bush wrote:
> >>> 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：
> >>>>>> On 5/2/22 8:35 PM, olcott wrote:
> >>>>>>> On 5/2/2022 5:47 PM, Mr Flibble wrote:
> >>>>>>>> On Mon, 2 May 2022 18:46:00 -0400
> >>>>>>>> Richard Damon wrote:
> >>>>>>>>
> >>>>>>>>> On 5/2/22 6:38 PM, Mr Flibble wrote:
> >>>>>>>>>> On Mon, 2 May 2022 18:32:16 -0400
> >>>>>>>>>> Richard Damon 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
> >>>>>>>>>
> >>>>>>>>> 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.
> >>>>>>>>
> >>>>>>>> If Wikipedia is wrong then correct it and have your corrections
> >>>>>>>> reviewed; until then please shut the fuck up.
> >>>>>>>>
> >>>>>>>> /Flibble
> >>>>>>>>
> >>>>>>>
> >>>>>>> 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
> >>>>>>>
> >>>>>>>
> >>>>>> 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.
> >>>>>
> >>>>> 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.
> >>>> 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.
> >>>
> >>> 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.
> >>>
> >> I have no idea what you mean.
> >
> > In other words you don't want to admit that this proves you are wrong.
> >
> 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.

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.

On Tuesday, 3 May 2022 at 20:05:39 UTC+1, 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:
> >
> >> On 5/3/2022 11:41 AM, Malcolm McLean wrote:
> >>> On Tuesday, 3 May 2022 at 15:33:45 UTC+1, olcott wrote:
> >>>> On 5/3/2022 6:08 AM, Malcolm McLean wrote:
> >>>>> On Tuesday, 3 May 2022 at 10:36:48 UTC+1, Mikko wrote:
> >>>>>> 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?
> >>>>>>
> >>>>> 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.
> >>
> >>>> Even infinitely recursive math expressions are semantically
> >>>> incorrect in that they can never be evaluated.
> >>>>
> >>> What's e ^ (PI * i) ?
> >>>
> >>> e is Euler's number.
> >>> PI is the ratio of the diameter of a circle to its circumference
> >>> i is the square root of -1.
> >>>
> >>
> >> I don't buy into the whole imaginary numbers game.
> >> We could imagine that 2 + 3 = 17 and call that an imaginary sum.
> >>
> >> That is not an infinitely recursive math expression.
> >> It is a math expression that can be evaluated on the basis of the
> >> numerical constants specified by e, PI and i. That it cannot be
> >> resolved to a finite string of digits does not make it invalid.
> >
> > It evaluates to -1, so is a finite string of digits. Google Euler's
> > Identity.
> >
> > /Flibble
> >
> 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.
>
> We might as well imagine that a cat is an office building and the ask:
> What should we feed this imaginary cat.
>
Zero doesn't have a physical representation. So Roman numbers didn't have
the concept. Negative numbers don't have a physical representation. Whilst
children generally accept that two negatives make a positive, justiifying this
is quite hard. Imaginary numbers are called "imaginary" because, again, they
don't have an obvious physical representation (it's now thought that maybe
some subatomic particles have imaginary mass).

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.

On 5/3/2022 3:51 PM, Malcolm McLean wrote:
> On Tuesday, 3 May 2022 at 20:05:39 UTC+1, 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:
>>>
>>>> On 5/3/2022 11:41 AM, Malcolm McLean wrote:
>>>>> On Tuesday, 3 May 2022 at 15:33:45 UTC+1, olcott wrote:
>>>>>> On 5/3/2022 6:08 AM, Malcolm McLean wrote:
>>>>>>> On Tuesday, 3 May 2022 at 10:36:48 UTC+1, Mikko wrote:
>>>>>>>> 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?
>>>>>>>>
>>>>>>> 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.
>>>>
>>>>>> Even infinitely recursive math expressions are semantically
>>>>>> incorrect in that they can never be evaluated.
>>>>>>
>>>>> What's e ^ (PI * i) ?
>>>>>
>>>>> e is Euler's number.
>>>>> PI is the ratio of the diameter of a circle to its circumference
>>>>> i is the square root of -1.
>>>>>
>>>>
>>>> I don't buy into the whole imaginary numbers game.
>>>> We could imagine that 2 + 3 = 17 and call that an imaginary sum.
>>>>
>>>> That is not an infinitely recursive math expression.
>>>> It is a math expression that can be evaluated on the basis of the
>>>> numerical constants specified by e, PI and i. That it cannot be
>>>> resolved to a finite string of digits does not make it invalid.
>>>
>>> It evaluates to -1, so is a finite string of digits. Google Euler's
>>> Identity.
>>>
>>> /Flibble
>>>
>> 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.
>>
>> We might as well imagine that a cat is an office building and the ask:
>> What should we feed this imaginary cat.
>>
> Zero doesn't have a physical representation. So Roman numbers didn't have
> the concept. Negative numbers don't have a physical representation. Whilst
> children generally accept that two negatives make a positive, justiifying this
> is quite hard. Imaginary numbers are called "imaginary" because, again, they
> don't have an obvious physical representation (it's now thought that maybe
> some subatomic particles have imaginary mass).
>
> 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.

May I try another by analogy: PO doesn't have a brain so he/it doesn't
exist. Thus, he is just a disembodied typist. Would this possibly
explain all the misquotes and misunderstandings that emanate from the
ghost terminal? I've been comparing him to a parrot but I now see that
is an insult to parrots with brains intact. But I still have an urge to
ask "Polly want a cracker?"
--
Jeff Barnett

On 5/3/2022 4:49 PM, Dennis Bush wrote:
> On Tuesday, May 3, 2022 at 12:39:49 PM UTC-4, olcott wrote:
>> On 5/3/2022 10:36 AM, Dennis Bush wrote:
>>> On Tuesday, May 3, 2022 at 11:19:33 AM UTC-4, olcott wrote:
>>>> On 5/3/2022 9:47 AM, Dennis Bush wrote:
>>>>> 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：
>>>>>>>> On 5/2/22 8:35 PM, olcott wrote:
>>>>>>>>> On 5/2/2022 5:47 PM, Mr Flibble wrote:
>>>>>>>>>> On Mon, 2 May 2022 18:46:00 -0400
>>>>>>>>>> Richard Damon wrote:
>>>>>>>>>>
>>>>>>>>>>> On 5/2/22 6:38 PM, Mr Flibble wrote:
>>>>>>>>>>>> On Mon, 2 May 2022 18:32:16 -0400
>>>>>>>>>>>> Richard Damon 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
>>>>>>>>>>>
>>>>>>>>>>> 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.
>>>>>>>>>>
>>>>>>>>>> If Wikipedia is wrong then correct it and have your corrections
>>>>>>>>>> reviewed; until then please shut the fuck up.
>>>>>>>>>>
>>>>>>>>>> /Flibble
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> 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
>>>>>>>>>
>>>>>>>>>
>>>>>>>> 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.
>>>>>>>
>>>>>>> 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.
>>>>>> 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.
>>>>>
>>>>> 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.
>>>>>
>>>> I have no idea what you mean.
>>>
>>> In other words you don't want to admit that this proves you are wrong.
>>>
>> 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.
>
> 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*.

Not at all because H(P,P) has itself as part of its input and Hb(P,P)
does not have itself as part of its input.

>
> 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.
>

Sure it does. The correctly simulated input to H(P,P) specifies
infinitely nested simulation where as correctly simulated input to
Hb(P,P) DOES NOT specify infinitely nested simulation.

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

On 5/3/2022 4:51 PM, Malcolm McLean wrote:
> On Tuesday, 3 May 2022 at 20:05:39 UTC+1, 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:
>>>
>>>> On 5/3/2022 11:41 AM, Malcolm McLean wrote:
>>>>> On Tuesday, 3 May 2022 at 15:33:45 UTC+1, olcott wrote:
>>>>>> On 5/3/2022 6:08 AM, Malcolm McLean wrote:
>>>>>>> On Tuesday, 3 May 2022 at 10:36:48 UTC+1, Mikko wrote:
>>>>>>>> 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?
>>>>>>>>
>>>>>>> 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.
>>>>
>>>>>> Even infinitely recursive math expressions are semantically
>>>>>> incorrect in that they can never be evaluated.
>>>>>>
>>>>> What's e ^ (PI * i) ?
>>>>>
>>>>> e is Euler's number.
>>>>> PI is the ratio of the diameter of a circle to its circumference
>>>>> i is the square root of -1.
>>>>>
>>>>
>>>> I don't buy into the whole imaginary numbers game.
>>>> We could imagine that 2 + 3 = 17 and call that an imaginary sum.
>>>>
>>>> That is not an infinitely recursive math expression.
>>>> It is a math expression that can be evaluated on the basis of the
>>>> numerical constants specified by e, PI and i. That it cannot be
>>>> resolved to a finite string of digits does not make it invalid.
>>>
>>> It evaluates to -1, so is a finite string of digits. Google Euler's
>>> Identity.
>>>
>>> /Flibble
>>>
>> 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.
>>
>> We might as well imagine that a cat is an office building and the ask:
>> What should we feed this imaginary cat.
>>
> Zero doesn't have a physical representation. So Roman numbers didn't have
> the concept. Negative numbers don't have a physical representation. Whilst
> children generally accept that two negatives make a positive, justiifying this
> is quite hard. Imaginary numbers are called "imaginary" because, again, they
> don't have an obvious physical representation (it's now thought that maybe
> some subatomic particles have imaginary mass).
>

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.

--
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 5/3/2022 5:06 PM, Jeff Barnett wrote:
> On 5/3/2022 3:51 PM, Malcolm McLean wrote:
>> On Tuesday, 3 May 2022 at 20:05:39 UTC+1, 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:
>>>>
>>>>> On 5/3/2022 11:41 AM, Malcolm McLean wrote:
>>>>>> On Tuesday, 3 May 2022 at 15:33:45 UTC+1, olcott wrote:
>>>>>>> On 5/3/2022 6:08 AM, Malcolm McLean wrote:
>>>>>>>> On Tuesday, 3 May 2022 at 10:36:48 UTC+1, Mikko wrote:
>>>>>>>>> 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?
>>>>>>>>>
>>>>>>>> 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.
>>>>>
>>>>>>> Even infinitely recursive math expressions are semantically
>>>>>>> incorrect in that they can never be evaluated.
>>>>>>>
>>>>>> What's e ^ (PI * i) ?
>>>>>>
>>>>>> e is Euler's number.
>>>>>> PI is the ratio of the diameter of a circle to its circumference
>>>>>> i is the square root of -1.
>>>>>>
>>>>>
>>>>> I don't buy into the whole imaginary numbers game.
>>>>> We could imagine that 2 + 3 = 17 and call that an imaginary sum.
>>>>>
>>>>> That is not an infinitely recursive math expression.
>>>>> It is a math expression that can be evaluated on the basis of the
>>>>> numerical constants specified by e, PI and i. That it cannot be
>>>>> resolved to a finite string of digits does not make it invalid.
>>>>
>>>> It evaluates to -1, so is a finite string of digits. Google Euler's
>>>> Identity.
>>>>
>>>> /Flibble
>>>>
>>> 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.
>>>
>>> We might as well imagine that a cat is an office building and the ask:
>>> What should we feed this imaginary cat.
>>>
>> Zero doesn't have a physical representation. So Roman numbers didn't have
>> the concept. Negative numbers don't have a physical representation.
>> Whilst
>> children generally accept that two negatives make a positive,
>> justiifying this
>> is quite hard. Imaginary numbers are called "imaginary" because,
>> again, they
>> don't have an obvious physical representation (it's now thought that
>> maybe
>> some subatomic particles have imaginary mass).
>>
>> 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.
>
> May I try another by analogy: PO doesn't have a brain so he/it doesn't
> exist. Thus, he is just a disembodied typist. Would this possibly
> explain all the misquotes and misunderstandings that emanate from the
> ghost terminal? I've been comparing him to a parrot but I now see that
> is an insult to parrots with brains intact. But I still have an urge to
> ask "Polly want a cracker?"

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.

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

Peter Olcott wrote:
> On 5/3/2022 1:59 PM, Mr Flibble wrote:
>> On Tue, 3 May 2022 11:57:39 -0500
>> olcott <polcott2@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]

On Tuesday, May 3, 2022 at 6:08:33 PM UTC-4, olcott wrote:
> On 5/3/2022 4:49 PM, Dennis Bush wrote:
> > On Tuesday, May 3, 2022 at 12:39:49 PM UTC-4, olcott wrote:
> >> On 5/3/2022 10:36 AM, Dennis Bush wrote:
> >>> On Tuesday, May 3, 2022 at 11:19:33 AM UTC-4, olcott wrote:
> >>>> On 5/3/2022 9:47 AM, Dennis Bush wrote:
> >>>>> 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：
> >>>>>>>> On 5/2/22 8:35 PM, olcott wrote:
> >>>>>>>>> On 5/2/2022 5:47 PM, Mr Flibble wrote:
> >>>>>>>>>> On Mon, 2 May 2022 18:46:00 -0400
> >>>>>>>>>> Richard Damon wrote:
> >>>>>>>>>>
> >>>>>>>>>>> On 5/2/22 6:38 PM, Mr Flibble wrote:
> >>>>>>>>>>>> On Mon, 2 May 2022 18:32:16 -0400
> >>>>>>>>>>>> Richard Damon 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
> >>>>>>>>>>>
> >>>>>>>>>>> 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.
> >>>>>>>>>>
> >>>>>>>>>> If Wikipedia is wrong then correct it and have your corrections
> >>>>>>>>>> reviewed; until then please shut the fuck up.
> >>>>>>>>>>
> >>>>>>>>>> /Flibble
> >>>>>>>>>>
> >>>>>>>>>
> >>>>>>>>> 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
> >>>>>>>>>
> >>>>>>>>>
> >>>>>>>> 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.
> >>>>>>>
> >>>>>>> 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.
> >>>>>> 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.
> >>>>>
> >>>>> 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.
> >>>>>
> >>>> I have no idea what you mean.
> >>>
> >>> In other words you don't want to admit that this proves you are wrong..
> >>>
> >> 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.
> >
> > 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*.
> Not at all because H(P,P) has itself as part of its input and Hb(P,P)
> does not have itself as part of its input.
> >
> > 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.
> >
> Sure it does. The correctly simulated input to H(P,P) specifies
> infinitely nested simulation where as correctly simulated input to
> Hb(P,P) DOES NOT specify infinitely nested simulation.
>
> How much longer are you going to continue the verified facts?
> This does make you look quite foolish or dishonest.
> > Ha(Pa,Pa) and Hb(Pa,Pa) have the *exact* same input.
> >
> The correctly simulated input to Ha(Pa,Pa) specifies infinitely nested
> simulation where as correctly simulated input to Hb(Pa,Pa) DOES NOT
> specify infinitely nested simulation.
>
> How much longer are you going to continue the verified facts?
> This does make you look quite foolish or dishonest.
> > 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.
> You only have to do a correct execution trace of Ha(Pa,Pa) and Hb(Pa,Pa)
> to see that:
>
> The correctly simulated input to Ha(Pa,Pa) specifies infinitely nested
> simulation where as correctly simulated input to Hb(Pa,Pa) DOES NOT
> specify infinitely nested simulation.
>
> How much longer are you going to continue the verified facts?
> This does make you look quite foolish or dishonest.
> >>>
> >>>>> 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.
> You only have to do a correct execution trace of Ha(Pa,Pa) and Hb(Pa,Pa)
> to see that:
>
> The correctly simulated input to Ha(Pa,Pa) specifies infinitely nested
> simulation where as correctly simulated input to Hb(Pa,Pa) DOES NOT
> specify infinitely nested simulation.
>
> How much longer are you going to continue the verified facts?
> This does make you look quite foolish or dishonest.
> >>>>> 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.
> You only have to do a correct execution trace of Ha(Pa,Pa) and Hb(Pa,Pa)
> to see that:
>
> The correctly simulated input to Ha(Pa,Pa) specifies infinitely nested
> simulation where as correctly simulated input to Hb(Pa,Pa) DOES NOT
> specify infinitely nested simulation.

Mr Flibble wrote:
> On Mon, 2 May 2022 20:40:13 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 5/2/22 7:30 PM, Mr Flibble wrote:
>>> On Mon, 2 May 2022 19:16:03 -0400
>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>
>>>> On 5/2/22 6:47 PM, Mr Flibble wrote:
>>>>> On Mon, 2 May 2022 18:46:00 -0400
>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>
>>>>>> On 5/2/22 6:38 PM, Mr Flibble wrote:
>>>>>>> On Mon, 2 May 2022 18:32:16 -0400
>>>>>>> Richard Damon <Richard@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
>>>>>>>
>>>>>>
>>>>>> 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.
>>>>>
>>>>> If Wikipedia is wrong then correct it and have your corrections
>>>>> reviewed; until then please shut the fuck up.
>>>>>
>>>>> /Flibble
>>>>>
>>>>
>>>> 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.
>>>
>>> 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
>>>
>>
>> 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.
>
> Wrong; the wording in the third paragraph suggests the prior
> paragraphs refer to the halting problem itself, i.e. its definition.
>
> Stop playing word games. The halting problem as defined in [Wikipedia,
> 2022] is erroneous as it contains a category error in the form of an
> erroneous infinite recursion. The fact that currently extant halting
> problem proofs are predicated on this erroneous infinite recursion
> tells us that the second paragraph *is* part of the halting problem
> definition and thus are invalid.
>
> /Flibble

You cranks are really unsufferable idiots...

On 5/3/2022 5:21 PM, Dennis Bush wrote:
> On Tuesday, May 3, 2022 at 6:08:33 PM UTC-4, olcott wrote:
>> On 5/3/2022 4:49 PM, Dennis Bush wrote:
>>> On Tuesday, May 3, 2022 at 12:39:49 PM UTC-4, olcott wrote:
>>>> On 5/3/2022 10:36 AM, Dennis Bush wrote:
>>>>> On Tuesday, May 3, 2022 at 11:19:33 AM UTC-4, olcott wrote:
>>>>>> On 5/3/2022 9:47 AM, Dennis Bush wrote:
>>>>>>> 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：
>>>>>>>>>> On 5/2/22 8:35 PM, olcott wrote:
>>>>>>>>>>> On 5/2/2022 5:47 PM, Mr Flibble wrote:
>>>>>>>>>>>> On Mon, 2 May 2022 18:46:00 -0400
>>>>>>>>>>>> Richard Damon wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> On 5/2/22 6:38 PM, Mr Flibble wrote:
>>>>>>>>>>>>>> On Mon, 2 May 2022 18:32:16 -0400
>>>>>>>>>>>>>> Richard Damon 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
>>>>>>>>>>>>>
>>>>>>>>>>>>> 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.
>>>>>>>>>>>>
>>>>>>>>>>>> If Wikipedia is wrong then correct it and have your corrections
>>>>>>>>>>>> reviewed; until then please shut the fuck up.
>>>>>>>>>>>>
>>>>>>>>>>>> /Flibble
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> 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
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>> 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.
>>>>>>>>>
>>>>>>>>> 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.
>>>>>>>> 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.
>>>>>>>
>>>>>>> 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.
>>>>>>>
>>>>>> I have no idea what you mean.
>>>>>
>>>>> In other words you don't want to admit that this proves you are wrong.
>>>>>
>>>> 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.
>>>
>>> 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*.
>> Not at all because H(P,P) has itself as part of its input and Hb(P,P)
>> does not have itself as part of its input.
>>>
>>> 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.
>>>
>> Sure it does. The correctly simulated input to H(P,P) specifies
>> infinitely nested simulation where as correctly simulated input to
>> Hb(P,P) DOES NOT specify infinitely nested simulation.
>>
>> How much longer are you going to continue the verified facts?
>> This does make you look quite foolish or dishonest.
>>> Ha(Pa,Pa) and Hb(Pa,Pa) have the *exact* same input.
>>>
>> The correctly simulated input to Ha(Pa,Pa) specifies infinitely nested
>> simulation where as correctly simulated input to Hb(Pa,Pa) DOES NOT
>> specify infinitely nested simulation.
>>
>> How much longer are you going to continue the verified facts?
>> This does make you look quite foolish or dishonest.
>>> 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.
>> You only have to do a correct execution trace of Ha(Pa,Pa) and Hb(Pa,Pa)
>> to see that:
>>
>> The correctly simulated input to Ha(Pa,Pa) specifies infinitely nested
>> simulation where as correctly simulated input to Hb(Pa,Pa) DOES NOT
>> specify infinitely nested simulation.
>>
>> How much longer are you going to continue the verified facts?
>> This does make you look quite foolish or dishonest.
>>>>>
>>>>>>> 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.
>> You only have to do a correct execution trace of Ha(Pa,Pa) and Hb(Pa,Pa)
>> to see that:
>>
>> The correctly simulated input to Ha(Pa,Pa) specifies infinitely nested
>> simulation where as correctly simulated input to Hb(Pa,Pa) DOES NOT
>> specify infinitely nested simulation.
>>
>> How much longer are you going to continue the verified facts?
>> This does make you look quite foolish or dishonest.
>>>>>>> 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.
>> You only have to do a correct execution trace of Ha(Pa,Pa) and Hb(Pa,Pa)
>> to see that:
>>
>> The correctly simulated input to Ha(Pa,Pa) specifies infinitely nested
>> simulation where as correctly simulated input to Hb(Pa,Pa) DOES NOT
>> specify infinitely nested simulation.
>
> Hb(Pa,Pa) proves that Ha(Pa,Pa) does NOT perform a correct simulation.
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.