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devel / comp.theory / Re: On recursion and infinite recursion (reprise #3)

SubjectAuthor
* On recursion and infinite recursion (reprise #3)Mr Flibble
+* On recursion and infinite recursion (reprise #3)olcott
|`* On recursion and infinite recursion (reprise #3)Ben
| `* On recursion and infinite recursion (reprise #3)olcott
|  `* On recursion and infinite recursion (reprise #3)Ben
|   `* On recursion and infinite recursion (reprise #3)olcott
|    `- On recursion and infinite recursion (reprise #3)Ben
+* On recursion and infinite recursion (reprise #3)Ben
|+* On recursion and infinite recursion (reprise #3)Mr Flibble
||+* On recursion and infinite recursion (reprise #3)Ben
|||`- On recursion and infinite recursion (reprise #3)Mr Flibble
||`- On recursion and infinite recursion (reprise #3)olcott
|`* On recursion and infinite recursion (reprise #3)olcott
| `* On recursion and infinite recursion (reprise #3)Ben
|  `* On recursion and infinite recursion (reprise #3)olcott
|   `- On recursion and infinite recursion (reprise #3)Ben
+* On recursion and infinite recursion (reprise #3)Richard Damon
|`* On recursion and infinite recursion (reprise #3)Mr Flibble
| +* On recursion and infinite recursion (reprise #3)Python
| |`- On recursion and infinite recursion (reprise #3)olcott
| `* On recursion and infinite recursion (reprise #3)Mikko
|  `* On recursion and infinite recursion (reprise #3)Mr Flibble
|   `* On recursion and infinite recursion (reprise #3)Mikko
|    `* On recursion and infinite recursion (reprise #3)Mr Flibble
|     `* On recursion and infinite recursion (reprise #3)Mikko
|      `* On recursion and infinite recursion (reprise #3)Mr Flibble
|       +* On recursion and infinite recursion (reprise #3)Mikko
|       |`* On recursion and infinite recursion (reprise #3)Mr Flibble
|       | +- On recursion and infinite recursion (reprise #3)olcott
|       | +* On recursion and infinite recursion (reprise #3)Ben
|       | |`* On recursion and infinite recursion (reprise #3)Mr Flibble
|       | | `* On recursion and infinite recursion (reprise #3)Richard Damon
|       | |  `* On recursion and infinite recursion (reprise #3)Mr Flibble
|       | |   `* On recursion and infinite recursion (reprise #3)Richard Damon
|       | |    `- On recursion and infinite recursion (reprise #3)Mr Flibble
|       | `* On recursion and infinite recursion (reprise #3)Mikko
|       |  +* On recursion and infinite recursion (reprise #3)Mr Flibble
|       |  |+* On recursion and infinite recursion (reprise #3)Mikko
|       |  ||`- On recursion and infinite recursion (reprise #3)Mr Flibble
|       |  |+* On recursion and infinite recursion (reprise #3)Richard Damon
|       |  ||`* On recursion and infinite recursion (reprise #3)Mr Flibble
|       |  || `* On recursion and infinite recursion (reprise #3)Richard Damon
|       |  ||  `* On recursion and infinite recursion (reprise #3)Mr Flibble
|       |  ||   +* On recursion and infinite recursion (reprise #3)Richard Damon
|       |  ||   |`* On recursion and infinite recursion (reprise #3)Mr Flibble
|       |  ||   | `* On recursion and infinite recursion (reprise #3)Richard Damon
|       |  ||   |  `* On recursion and infinite recursion (reprise #3)Mr Flibble
|       |  ||   |   `* On recursion and infinite recursion (reprise #3)Richard Damon
|       |  ||   |    `* On recursion and infinite recursion (reprise #3)Mr Flibble
|       |  ||   |     +- On recursion and infinite recursion (reprise #3)Ben
|       |  ||   |     +- On recursion and infinite recursion (reprise #3)Richard Damon
|       |  ||   |     `* On recursion and infinite recursion (reprise #3)olcott
|       |  ||   |      `- On recursion and infinite recursion (reprise #3)Richard Damon
|       |  ||   `* On recursion and infinite recursion (reprise #3)olcott
|       |  ||    +* On recursion and infinite recursion (reprise #3)Mr Flibble
|       |  ||    |`* On recursion and infinite recursion (reprise #3)olcott
|       |  ||    | `- On recursion and infinite recursion (reprise #3)Richard Damon
|       |  ||    `- On recursion and infinite recursion (reprise #3)Richard Damon
|       |  |`* On recursion and infinite recursion (reprise #3)olcott
|       |  | `* On recursion and infinite recursion (reprise #3)Mr Flibble
|       |  |  `- On recursion and infinite recursion (reprise #3)olcott
|       |  `* On recursion and infinite recursion (reprise #3)olcott
|       |   `* On recursion and infinite recursion (reprise #3)Mikko
|       |    `* On recursion and infinite recursion (reprise #3)olcott
|       |     `- On recursion and infinite recursion (reprise #3)Richard Damon
|       `- On recursion and infinite recursion (reprise #3)olcott
`- On recursion and infinite recursion (reprise #3)Mikko

Pages:123
Re: On recursion and infinite recursion (reprise #3)

<t55uvv$fpf$1@dont-email.me>

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From: mikko.le...@iki.fi (Mikko)
Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
Date: Sat, 7 May 2022 17:16:31 +0300
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 by: Mikko - Sat, 7 May 2022 14:16 UTC

On 2022-05-07 14:06:12 +0000, Mr Flibble said:

> On Sat, 7 May 2022 16:59:55 +0300
> Mikko <mikko.levanto@iki.fi> wrote:
>
>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
>>
>>> On Sat, 7 May 2022 12:52:58 +0300
>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>
>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
>>>>
>>>>> The decider could never be compiled and run in the first place due
>>>>> to the category error in the definition of the proof.
>>>>
>>>> An error in the definition of the proof does not prevent
>>>> compilation and execution of the program.
>>>
>>> In this case the error in the definition of the proof
>>
>> No part of the proof is identified as errorneous, so the rest
>> is irrelevant. Anyway,
>>
>>> does prevent compilation unless
>>
>> There is no option here, so the "unless" is void.
>>
>>> the decider is made part of the program that is being decided
>>
>> The decider is made a part of the program discussed in the proof.
>>
>>> in which case we get a function call-like infinite recursion
>>> instead
>>
>> We get or we don't get, depending on how the halt decider candidate
>> attempts to decide.
>>
>>> (as described by Pete Olcott) and we are attempting (and failing)
>>> to decide if a procedure halts rather than if a program halts.
>>
>> In any case, the decider candidate fails to give the correct answer,
>> and therefore is not a halt decider. Note that this is correctly
>> inferred in the proof. Therefore either the conclusion of the proof
>> is correct or you are wrong.
>
> No answer is ever given due to the infinite recursion.

True about some candidates, false about others. For example, a
candidate that always says "no" is not infinitely recursive but
is wrong in the particular case discussed in the proof. But anyway,
you have confirmed the conclusion of the proof.

Mikko

Re: On recursion and infinite recursion (reprise #3)

<20220507152017.00000990@reddwarf.jmc>

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From: flib...@reddwarf.jmc (Mr Flibble)
Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
Message-ID: <20220507152017.00000990@reddwarf.jmc>
References: <20220505175034.0000253c@reddwarf.jmc>
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 by: Mr Flibble - Sat, 7 May 2022 14:20 UTC

On Sat, 7 May 2022 17:16:31 +0300
Mikko <mikko.levanto@iki.fi> wrote:

> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
>
> > On Sat, 7 May 2022 16:59:55 +0300
> > Mikko <mikko.levanto@iki.fi> wrote:
> >
> >> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
> >>
> >>> On Sat, 7 May 2022 12:52:58 +0300
> >>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>
> >>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
> >>>>
> >>>>> The decider could never be compiled and run in the first place
> >>>>> due to the category error in the definition of the proof.
> >>>>
> >>>> An error in the definition of the proof does not prevent
> >>>> compilation and execution of the program.
> >>>
> >>> In this case the error in the definition of the proof
> >>
> >> No part of the proof is identified as errorneous, so the rest
> >> is irrelevant. Anyway,
> >>
> >>> does prevent compilation unless
> >>
> >> There is no option here, so the "unless" is void.
> >>
> >>> the decider is made part of the program that is being decided
> >>
> >> The decider is made a part of the program discussed in the proof.
> >>
> >>> in which case we get a function call-like infinite recursion
> >>> instead
> >>
> >> We get or we don't get, depending on how the halt decider candidate
> >> attempts to decide.
> >>
> >>> (as described by Pete Olcott) and we are attempting (and failing)
> >>> to decide if a procedure halts rather than if a program halts.
> >>
> >> In any case, the decider candidate fails to give the correct
> >> answer, and therefore is not a halt decider. Note that this is
> >> correctly inferred in the proof. Therefore either the conclusion
> >> of the proof is correct or you are wrong.
> >
> > No answer is ever given due to the infinite recursion.
>
> True about some candidates, false about others. For example, a
> candidate that always says "no" is not infinitely recursive but
> is wrong in the particular case discussed in the proof. But anyway,
> you have confirmed the conclusion of the proof.

False. The proof is not cognisant of the presence of the infinite
recursion. The decider can never return an answer of halts/doesn't
halt due to the infinite recursion.

/Flibble

Re: On recursion and infinite recursion (reprise #3)

<t562a4$9ao$1@dont-email.me>

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From: mikko.le...@iki.fi (Mikko)
Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
Date: Sat, 7 May 2022 18:13:08 +0300
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 by: Mikko - Sat, 7 May 2022 15:13 UTC

On 2022-05-07 14:20:17 +0000, Mr Flibble said:

> On Sat, 7 May 2022 17:16:31 +0300
> Mikko <mikko.levanto@iki.fi> wrote:
>
>> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
>>
>>> On Sat, 7 May 2022 16:59:55 +0300
>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>
>>>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
>>>>
>>>>> On Sat, 7 May 2022 12:52:58 +0300
>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>
>>>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
>>>>>>
>>>>>>> The decider could never be compiled and run in the first place
>>>>>>> due to the category error in the definition of the proof.
>>>>>>
>>>>>> An error in the definition of the proof does not prevent
>>>>>> compilation and execution of the program.
>>>>>
>>>>> In this case the error in the definition of the proof
>>>>
>>>> No part of the proof is identified as errorneous, so the rest
>>>> is irrelevant. Anyway,
>>>>
>>>>> does prevent compilation unless
>>>>
>>>> There is no option here, so the "unless" is void.
>>>>
>>>>> the decider is made part of the program that is being decided
>>>>
>>>> The decider is made a part of the program discussed in the proof.
>>>>
>>>>> in which case we get a function call-like infinite recursion
>>>>> instead
>>>>
>>>> We get or we don't get, depending on how the halt decider candidate
>>>> attempts to decide.
>>>>
>>>>> (as described by Pete Olcott) and we are attempting (and failing)
>>>>> to decide if a procedure halts rather than if a program halts.
>>>>
>>>> In any case, the decider candidate fails to give the correct
>>>> answer, and therefore is not a halt decider. Note that this is
>>>> correctly inferred in the proof. Therefore either the conclusion
>>>> of the proof is correct or you are wrong.
>>>
>>> No answer is ever given due to the infinite recursion.
>>
>> True about some candidates, false about others. For example, a
>> candidate that always says "no" is not infinitely recursive but
>> is wrong in the particular case discussed in the proof. But anyway,
>> you have confirmed the conclusion of the proof.
>
> False. The proof is not cognisant of the presence of the infinite
> recursion. The decider can never return an answer of halts/doesn't
> halt due to the infinite recursion.

The correct term is decider candidate. It is not a decider because
the infinite recursion prevents it from halting in finite time.

You cannot call the proof incorrect merely because it agrees with you.

Mikko

Re: On recursion and infinite recursion (reprise #3)

<20220507161901.00002e54@reddwarf.jmc>

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From: flib...@reddwarf.jmc (Mr Flibble)
Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
Message-ID: <20220507161901.00002e54@reddwarf.jmc>
References: <20220505175034.0000253c@reddwarf.jmc> <5O%cK.17$wYy9.0@fx11.iad> <20220506150253.00007c7f@reddwarf.jmc> <t55fhq$u5e$1@dont-email.me> <20220507134250.00007acc@reddwarf.jmc> <t55u0r$7lf$1@dont-email.me> <20220507150612.00003fab@reddwarf.jmc> <t55uvv$fpf$1@dont-email.me> <20220507152017.00000990@reddwarf.jmc> <t562a4$9ao$1@dont-email.me>
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 by: Mr Flibble - Sat, 7 May 2022 15:19 UTC

On Sat, 7 May 2022 18:13:08 +0300
Mikko <mikko.levanto@iki.fi> wrote:

> On 2022-05-07 14:20:17 +0000, Mr Flibble said:
>
> > On Sat, 7 May 2022 17:16:31 +0300
> > Mikko <mikko.levanto@iki.fi> wrote:
> >
> >> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
> >>
> >>> On Sat, 7 May 2022 16:59:55 +0300
> >>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>
> >>>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
> >>>>
> >>>>> On Sat, 7 May 2022 12:52:58 +0300
> >>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>
> >>>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
> >>>>>>
> >>>>>>> The decider could never be compiled and run in the first place
> >>>>>>> due to the category error in the definition of the proof.
> >>>>>>
> >>>>>> An error in the definition of the proof does not prevent
> >>>>>> compilation and execution of the program.
> >>>>>
> >>>>> In this case the error in the definition of the proof
> >>>>
> >>>> No part of the proof is identified as errorneous, so the rest
> >>>> is irrelevant. Anyway,
> >>>>
> >>>>> does prevent compilation unless
> >>>>
> >>>> There is no option here, so the "unless" is void.
> >>>>
> >>>>> the decider is made part of the program that is being decided
> >>>>
> >>>> The decider is made a part of the program discussed in the proof.
> >>>>
> >>>>> in which case we get a function call-like infinite recursion
> >>>>> instead
> >>>>
> >>>> We get or we don't get, depending on how the halt decider
> >>>> candidate attempts to decide.
> >>>>
> >>>>> (as described by Pete Olcott) and we are attempting (and
> >>>>> failing) to decide if a procedure halts rather than if a
> >>>>> program halts.
> >>>>
> >>>> In any case, the decider candidate fails to give the correct
> >>>> answer, and therefore is not a halt decider. Note that this is
> >>>> correctly inferred in the proof. Therefore either the conclusion
> >>>> of the proof is correct or you are wrong.
> >>>
> >>> No answer is ever given due to the infinite recursion.
> >>
> >> True about some candidates, false about others. For example, a
> >> candidate that always says "no" is not infinitely recursive but
> >> is wrong in the particular case discussed in the proof. But anyway,
> >> you have confirmed the conclusion of the proof.
> >
> > False. The proof is not cognisant of the presence of the infinite
> > recursion. The decider can never return an answer of halts/doesn't
> > halt due to the infinite recursion.
>
> The correct term is decider candidate. It is not a decider because
> the infinite recursion prevents it from halting in finite time.
>
> You cannot call the proof incorrect merely because it agrees with you.

Try actually reading what Strachey wrote: a contradiction arises based
on the result of evaluating T[P] however T[P] is never evaluated due to
the infinite recursion: a fact ignored by the proof.

/Flibble

Re: On recursion and infinite recursion (reprise #3)

<t56585$uq$1@dont-email.me>

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From: polco...@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
Date: Sat, 7 May 2022 11:03:13 -0500
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 by: olcott - Sat, 7 May 2022 16:03 UTC

On 5/7/2022 10:19 AM, Mr Flibble wrote:
> On Sat, 7 May 2022 18:13:08 +0300
> Mikko <mikko.levanto@iki.fi> wrote:
>
>> On 2022-05-07 14:20:17 +0000, Mr Flibble said:
>>
>>> On Sat, 7 May 2022 17:16:31 +0300
>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>
>>>> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
>>>>
>>>>> On Sat, 7 May 2022 16:59:55 +0300
>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>
>>>>>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
>>>>>>
>>>>>>> On Sat, 7 May 2022 12:52:58 +0300
>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>
>>>>>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
>>>>>>>>
>>>>>>>>> The decider could never be compiled and run in the first place
>>>>>>>>> due to the category error in the definition of the proof.
>>>>>>>>
>>>>>>>> An error in the definition of the proof does not prevent
>>>>>>>> compilation and execution of the program.
>>>>>>>
>>>>>>> In this case the error in the definition of the proof
>>>>>>
>>>>>> No part of the proof is identified as errorneous, so the rest
>>>>>> is irrelevant. Anyway,
>>>>>>
>>>>>>> does prevent compilation unless
>>>>>>
>>>>>> There is no option here, so the "unless" is void.
>>>>>>
>>>>>>> the decider is made part of the program that is being decided
>>>>>>
>>>>>> The decider is made a part of the program discussed in the proof.
>>>>>>
>>>>>>> in which case we get a function call-like infinite recursion
>>>>>>> instead
>>>>>>
>>>>>> We get or we don't get, depending on how the halt decider
>>>>>> candidate attempts to decide.
>>>>>>
>>>>>>> (as described by Pete Olcott) and we are attempting (and
>>>>>>> failing) to decide if a procedure halts rather than if a
>>>>>>> program halts.
>>>>>>
>>>>>> In any case, the decider candidate fails to give the correct
>>>>>> answer, and therefore is not a halt decider. Note that this is
>>>>>> correctly inferred in the proof. Therefore either the conclusion
>>>>>> of the proof is correct or you are wrong.
>>>>>
>>>>> No answer is ever given due to the infinite recursion.
>>>>
>>>> True about some candidates, false about others. For example, a
>>>> candidate that always says "no" is not infinitely recursive but
>>>> is wrong in the particular case discussed in the proof. But anyway,
>>>> you have confirmed the conclusion of the proof.
>>>
>>> False. The proof is not cognisant of the presence of the infinite
>>> recursion. The decider can never return an answer of halts/doesn't
>>> halt due to the infinite recursion.
>>
>> The correct term is decider candidate. It is not a decider because
>> the infinite recursion prevents it from halting in finite time.
>>
>> You cannot call the proof incorrect merely because it agrees with you.
>
> Try actually reading what Strachey wrote: a contradiction arises based
> on the result of evaluating T[P] however T[P] is never evaluated due to
> the infinite recursion: a fact ignored by the proof.
>
> /Flibble
>

Yes and a fact that is documented that I first discovered in 2016.

it looks like the original specification provided in the Linz text may
be infinitely recursive in that each TM requires its own input.

https://www.researchgate.net/publication/307509556_Self_Modifying_Turing_Machine_SMTM_Solution_to_the_Halting_Problem_concrete_example

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

Re: On recursion and infinite recursion (reprise #3)

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From: polco...@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
Date: Sat, 7 May 2022 11:06:15 -0500
Organization: A noiseless patient Spider
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 by: olcott - Sat, 7 May 2022 16:06 UTC

On 5/7/2022 9:20 AM, Mr Flibble wrote:
> On Sat, 7 May 2022 17:16:31 +0300
> Mikko <mikko.levanto@iki.fi> wrote:
>
>> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
>>
>>> On Sat, 7 May 2022 16:59:55 +0300
>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>
>>>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
>>>>
>>>>> On Sat, 7 May 2022 12:52:58 +0300
>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>
>>>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
>>>>>>
>>>>>>> The decider could never be compiled and run in the first place
>>>>>>> due to the category error in the definition of the proof.
>>>>>>
>>>>>> An error in the definition of the proof does not prevent
>>>>>> compilation and execution of the program.
>>>>>
>>>>> In this case the error in the definition of the proof
>>>>
>>>> No part of the proof is identified as errorneous, so the rest
>>>> is irrelevant. Anyway,
>>>>
>>>>> does prevent compilation unless
>>>>
>>>> There is no option here, so the "unless" is void.
>>>>
>>>>> the decider is made part of the program that is being decided
>>>>
>>>> The decider is made a part of the program discussed in the proof.
>>>>
>>>>> in which case we get a function call-like infinite recursion
>>>>> instead
>>>>
>>>> We get or we don't get, depending on how the halt decider candidate
>>>> attempts to decide.
>>>>
>>>>> (as described by Pete Olcott) and we are attempting (and failing)
>>>>> to decide if a procedure halts rather than if a program halts.
>>>>
>>>> In any case, the decider candidate fails to give the correct
>>>> answer, and therefore is not a halt decider. Note that this is
>>>> correctly inferred in the proof. Therefore either the conclusion
>>>> of the proof is correct or you are wrong.
>>>
>>> No answer is ever given due to the infinite recursion.
>>
>> True about some candidates, false about others. For example, a
>> candidate that always says "no" is not infinitely recursive but
>> is wrong in the particular case discussed in the proof. But anyway,
>> you have confirmed the conclusion of the proof.
>
> False. The proof is not cognisant of the presence of the infinite
> recursion. The decider can never return an answer of halts/doesn't
> halt due to the infinite recursion.
>
> /Flibble
>

The decider simulates its input in debug step mode and when it sees that
it its input is stuck in infinitely nested simulation it aborts this
simulation and rejects this input. This is all fully elaborated in my
five papers.

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

Re: On recursion and infinite recursion (reprise #3)

<87pmkpvuce.fsf@bsb.me.uk>

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Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
Date: Sat, 07 May 2022 23:41:37 +0100
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 by: Ben - Sat, 7 May 2022 22:41 UTC

Mr Flibble <flibble@reddwarf.jmc> writes:

> Try actually reading what Strachey wrote: a contradiction arises based
> on the result of evaluating T[P] however T[P] is never evaluated due to
> the infinite recursion: a fact ignored by the proof.

It's a central part of the argument. If T[X] does not always return in
finite time, T fails to be a halt decider. If the call to T[P] results
in non-terminating recursion, then T has been badly written. There are
lots of ways the attempt to implement T can fail.

--
Ben.

Re: On recursion and infinite recursion (reprise #3)

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From: flib...@reddwarf.jmc (Mr Flibble)
Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
Message-ID: <20220507234312.00005179@reddwarf.jmc>
References: <20220505175034.0000253c@reddwarf.jmc>
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 by: Mr Flibble - Sat, 7 May 2022 22:43 UTC

On Sat, 07 May 2022 23:41:37 +0100
Ben <ben.usenet@bsb.me.uk> wrote:

> Mr Flibble <flibble@reddwarf.jmc> writes:
>
> > Try actually reading what Strachey wrote: a contradiction arises
> > based on the result of evaluating T[P] however T[P] is never
> > evaluated due to the infinite recursion: a fact ignored by the
> > proof.
>
> It's a central part of the argument. If T[X] does not always return
> in finite time, T fails to be a halt decider. If the call to T[P]
> results in non-terminating recursion, then T has been badly written.
> There are lots of ways the attempt to implement T can fail.
You are completely and utterly missing the point. You are completely
and utterly oblivious to the nature of the infinite recursion.

/Flibble

Re: On recursion and infinite recursion (reprise #3)

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 by: Richard Damon - Sat, 7 May 2022 23:59 UTC

On 5/7/22 6:43 PM, Mr Flibble wrote:
> On Sat, 07 May 2022 23:41:37 +0100
> Ben <ben.usenet@bsb.me.uk> wrote:
>
>> Mr Flibble <flibble@reddwarf.jmc> writes:
>>
>>> Try actually reading what Strachey wrote: a contradiction arises
>>> based on the result of evaluating T[P] however T[P] is never
>>> evaluated due to the infinite recursion: a fact ignored by the
>>> proof.
>>
>> It's a central part of the argument. If T[X] does not always return
>> in finite time, T fails to be a halt decider. If the call to T[P]
>> results in non-terminating recursion, then T has been badly written.
>> There are lots of ways the attempt to implement T can fail.
>
> You are completely and utterly missing the point. You are completely
> and utterly oblivious to the nature of the infinite recursion.
>
> /Flibble
>

No, YOU are missing that if T succombs to this sort of infinite
recursion, then it has failed to meet the requirements of being a
decider to answer in finite time.

Infinite Recursion can not exist in finite time, so T, if it is a
decider, can't get infinitely recursive with ANY input, and the mere
fact that some input can make it so, proves that the candidate T fails
its test.

Re: On recursion and infinite recursion (reprise #3)

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From: flib...@reddwarf.jmc (Mr Flibble)
Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
Message-ID: <20220508010142.00007bf1@reddwarf.jmc>
References: <20220505175034.0000253c@reddwarf.jmc> <5O%cK.17$wYy9.0@fx11.iad> <20220506150253.00007c7f@reddwarf.jmc> <t55fhq$u5e$1@dont-email.me> <20220507134250.00007acc@reddwarf.jmc> <t55u0r$7lf$1@dont-email.me> <20220507150612.00003fab@reddwarf.jmc> <t55uvv$fpf$1@dont-email.me> <20220507152017.00000990@reddwarf.jmc> <t562a4$9ao$1@dont-email.me> <20220507161901.00002e54@reddwarf.jmc> <87pmkpvuce.fsf@bsb.me.uk> <20220507234312.00005179@reddwarf.jmc> <RJDdK.19199$h6X.16518@fx04.iad>
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 by: Mr Flibble - Sun, 8 May 2022 00:01 UTC

On Sat, 7 May 2022 19:59:45 -0400
Richard Damon <Richard@Damon-Family.org> wrote:

> On 5/7/22 6:43 PM, Mr Flibble wrote:
> > On Sat, 07 May 2022 23:41:37 +0100
> > Ben <ben.usenet@bsb.me.uk> wrote:
> >
> >> Mr Flibble <flibble@reddwarf.jmc> writes:
> >>
> >>> Try actually reading what Strachey wrote: a contradiction arises
> >>> based on the result of evaluating T[P] however T[P] is never
> >>> evaluated due to the infinite recursion: a fact ignored by the
> >>> proof.
> >>
> >> It's a central part of the argument. If T[X] does not always
> >> return in finite time, T fails to be a halt decider. If the call
> >> to T[P] results in non-terminating recursion, then T has been
> >> badly written. There are lots of ways the attempt to implement T
> >> can fail.
> >
> > You are completely and utterly missing the point. You are completely
> > and utterly oblivious to the nature of the infinite recursion.
> >
> > /Flibble
> >
>
> No, YOU are missing that if T succombs to this sort of infinite
> recursion, then it has failed to meet the requirements of being a
> decider to answer in finite time.
>
> Infinite Recursion can not exist in finite time, so T, if it is a
> decider, can't get infinitely recursive with ANY input, and the mere
> fact that some input can make it so, proves that the candidate T
> fails its test.

You also are completely and utterly missing the point. You also are
completely and utterly oblivious to the nature of the infinite
recursion.

/Flibble

Re: On recursion and infinite recursion (reprise #3)

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 by: Richard Damon - Sun, 8 May 2022 00:18 UTC

On 5/7/22 8:01 PM, Mr Flibble wrote:
> On Sat, 7 May 2022 19:59:45 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 5/7/22 6:43 PM, Mr Flibble wrote:
>>> On Sat, 07 May 2022 23:41:37 +0100
>>> Ben <ben.usenet@bsb.me.uk> wrote:
>>>
>>>> Mr Flibble <flibble@reddwarf.jmc> writes:
>>>>
>>>>> Try actually reading what Strachey wrote: a contradiction arises
>>>>> based on the result of evaluating T[P] however T[P] is never
>>>>> evaluated due to the infinite recursion: a fact ignored by the
>>>>> proof.
>>>>
>>>> It's a central part of the argument. If T[X] does not always
>>>> return in finite time, T fails to be a halt decider. If the call
>>>> to T[P] results in non-terminating recursion, then T has been
>>>> badly written. There are lots of ways the attempt to implement T
>>>> can fail.
>>>
>>> You are completely and utterly missing the point. You are completely
>>> and utterly oblivious to the nature of the infinite recursion.
>>>
>>> /Flibble
>>>
>>
>> No, YOU are missing that if T succombs to this sort of infinite
>> recursion, then it has failed to meet the requirements of being a
>> decider to answer in finite time.
>>
>> Infinite Recursion can not exist in finite time, so T, if it is a
>> decider, can't get infinitely recursive with ANY input, and the mere
>> fact that some input can make it so, proves that the candidate T
>> fails its test.
>
> You also are completely and utterly missing the point. You also are
> completely and utterly oblivious to the nature of the infinite
> recursion.
>
> /Flibble
>

Then try to expalin it in well defined terms.

You won't be able to in a way that makes the proof invalid.

YOU are the one that doesn't understand the "Point", the "reveled"
"infinite recursion" is just proving that the simulation method to try
and solve that halting problem was doomed from the start, and that such
a "decider" can't work.

Re: On recursion and infinite recursion (reprise #3)

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From: flib...@reddwarf.jmc (Mr Flibble)
Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
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 by: Mr Flibble - Sun, 8 May 2022 00:55 UTC

On Sat, 7 May 2022 20:18:21 -0400
Richard Damon <Richard@Damon-Family.org> wrote:

> On 5/7/22 8:01 PM, Mr Flibble wrote:
> > On Sat, 7 May 2022 19:59:45 -0400
> > Richard Damon <Richard@Damon-Family.org> wrote:
> >
> >> On 5/7/22 6:43 PM, Mr Flibble wrote:
> >>> On Sat, 07 May 2022 23:41:37 +0100
> >>> Ben <ben.usenet@bsb.me.uk> wrote:
> >>>
> >>>> Mr Flibble <flibble@reddwarf.jmc> writes:
> >>>>
> >>>>> Try actually reading what Strachey wrote: a contradiction arises
> >>>>> based on the result of evaluating T[P] however T[P] is never
> >>>>> evaluated due to the infinite recursion: a fact ignored by the
> >>>>> proof.
> >>>>
> >>>> It's a central part of the argument. If T[X] does not always
> >>>> return in finite time, T fails to be a halt decider. If the call
> >>>> to T[P] results in non-terminating recursion, then T has been
> >>>> badly written. There are lots of ways the attempt to implement T
> >>>> can fail.
> >>>
> >>> You are completely and utterly missing the point. You are
> >>> completely and utterly oblivious to the nature of the infinite
> >>> recursion.
> >>>
> >>> /Flibble
> >>>
> >>
> >> No, YOU are missing that if T succombs to this sort of infinite
> >> recursion, then it has failed to meet the requirements of being a
> >> decider to answer in finite time.
> >>
> >> Infinite Recursion can not exist in finite time, so T, if it is a
> >> decider, can't get infinitely recursive with ANY input, and the
> >> mere fact that some input can make it so, proves that the
> >> candidate T fails its test.
> >
> > You also are completely and utterly missing the point. You also are
> > completely and utterly oblivious to the nature of the infinite
> > recursion.
> >
> > /Flibble
> >
>
> Then try to expalin it in well defined terms.
>
> You won't be able to in a way that makes the proof invalid.
>
> YOU are the one that doesn't understand the "Point", the "reveled"
> "infinite recursion" is just proving that the simulation method to
> try and solve that halting problem was doomed from the start, and
> that such a "decider" can't work.

You are confusing me with Olcott: I have never mentioned the simulation
method. It is obvious that you are still missing the point.

/Flibble

Re: On recursion and infinite recursion (reprise #3)

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From: mikko.le...@iki.fi (Mikko)
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Subject: Re: On recursion and infinite recursion (reprise #3)
Date: Sun, 8 May 2022 11:31:08 +0300
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 by: Mikko - Sun, 8 May 2022 08:31 UTC

On 2022-05-07 15:19:01 +0000, Mr Flibble said:

> On Sat, 7 May 2022 18:13:08 +0300
> Mikko <mikko.levanto@iki.fi> wrote:
>
>> On 2022-05-07 14:20:17 +0000, Mr Flibble said:
>>
>>> On Sat, 7 May 2022 17:16:31 +0300
>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>
>>>> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
>>>>
>>>>> On Sat, 7 May 2022 16:59:55 +0300
>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>
>>>>>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
>>>>>>
>>>>>>> On Sat, 7 May 2022 12:52:58 +0300
>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>
>>>>>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
>>>>>>>>
>>>>>>>>> The decider could never be compiled and run in the first place
>>>>>>>>> due to the category error in the definition of the proof.
>>>>>>>>
>>>>>>>> An error in the definition of the proof does not prevent
>>>>>>>> compilation and execution of the program.
>>>>>>>
>>>>>>> In this case the error in the definition of the proof
>>>>>>
>>>>>> No part of the proof is identified as errorneous, so the rest
>>>>>> is irrelevant. Anyway,
>>>>>>
>>>>>>> does prevent compilation unless
>>>>>>
>>>>>> There is no option here, so the "unless" is void.
>>>>>>
>>>>>>> the decider is made part of the program that is being decided
>>>>>>
>>>>>> The decider is made a part of the program discussed in the proof.
>>>>>>
>>>>>>> in which case we get a function call-like infinite recursion
>>>>>>> instead
>>>>>>
>>>>>> We get or we don't get, depending on how the halt decider
>>>>>> candidate attempts to decide.
>>>>>>
>>>>>>> (as described by Pete Olcott) and we are attempting (and
>>>>>>> failing) to decide if a procedure halts rather than if a
>>>>>>> program halts.
>>>>>>
>>>>>> In any case, the decider candidate fails to give the correct
>>>>>> answer, and therefore is not a halt decider. Note that this is
>>>>>> correctly inferred in the proof. Therefore either the conclusion
>>>>>> of the proof is correct or you are wrong.
>>>>>
>>>>> No answer is ever given due to the infinite recursion.
>>>>
>>>> True about some candidates, false about others. For example, a
>>>> candidate that always says "no" is not infinitely recursive but
>>>> is wrong in the particular case discussed in the proof. But anyway,
>>>> you have confirmed the conclusion of the proof.
>>>
>>> False. The proof is not cognisant of the presence of the infinite
>>> recursion. The decider can never return an answer of halts/doesn't
>>> halt due to the infinite recursion.
>>
>> The correct term is decider candidate. It is not a decider because
>> the infinite recursion prevents it from halting in finite time.
>>
>> You cannot call the proof incorrect merely because it agrees with you.
>
> Try actually reading what Strachey wrote: a contradiction arises based
> on the result of evaluating T[P] however T[P] is never evaluated due to
> the infinite recursion: a fact ignored by the proof.

If you can prove that the program does give the correct result in that
case you have proven Strachey wrong. Otherwise you haven't.

Mikko

Re: On recursion and infinite recursion (reprise #3)

<20220508094913.00002e7c@reddwarf.jmc>

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From: flib...@reddwarf.jmc (Mr Flibble)
Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
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 by: Mr Flibble - Sun, 8 May 2022 08:49 UTC

On Sun, 8 May 2022 11:31:08 +0300
Mikko <mikko.levanto@iki.fi> wrote:

> On 2022-05-07 15:19:01 +0000, Mr Flibble said:
>
> > On Sat, 7 May 2022 18:13:08 +0300
> > Mikko <mikko.levanto@iki.fi> wrote:
> >
> >> On 2022-05-07 14:20:17 +0000, Mr Flibble said:
> >>
> >>> On Sat, 7 May 2022 17:16:31 +0300
> >>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>
> >>>> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
> >>>>
> >>>>> On Sat, 7 May 2022 16:59:55 +0300
> >>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>
> >>>>>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
> >>>>>>
> >>>>>>> On Sat, 7 May 2022 12:52:58 +0300
> >>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>>>
> >>>>>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
> >>>>>>>>
> >>>>>>>>> The decider could never be compiled and run in the first
> >>>>>>>>> place due to the category error in the definition of the
> >>>>>>>>> proof.
> >>>>>>>>
> >>>>>>>> An error in the definition of the proof does not prevent
> >>>>>>>> compilation and execution of the program.
> >>>>>>>
> >>>>>>> In this case the error in the definition of the proof
> >>>>>>
> >>>>>> No part of the proof is identified as errorneous, so the rest
> >>>>>> is irrelevant. Anyway,
> >>>>>>
> >>>>>>> does prevent compilation unless
> >>>>>>
> >>>>>> There is no option here, so the "unless" is void.
> >>>>>>
> >>>>>>> the decider is made part of the program that is being decided
> >>>>>>>
> >>>>>>
> >>>>>> The decider is made a part of the program discussed in the
> >>>>>> proof.
> >>>>>>> in which case we get a function call-like infinite recursion
> >>>>>>> instead
> >>>>>>
> >>>>>> We get or we don't get, depending on how the halt decider
> >>>>>> candidate attempts to decide.
> >>>>>>
> >>>>>>> (as described by Pete Olcott) and we are attempting (and
> >>>>>>> failing) to decide if a procedure halts rather than if a
> >>>>>>> program halts.
> >>>>>>
> >>>>>> In any case, the decider candidate fails to give the correct
> >>>>>> answer, and therefore is not a halt decider. Note that this is
> >>>>>> correctly inferred in the proof. Therefore either the
> >>>>>> conclusion of the proof is correct or you are wrong.
> >>>>>
> >>>>> No answer is ever given due to the infinite recursion.
> >>>>
> >>>> True about some candidates, false about others. For example, a
> >>>> candidate that always says "no" is not infinitely recursive but
> >>>> is wrong in the particular case discussed in the proof. But
> >>>> anyway, you have confirmed the conclusion of the proof.
> >>>
> >>> False. The proof is not cognisant of the presence of the infinite
> >>> recursion. The decider can never return an answer of
> >>> halts/doesn't halt due to the infinite recursion.
> >>
> >> The correct term is decider candidate. It is not a decider because
> >> the infinite recursion prevents it from halting in finite time.
> >>
> >> You cannot call the proof incorrect merely because it agrees with
> >> you.
> >
> > Try actually reading what Strachey wrote: a contradiction arises
> > based on the result of evaluating T[P] however T[P] is never
> > evaluated due to the infinite recursion: a fact ignored by the
> > proof.
>
> If you can prove that the program does give the correct result in that
> case you have proven Strachey wrong. Otherwise you haven't.

Strachey is wrong because he neglected to account for the infinite
recursion; this should be obvious to anyone who has actually read and
understood what Strachey wrote: it seems that you haven't.

/Flibble

Re: On recursion and infinite recursion (reprise #3)

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Subject: Re: On recursion and infinite recursion (reprise #3)
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 by: Mikko - Sun, 8 May 2022 10:00 UTC

On 2022-05-08 08:49:13 +0000, Mr Flibble said:

> Strachey is wrong because he neglected to account for the infinite
> recursion; this should be obvious to anyone who has actually read and
> understood what Strachey wrote: it seems that you haven't.

Don't forget that Strachey is not presenting a new proof but merely
offering another point of view to an already existing proof. Therefore
it is not important that his presentantion be complete. More important
is that it covers the main idea in a way that is easy to understand.

Mikko

Re: On recursion and infinite recursion (reprise #3)

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 by: Richard Damon - Sun, 8 May 2022 11:45 UTC

On 5/8/22 4:49 AM, Mr Flibble wrote:
> On Sun, 8 May 2022 11:31:08 +0300
> Mikko <mikko.levanto@iki.fi> wrote:
>
>> On 2022-05-07 15:19:01 +0000, Mr Flibble said:
>>
>>> On Sat, 7 May 2022 18:13:08 +0300
>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>
>>>> On 2022-05-07 14:20:17 +0000, Mr Flibble said:
>>>>
>>>>> On Sat, 7 May 2022 17:16:31 +0300
>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>
>>>>>> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
>>>>>>
>>>>>>> On Sat, 7 May 2022 16:59:55 +0300
>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>
>>>>>>>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
>>>>>>>>
>>>>>>>>> On Sat, 7 May 2022 12:52:58 +0300
>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>>>
>>>>>>>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
>>>>>>>>>>
>>>>>>>>>>> The decider could never be compiled and run in the first
>>>>>>>>>>> place due to the category error in the definition of the
>>>>>>>>>>> proof.
>>>>>>>>>>
>>>>>>>>>> An error in the definition of the proof does not prevent
>>>>>>>>>> compilation and execution of the program.
>>>>>>>>>
>>>>>>>>> In this case the error in the definition of the proof
>>>>>>>>
>>>>>>>> No part of the proof is identified as errorneous, so the rest
>>>>>>>> is irrelevant. Anyway,
>>>>>>>>
>>>>>>>>> does prevent compilation unless
>>>>>>>>
>>>>>>>> There is no option here, so the "unless" is void.
>>>>>>>>
>>>>>>>>> the decider is made part of the program that is being decided
>>>>>>>>>
>>>>>>>>
>>>>>>>> The decider is made a part of the program discussed in the
>>>>>>>> proof.
>>>>>>>>> in which case we get a function call-like infinite recursion
>>>>>>>>> instead
>>>>>>>>
>>>>>>>> We get or we don't get, depending on how the halt decider
>>>>>>>> candidate attempts to decide.
>>>>>>>>
>>>>>>>>> (as described by Pete Olcott) and we are attempting (and
>>>>>>>>> failing) to decide if a procedure halts rather than if a
>>>>>>>>> program halts.
>>>>>>>>
>>>>>>>> In any case, the decider candidate fails to give the correct
>>>>>>>> answer, and therefore is not a halt decider. Note that this is
>>>>>>>> correctly inferred in the proof. Therefore either the
>>>>>>>> conclusion of the proof is correct or you are wrong.
>>>>>>>
>>>>>>> No answer is ever given due to the infinite recursion.
>>>>>>
>>>>>> True about some candidates, false about others. For example, a
>>>>>> candidate that always says "no" is not infinitely recursive but
>>>>>> is wrong in the particular case discussed in the proof. But
>>>>>> anyway, you have confirmed the conclusion of the proof.
>>>>>
>>>>> False. The proof is not cognisant of the presence of the infinite
>>>>> recursion. The decider can never return an answer of
>>>>> halts/doesn't halt due to the infinite recursion.
>>>>
>>>> The correct term is decider candidate. It is not a decider because
>>>> the infinite recursion prevents it from halting in finite time.
>>>>
>>>> You cannot call the proof incorrect merely because it agrees with
>>>> you.
>>>
>>> Try actually reading what Strachey wrote: a contradiction arises
>>> based on the result of evaluating T[P] however T[P] is never
>>> evaluated due to the infinite recursion: a fact ignored by the
>>> proof.
>>
>> If you can prove that the program does give the correct result in that
>> case you have proven Strachey wrong. Otherwise you haven't.
>
> Strachey is wrong because he neglected to account for the infinite
> recursion; this should be obvious to anyone who has actually read and
> understood what Strachey wrote: it seems that you haven't.
>
> /Flibble
>

No, because if T gets stuck in an infinite recursion, it is wrong,
because it failed to answer in finite time.

If T does answer in finite time, then there never was an infinite recursion.

All you have shown is that there exists a WRONG method to build a T that
gets stuck, Like a T that needs to run its input to completion to answer
about it.

Re: On recursion and infinite recursion (reprise #3)

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From: flib...@reddwarf.jmc (Mr Flibble)
Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
Message-ID: <20220508125704.00001b8f@reddwarf.jmc>
References: <20220505175034.0000253c@reddwarf.jmc>
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 by: Mr Flibble - Sun, 8 May 2022 11:57 UTC

On Sun, 8 May 2022 13:00:10 +0300
Mikko <mikko.levanto@iki.fi> wrote:

> On 2022-05-08 08:49:13 +0000, Mr Flibble said:
>
> > Strachey is wrong because he neglected to account for the infinite
> > recursion; this should be obvious to anyone who has actually read
> > and understood what Strachey wrote: it seems that you haven't.
>
> Don't forget that Strachey is not presenting a new proof but merely
> offering another point of view to an already existing proof. Therefore
> it is not important that his presentantion be complete. More important
> is that it covers the main idea in a way that is easy to understand.

It is a pretty stunning omission. No, Occams's Razor suggests it is an
oversight. The proof is invalid.

/Flibble

Re: On recursion and infinite recursion (reprise #3)

<20220508130212.00007ab6@reddwarf.jmc>

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From: flib...@reddwarf.jmc (Mr Flibble)
Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
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References: <20220505175034.0000253c@reddwarf.jmc>
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 by: Mr Flibble - Sun, 8 May 2022 12:02 UTC

On Sun, 8 May 2022 07:45:52 -0400
Richard Damon <Richard@Damon-Family.org> wrote:

> On 5/8/22 4:49 AM, Mr Flibble wrote:
> > On Sun, 8 May 2022 11:31:08 +0300
> > Mikko <mikko.levanto@iki.fi> wrote:
> >
> >> On 2022-05-07 15:19:01 +0000, Mr Flibble said:
> >>
> >>> On Sat, 7 May 2022 18:13:08 +0300
> >>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>
> >>>> On 2022-05-07 14:20:17 +0000, Mr Flibble said:
> >>>>
> >>>>> On Sat, 7 May 2022 17:16:31 +0300
> >>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>
> >>>>>> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
> >>>>>>
> >>>>>>> On Sat, 7 May 2022 16:59:55 +0300
> >>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>>>
> >>>>>>>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
> >>>>>>>>
> >>>>>>>>> On Sat, 7 May 2022 12:52:58 +0300
> >>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>>>>>
> >>>>>>>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
> >>>>>>>>>>
> >>>>>>>>>>> The decider could never be compiled and run in the first
> >>>>>>>>>>> place due to the category error in the definition of the
> >>>>>>>>>>> proof.
> >>>>>>>>>>
> >>>>>>>>>> An error in the definition of the proof does not prevent
> >>>>>>>>>> compilation and execution of the program.
> >>>>>>>>>
> >>>>>>>>> In this case the error in the definition of the proof
> >>>>>>>>
> >>>>>>>> No part of the proof is identified as errorneous, so the rest
> >>>>>>>> is irrelevant. Anyway,
> >>>>>>>>
> >>>>>>>>> does prevent compilation unless
> >>>>>>>>
> >>>>>>>> There is no option here, so the "unless" is void.
> >>>>>>>>
> >>>>>>>>> the decider is made part of the program that is being
> >>>>>>>>> decided
> >>>>>>>>
> >>>>>>>> The decider is made a part of the program discussed in the
> >>>>>>>> proof.
> >>>>>>>>> in which case we get a function call-like infinite recursion
> >>>>>>>>> instead
> >>>>>>>>
> >>>>>>>> We get or we don't get, depending on how the halt decider
> >>>>>>>> candidate attempts to decide.
> >>>>>>>>
> >>>>>>>>> (as described by Pete Olcott) and we are attempting (and
> >>>>>>>>> failing) to decide if a procedure halts rather than if a
> >>>>>>>>> program halts.
> >>>>>>>>
> >>>>>>>> In any case, the decider candidate fails to give the correct
> >>>>>>>> answer, and therefore is not a halt decider. Note that this
> >>>>>>>> is correctly inferred in the proof. Therefore either the
> >>>>>>>> conclusion of the proof is correct or you are wrong.
> >>>>>>>
> >>>>>>> No answer is ever given due to the infinite recursion.
> >>>>>>
> >>>>>> True about some candidates, false about others. For example, a
> >>>>>> candidate that always says "no" is not infinitely recursive but
> >>>>>> is wrong in the particular case discussed in the proof. But
> >>>>>> anyway, you have confirmed the conclusion of the proof.
> >>>>>
> >>>>> False. The proof is not cognisant of the presence of the
> >>>>> infinite recursion. The decider can never return an answer of
> >>>>> halts/doesn't halt due to the infinite recursion.
> >>>>
> >>>> The correct term is decider candidate. It is not a decider
> >>>> because the infinite recursion prevents it from halting in
> >>>> finite time.
> >>>>
> >>>> You cannot call the proof incorrect merely because it agrees with
> >>>> you.
> >>>
> >>> Try actually reading what Strachey wrote: a contradiction arises
> >>> based on the result of evaluating T[P] however T[P] is never
> >>> evaluated due to the infinite recursion: a fact ignored by the
> >>> proof.
> >>
> >> If you can prove that the program does give the correct result in
> >> that case you have proven Strachey wrong. Otherwise you haven't.
> >
> > Strachey is wrong because he neglected to account for the infinite
> > recursion; this should be obvious to anyone who has actually read
> > and understood what Strachey wrote: it seems that you haven't.
> >
> > /Flibble
> >
>
> No, because if T gets stuck in an infinite recursion, it is wrong,
> because it failed to answer in finite time.
>
> If T does answer in finite time, then there never was an infinite
> recursion.
>
> All you have shown is that there exists a WRONG method to build a T
> that gets stuck, Like a T that needs to run its input to completion
> to answer about it.

I have shown that [Strachey, 1965] contains an infinite recursion and
is thus invalid.

/Flibble

Re: On recursion and infinite recursion (reprise #3)

<iKOdK.8684$t72a.8070@fx10.iad>

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https://www.novabbs.com/devel/article-flat.php?id=31940&group=comp.theory#31940

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<20220507152017.00000990@reddwarf.jmc> <t562a4$9ao$1@dont-email.me>
<20220507161901.00002e54@reddwarf.jmc> <t57v4c$7do$1@dont-email.me>
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 by: Richard Damon - Sun, 8 May 2022 12:31 UTC

On 5/8/22 8:02 AM, Mr Flibble wrote:
> On Sun, 8 May 2022 07:45:52 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 5/8/22 4:49 AM, Mr Flibble wrote:
>>> On Sun, 8 May 2022 11:31:08 +0300
>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>
>>>> On 2022-05-07 15:19:01 +0000, Mr Flibble said:
>>>>
>>>>> On Sat, 7 May 2022 18:13:08 +0300
>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>
>>>>>> On 2022-05-07 14:20:17 +0000, Mr Flibble said:
>>>>>>
>>>>>>> On Sat, 7 May 2022 17:16:31 +0300
>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>
>>>>>>>> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
>>>>>>>>
>>>>>>>>> On Sat, 7 May 2022 16:59:55 +0300
>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>>>
>>>>>>>>>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
>>>>>>>>>>
>>>>>>>>>>> On Sat, 7 May 2022 12:52:58 +0300
>>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
>>>>>>>>>>>>
>>>>>>>>>>>>> The decider could never be compiled and run in the first
>>>>>>>>>>>>> place due to the category error in the definition of the
>>>>>>>>>>>>> proof.
>>>>>>>>>>>>
>>>>>>>>>>>> An error in the definition of the proof does not prevent
>>>>>>>>>>>> compilation and execution of the program.
>>>>>>>>>>>
>>>>>>>>>>> In this case the error in the definition of the proof
>>>>>>>>>>
>>>>>>>>>> No part of the proof is identified as errorneous, so the rest
>>>>>>>>>> is irrelevant. Anyway,
>>>>>>>>>>
>>>>>>>>>>> does prevent compilation unless
>>>>>>>>>>
>>>>>>>>>> There is no option here, so the "unless" is void.
>>>>>>>>>>
>>>>>>>>>>> the decider is made part of the program that is being
>>>>>>>>>>> decided
>>>>>>>>>>
>>>>>>>>>> The decider is made a part of the program discussed in the
>>>>>>>>>> proof.
>>>>>>>>>>> in which case we get a function call-like infinite recursion
>>>>>>>>>>> instead
>>>>>>>>>>
>>>>>>>>>> We get or we don't get, depending on how the halt decider
>>>>>>>>>> candidate attempts to decide.
>>>>>>>>>>
>>>>>>>>>>> (as described by Pete Olcott) and we are attempting (and
>>>>>>>>>>> failing) to decide if a procedure halts rather than if a
>>>>>>>>>>> program halts.
>>>>>>>>>>
>>>>>>>>>> In any case, the decider candidate fails to give the correct
>>>>>>>>>> answer, and therefore is not a halt decider. Note that this
>>>>>>>>>> is correctly inferred in the proof. Therefore either the
>>>>>>>>>> conclusion of the proof is correct or you are wrong.
>>>>>>>>>
>>>>>>>>> No answer is ever given due to the infinite recursion.
>>>>>>>>
>>>>>>>> True about some candidates, false about others. For example, a
>>>>>>>> candidate that always says "no" is not infinitely recursive but
>>>>>>>> is wrong in the particular case discussed in the proof. But
>>>>>>>> anyway, you have confirmed the conclusion of the proof.
>>>>>>>
>>>>>>> False. The proof is not cognisant of the presence of the
>>>>>>> infinite recursion. The decider can never return an answer of
>>>>>>> halts/doesn't halt due to the infinite recursion.
>>>>>>
>>>>>> The correct term is decider candidate. It is not a decider
>>>>>> because the infinite recursion prevents it from halting in
>>>>>> finite time.
>>>>>>
>>>>>> You cannot call the proof incorrect merely because it agrees with
>>>>>> you.
>>>>>
>>>>> Try actually reading what Strachey wrote: a contradiction arises
>>>>> based on the result of evaluating T[P] however T[P] is never
>>>>> evaluated due to the infinite recursion: a fact ignored by the
>>>>> proof.
>>>>
>>>> If you can prove that the program does give the correct result in
>>>> that case you have proven Strachey wrong. Otherwise you haven't.
>>>
>>> Strachey is wrong because he neglected to account for the infinite
>>> recursion; this should be obvious to anyone who has actually read
>>> and understood what Strachey wrote: it seems that you haven't.
>>>
>>> /Flibble
>>>
>>
>> No, because if T gets stuck in an infinite recursion, it is wrong,
>> because it failed to answer in finite time.
>>
>> If T does answer in finite time, then there never was an infinite
>> recursion.
>>
>> All you have shown is that there exists a WRONG method to build a T
>> that gets stuck, Like a T that needs to run its input to completion
>> to answer about it.
>
> I have shown that [Strachey, 1965] contains an infinite recursion and
> is thus invalid.
>
> /Flibble
>

No, you haven't. All you have shown is that one way to attempt to make
the program that Strachev says doesn't exist, fails due to getting
caught in infinite recursion.

That just helps confirm Strachev, not refute it.

Re: On recursion and infinite recursion (reprise #3)

<20220508133909.00007b6b@reddwarf.jmc>

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From: flib...@reddwarf.jmc (Mr Flibble)
Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
Message-ID: <20220508133909.00007b6b@reddwarf.jmc>
References: <20220505175034.0000253c@reddwarf.jmc> <5O%cK.17$wYy9.0@fx11.iad> <20220506150253.00007c7f@reddwarf.jmc> <t55fhq$u5e$1@dont-email.me> <20220507134250.00007acc@reddwarf.jmc> <t55u0r$7lf$1@dont-email.me> <20220507150612.00003fab@reddwarf.jmc> <t55uvv$fpf$1@dont-email.me> <20220507152017.00000990@reddwarf.jmc> <t562a4$9ao$1@dont-email.me> <20220507161901.00002e54@reddwarf.jmc> <t57v4c$7do$1@dont-email.me> <20220508094913.00002e7c@reddwarf.jmc> <Q3OdK.15730$Bm21.3438@fx07.iad> <20220508130212.00007ab6@reddwarf.jmc> <iKOdK.8684$t72a.8070@fx10.iad>
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 by: Mr Flibble - Sun, 8 May 2022 12:39 UTC

On Sun, 8 May 2022 08:31:10 -0400
Richard Damon <Richard@Damon-Family.org> wrote:

> On 5/8/22 8:02 AM, Mr Flibble wrote:
> > On Sun, 8 May 2022 07:45:52 -0400
> > Richard Damon <Richard@Damon-Family.org> wrote:
> >
> >> On 5/8/22 4:49 AM, Mr Flibble wrote:
> >>> On Sun, 8 May 2022 11:31:08 +0300
> >>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>
> >>>> On 2022-05-07 15:19:01 +0000, Mr Flibble said:
> >>>>
> >>>>> On Sat, 7 May 2022 18:13:08 +0300
> >>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>
> >>>>>> On 2022-05-07 14:20:17 +0000, Mr Flibble said:
> >>>>>>
> >>>>>>> On Sat, 7 May 2022 17:16:31 +0300
> >>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>>>
> >>>>>>>> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
> >>>>>>>>
> >>>>>>>>> On Sat, 7 May 2022 16:59:55 +0300
> >>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>>>>>
> >>>>>>>>>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
> >>>>>>>>>>
> >>>>>>>>>>> On Sat, 7 May 2022 12:52:58 +0300
> >>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>>>>>>>
> >>>>>>>>>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
> >>>>>>>>>>>>
> >>>>>>>>>>>>> The decider could never be compiled and run in the first
> >>>>>>>>>>>>> place due to the category error in the definition of the
> >>>>>>>>>>>>> proof.
> >>>>>>>>>>>>
> >>>>>>>>>>>> An error in the definition of the proof does not prevent
> >>>>>>>>>>>> compilation and execution of the program.
> >>>>>>>>>>>
> >>>>>>>>>>> In this case the error in the definition of the proof
> >>>>>>>>>>
> >>>>>>>>>> No part of the proof is identified as errorneous, so the
> >>>>>>>>>> rest is irrelevant. Anyway,
> >>>>>>>>>>
> >>>>>>>>>>> does prevent compilation unless
> >>>>>>>>>>
> >>>>>>>>>> There is no option here, so the "unless" is void.
> >>>>>>>>>>
> >>>>>>>>>>> the decider is made part of the program that is being
> >>>>>>>>>>> decided
> >>>>>>>>>>
> >>>>>>>>>> The decider is made a part of the program discussed in the
> >>>>>>>>>> proof.
> >>>>>>>>>>> in which case we get a function call-like infinite
> >>>>>>>>>>> recursion instead
> >>>>>>>>>>
> >>>>>>>>>> We get or we don't get, depending on how the halt decider
> >>>>>>>>>> candidate attempts to decide.
> >>>>>>>>>>
> >>>>>>>>>>> (as described by Pete Olcott) and we are attempting (and
> >>>>>>>>>>> failing) to decide if a procedure halts rather than if a
> >>>>>>>>>>> program halts.
> >>>>>>>>>>
> >>>>>>>>>> In any case, the decider candidate fails to give the
> >>>>>>>>>> correct answer, and therefore is not a halt decider. Note
> >>>>>>>>>> that this is correctly inferred in the proof. Therefore
> >>>>>>>>>> either the conclusion of the proof is correct or you are
> >>>>>>>>>> wrong.
> >>>>>>>>>
> >>>>>>>>> No answer is ever given due to the infinite recursion.
> >>>>>>>>
> >>>>>>>> True about some candidates, false about others. For example,
> >>>>>>>> a candidate that always says "no" is not infinitely
> >>>>>>>> recursive but is wrong in the particular case discussed in
> >>>>>>>> the proof. But anyway, you have confirmed the conclusion of
> >>>>>>>> the proof.
> >>>>>>>
> >>>>>>> False. The proof is not cognisant of the presence of the
> >>>>>>> infinite recursion. The decider can never return an answer of
> >>>>>>> halts/doesn't halt due to the infinite recursion.
> >>>>>>
> >>>>>> The correct term is decider candidate. It is not a decider
> >>>>>> because the infinite recursion prevents it from halting in
> >>>>>> finite time.
> >>>>>>
> >>>>>> You cannot call the proof incorrect merely because it agrees
> >>>>>> with you.
> >>>>>
> >>>>> Try actually reading what Strachey wrote: a contradiction arises
> >>>>> based on the result of evaluating T[P] however T[P] is never
> >>>>> evaluated due to the infinite recursion: a fact ignored by the
> >>>>> proof.
> >>>>
> >>>> If you can prove that the program does give the correct result in
> >>>> that case you have proven Strachey wrong. Otherwise you haven't.
> >>>>
> >>>
> >>> Strachey is wrong because he neglected to account for the infinite
> >>> recursion; this should be obvious to anyone who has actually read
> >>> and understood what Strachey wrote: it seems that you haven't.
> >>>
> >>> /Flibble
> >>>
> >>
> >> No, because if T gets stuck in an infinite recursion, it is wrong,
> >> because it failed to answer in finite time.
> >>
> >> If T does answer in finite time, then there never was an infinite
> >> recursion.
> >>
> >> All you have shown is that there exists a WRONG method to build a T
> >> that gets stuck, Like a T that needs to run its input to completion
> >> to answer about it.
> >
> > I have shown that [Strachey, 1965] contains an infinite recursion
> > and is thus invalid.
> >
> > /Flibble
> >
>
> No, you haven't. All you have shown is that one way to attempt to
> make the program that Strachev says doesn't exist, fails due to
> getting caught in infinite recursion.
>
> That just helps confirm Strachev, not refute it.

Nope. Strachey's proof is based on a contradiction relating to
evaluating the result of T[P] however T[P] can never be evaluated if
there is an infinite recursion.

/Flibble

Re: On recursion and infinite recursion (reprise #3)

<9ITdK.2055$wYy9.1130@fx11.iad>

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 by: Richard Damon - Sun, 8 May 2022 18:10 UTC

On 5/8/22 8:39 AM, Mr Flibble wrote:
> On Sun, 8 May 2022 08:31:10 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 5/8/22 8:02 AM, Mr Flibble wrote:
>>> On Sun, 8 May 2022 07:45:52 -0400
>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>
>>>> On 5/8/22 4:49 AM, Mr Flibble wrote:
>>>>> On Sun, 8 May 2022 11:31:08 +0300
>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>
>>>>>> On 2022-05-07 15:19:01 +0000, Mr Flibble said:
>>>>>>
>>>>>>> On Sat, 7 May 2022 18:13:08 +0300
>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>
>>>>>>>> On 2022-05-07 14:20:17 +0000, Mr Flibble said:
>>>>>>>>
>>>>>>>>> On Sat, 7 May 2022 17:16:31 +0300
>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>>>
>>>>>>>>>> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
>>>>>>>>>>
>>>>>>>>>>> On Sat, 7 May 2022 16:59:55 +0300
>>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
>>>>>>>>>>>>
>>>>>>>>>>>>> On Sat, 7 May 2022 12:52:58 +0300
>>>>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> The decider could never be compiled and run in the first
>>>>>>>>>>>>>>> place due to the category error in the definition of the
>>>>>>>>>>>>>>> proof.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> An error in the definition of the proof does not prevent
>>>>>>>>>>>>>> compilation and execution of the program.
>>>>>>>>>>>>>
>>>>>>>>>>>>> In this case the error in the definition of the proof
>>>>>>>>>>>>
>>>>>>>>>>>> No part of the proof is identified as errorneous, so the
>>>>>>>>>>>> rest is irrelevant. Anyway,
>>>>>>>>>>>>
>>>>>>>>>>>>> does prevent compilation unless
>>>>>>>>>>>>
>>>>>>>>>>>> There is no option here, so the "unless" is void.
>>>>>>>>>>>>
>>>>>>>>>>>>> the decider is made part of the program that is being
>>>>>>>>>>>>> decided
>>>>>>>>>>>>
>>>>>>>>>>>> The decider is made a part of the program discussed in the
>>>>>>>>>>>> proof.
>>>>>>>>>>>>> in which case we get a function call-like infinite
>>>>>>>>>>>>> recursion instead
>>>>>>>>>>>>
>>>>>>>>>>>> We get or we don't get, depending on how the halt decider
>>>>>>>>>>>> candidate attempts to decide.
>>>>>>>>>>>>
>>>>>>>>>>>>> (as described by Pete Olcott) and we are attempting (and
>>>>>>>>>>>>> failing) to decide if a procedure halts rather than if a
>>>>>>>>>>>>> program halts.
>>>>>>>>>>>>
>>>>>>>>>>>> In any case, the decider candidate fails to give the
>>>>>>>>>>>> correct answer, and therefore is not a halt decider. Note
>>>>>>>>>>>> that this is correctly inferred in the proof. Therefore
>>>>>>>>>>>> either the conclusion of the proof is correct or you are
>>>>>>>>>>>> wrong.
>>>>>>>>>>>
>>>>>>>>>>> No answer is ever given due to the infinite recursion.
>>>>>>>>>>
>>>>>>>>>> True about some candidates, false about others. For example,
>>>>>>>>>> a candidate that always says "no" is not infinitely
>>>>>>>>>> recursive but is wrong in the particular case discussed in
>>>>>>>>>> the proof. But anyway, you have confirmed the conclusion of
>>>>>>>>>> the proof.
>>>>>>>>>
>>>>>>>>> False. The proof is not cognisant of the presence of the
>>>>>>>>> infinite recursion. The decider can never return an answer of
>>>>>>>>> halts/doesn't halt due to the infinite recursion.
>>>>>>>>
>>>>>>>> The correct term is decider candidate. It is not a decider
>>>>>>>> because the infinite recursion prevents it from halting in
>>>>>>>> finite time.
>>>>>>>>
>>>>>>>> You cannot call the proof incorrect merely because it agrees
>>>>>>>> with you.
>>>>>>>
>>>>>>> Try actually reading what Strachey wrote: a contradiction arises
>>>>>>> based on the result of evaluating T[P] however T[P] is never
>>>>>>> evaluated due to the infinite recursion: a fact ignored by the
>>>>>>> proof.
>>>>>>
>>>>>> If you can prove that the program does give the correct result in
>>>>>> that case you have proven Strachey wrong. Otherwise you haven't.
>>>>>>
>>>>>
>>>>> Strachey is wrong because he neglected to account for the infinite
>>>>> recursion; this should be obvious to anyone who has actually read
>>>>> and understood what Strachey wrote: it seems that you haven't.
>>>>>
>>>>> /Flibble
>>>>>
>>>>
>>>> No, because if T gets stuck in an infinite recursion, it is wrong,
>>>> because it failed to answer in finite time.
>>>>
>>>> If T does answer in finite time, then there never was an infinite
>>>> recursion.
>>>>
>>>> All you have shown is that there exists a WRONG method to build a T
>>>> that gets stuck, Like a T that needs to run its input to completion
>>>> to answer about it.
>>>
>>> I have shown that [Strachey, 1965] contains an infinite recursion
>>> and is thus invalid.
>>>
>>> /Flibble
>>>
>>
>> No, you haven't. All you have shown is that one way to attempt to
>> make the program that Strachev says doesn't exist, fails due to
>> getting caught in infinite recursion.
>>
>> That just helps confirm Strachev, not refute it.
>
> Nope. Strachey's proof is based on a contradiction relating to
> evaluating the result of T[P] however T[P] can never be evaluated if
> there is an infinite recursion.
>
> /Flibble
>

Nope, because if T actually meets the requirement to be able to take ANY
program and answer, then it need to be able to take a program that uses
a copy of T in it, as that is a valid program.

If T gets into an infinite recursion on such a program, it is that *T*
fails to meet the requirements, not that the such a program is invalid
to give to T.

This means that the program T is Strachey's proof can't just execute its
input to get the needed answer, as that makes it (T) not meet its
requirements.

Thus, the "recursion" error isn't in the proof of impossibility, but in
the candidate T that is attempting to be the counter example for the proof.

Recursion is allowed in programming and in math. If you try to just ban
it, you will find you logic can't handle what it needs to.


Click here to read the complete article
Re: On recursion and infinite recursion (reprise #3)

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https://www.novabbs.com/devel/article-flat.php?id=31951&group=comp.theory#31951

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From: flib...@reddwarf.jmc (Mr Flibble)
Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
Message-ID: <20220508200637.0000491c@reddwarf.jmc>
References: <20220505175034.0000253c@reddwarf.jmc> <5O%cK.17$wYy9.0@fx11.iad> <20220506150253.00007c7f@reddwarf.jmc> <t55fhq$u5e$1@dont-email.me> <20220507134250.00007acc@reddwarf.jmc> <t55u0r$7lf$1@dont-email.me> <20220507150612.00003fab@reddwarf.jmc> <t55uvv$fpf$1@dont-email.me> <20220507152017.00000990@reddwarf.jmc> <t562a4$9ao$1@dont-email.me> <20220507161901.00002e54@reddwarf.jmc> <t57v4c$7do$1@dont-email.me> <20220508094913.00002e7c@reddwarf.jmc> <Q3OdK.15730$Bm21.3438@fx07.iad> <20220508130212.00007ab6@reddwarf.jmc> <iKOdK.8684$t72a.8070@fx10.iad> <20220508133909.00007b6b@reddwarf.jmc> <9ITdK.2055$wYy9.1130@fx11.iad>
Organization: Jupiter Mining Corp
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Date: Sun, 8 May 2022 20:06:37 +0100
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 by: Mr Flibble - Sun, 8 May 2022 19:06 UTC

On Sun, 8 May 2022 14:10:13 -0400
Richard Damon <Richard@Damon-Family.org> wrote:

> On 5/8/22 8:39 AM, Mr Flibble wrote:
> > On Sun, 8 May 2022 08:31:10 -0400
> > Richard Damon <Richard@Damon-Family.org> wrote:
> >
> >> On 5/8/22 8:02 AM, Mr Flibble wrote:
> >>> On Sun, 8 May 2022 07:45:52 -0400
> >>> Richard Damon <Richard@Damon-Family.org> wrote:
> >>>
> >>>> On 5/8/22 4:49 AM, Mr Flibble wrote:
> >>>>> On Sun, 8 May 2022 11:31:08 +0300
> >>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>
> >>>>>> On 2022-05-07 15:19:01 +0000, Mr Flibble said:
> >>>>>>
> >>>>>>> On Sat, 7 May 2022 18:13:08 +0300
> >>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>>>
> >>>>>>>> On 2022-05-07 14:20:17 +0000, Mr Flibble said:
> >>>>>>>>
> >>>>>>>>> On Sat, 7 May 2022 17:16:31 +0300
> >>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>>>>>
> >>>>>>>>>> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
> >>>>>>>>>>
> >>>>>>>>>>> On Sat, 7 May 2022 16:59:55 +0300
> >>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>>>>>>>
> >>>>>>>>>>>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
> >>>>>>>>>>>>
> >>>>>>>>>>>>> On Sat, 7 May 2022 12:52:58 +0300
> >>>>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>>>>>>>>>
> >>>>>>>>>>>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>>> The decider could never be compiled and run in the
> >>>>>>>>>>>>>>> first place due to the category error in the
> >>>>>>>>>>>>>>> definition of the proof.
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> An error in the definition of the proof does not
> >>>>>>>>>>>>>> prevent compilation and execution of the program.
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> In this case the error in the definition of the proof
> >>>>>>>>>>>>
> >>>>>>>>>>>> No part of the proof is identified as errorneous, so the
> >>>>>>>>>>>> rest is irrelevant. Anyway,
> >>>>>>>>>>>>
> >>>>>>>>>>>>> does prevent compilation unless
> >>>>>>>>>>>>
> >>>>>>>>>>>> There is no option here, so the "unless" is void.
> >>>>>>>>>>>>
> >>>>>>>>>>>>> the decider is made part of the program that is being
> >>>>>>>>>>>>> decided
> >>>>>>>>>>>>
> >>>>>>>>>>>> The decider is made a part of the program discussed in
> >>>>>>>>>>>> the proof.
> >>>>>>>>>>>>> in which case we get a function call-like infinite
> >>>>>>>>>>>>> recursion instead
> >>>>>>>>>>>>
> >>>>>>>>>>>> We get or we don't get, depending on how the halt decider
> >>>>>>>>>>>> candidate attempts to decide.
> >>>>>>>>>>>>
> >>>>>>>>>>>>> (as described by Pete Olcott) and we are attempting (and
> >>>>>>>>>>>>> failing) to decide if a procedure halts rather than if a
> >>>>>>>>>>>>> program halts.
> >>>>>>>>>>>>
> >>>>>>>>>>>> In any case, the decider candidate fails to give the
> >>>>>>>>>>>> correct answer, and therefore is not a halt decider. Note
> >>>>>>>>>>>> that this is correctly inferred in the proof. Therefore
> >>>>>>>>>>>> either the conclusion of the proof is correct or you are
> >>>>>>>>>>>> wrong.
> >>>>>>>>>>>
> >>>>>>>>>>> No answer is ever given due to the infinite recursion.
> >>>>>>>>>>
> >>>>>>>>>> True about some candidates, false about others. For
> >>>>>>>>>> example, a candidate that always says "no" is not
> >>>>>>>>>> infinitely recursive but is wrong in the particular case
> >>>>>>>>>> discussed in the proof. But anyway, you have confirmed the
> >>>>>>>>>> conclusion of the proof.
> >>>>>>>>>
> >>>>>>>>> False. The proof is not cognisant of the presence of the
> >>>>>>>>> infinite recursion. The decider can never return an answer
> >>>>>>>>> of halts/doesn't halt due to the infinite recursion.
> >>>>>>>>
> >>>>>>>> The correct term is decider candidate. It is not a decider
> >>>>>>>> because the infinite recursion prevents it from halting in
> >>>>>>>> finite time.
> >>>>>>>>
> >>>>>>>> You cannot call the proof incorrect merely because it agrees
> >>>>>>>> with you.
> >>>>>>>
> >>>>>>> Try actually reading what Strachey wrote: a contradiction
> >>>>>>> arises based on the result of evaluating T[P] however T[P] is
> >>>>>>> never evaluated due to the infinite recursion: a fact ignored
> >>>>>>> by the proof.
> >>>>>>
> >>>>>> If you can prove that the program does give the correct result
> >>>>>> in that case you have proven Strachey wrong. Otherwise you
> >>>>>> haven't.
> >>>>>
> >>>>> Strachey is wrong because he neglected to account for the
> >>>>> infinite recursion; this should be obvious to anyone who has
> >>>>> actually read and understood what Strachey wrote: it seems that
> >>>>> you haven't.
> >>>>>
> >>>>> /Flibble
> >>>>>
> >>>>
> >>>> No, because if T gets stuck in an infinite recursion, it is
> >>>> wrong, because it failed to answer in finite time.
> >>>>
> >>>> If T does answer in finite time, then there never was an infinite
> >>>> recursion.
> >>>>
> >>>> All you have shown is that there exists a WRONG method to build
> >>>> a T that gets stuck, Like a T that needs to run its input to
> >>>> completion to answer about it.
> >>>
> >>> I have shown that [Strachey, 1965] contains an infinite recursion
> >>> and is thus invalid.
> >>>
> >>> /Flibble
> >>>
> >>
> >> No, you haven't. All you have shown is that one way to attempt to
> >> make the program that Strachev says doesn't exist, fails due to
> >> getting caught in infinite recursion.
> >>
> >> That just helps confirm Strachev, not refute it.
> >
> > Nope. Strachey's proof is based on a contradiction relating to
> > evaluating the result of T[P] however T[P] can never be evaluated if
> > there is an infinite recursion.
> >
> > /Flibble
> >
>
> Nope, because if T actually meets the requirement to be able to take
> ANY program and answer, then it need to be able to take a program
> that uses a copy of T in it, as that is a valid program.
>
> If T gets into an infinite recursion on such a program, it is that
> *T* fails to meet the requirements, not that the such a program is
> invalid to give to T.
>
> This means that the program T is Strachey's proof can't just execute
> its input to get the needed answer, as that makes it (T) not meet its
> requirements.
>
> Thus, the "recursion" error isn't in the proof of impossibility, but
> in the candidate T that is attempting to be the counter example for
> the proof.
>
> Recursion is allowed in programming and in math. If you try to just
> ban it, you will find you logic can't handle what it needs to.


Click here to read the complete article
Re: On recursion and infinite recursion (reprise #3)

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References: <20220505175034.0000253c@reddwarf.jmc> <5O%cK.17$wYy9.0@fx11.iad>
<20220506150253.00007c7f@reddwarf.jmc> <t55fhq$u5e$1@dont-email.me>
<20220507134250.00007acc@reddwarf.jmc> <t55u0r$7lf$1@dont-email.me>
<20220507150612.00003fab@reddwarf.jmc> <t55uvv$fpf$1@dont-email.me>
<20220507152017.00000990@reddwarf.jmc> <t562a4$9ao$1@dont-email.me>
<20220507161901.00002e54@reddwarf.jmc> <t57v4c$7do$1@dont-email.me>
<20220508094913.00002e7c@reddwarf.jmc> <Q3OdK.15730$Bm21.3438@fx07.iad>
<20220508130212.00007ab6@reddwarf.jmc> <iKOdK.8684$t72a.8070@fx10.iad>
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<20220508200637.0000491c@reddwarf.jmc>
From: Rich...@Damon-Family.org (Richard Damon)
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Date: Sun, 8 May 2022 15:39:32 -0400
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 by: Richard Damon - Sun, 8 May 2022 19:39 UTC

On 5/8/22 3:06 PM, Mr Flibble wrote:
> On Sun, 8 May 2022 14:10:13 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 5/8/22 8:39 AM, Mr Flibble wrote:
>>> On Sun, 8 May 2022 08:31:10 -0400
>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>
>>>> On 5/8/22 8:02 AM, Mr Flibble wrote:
>>>>> On Sun, 8 May 2022 07:45:52 -0400
>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>
>>>>>> On 5/8/22 4:49 AM, Mr Flibble wrote:
>>>>>>> On Sun, 8 May 2022 11:31:08 +0300
>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>
>>>>>>>> On 2022-05-07 15:19:01 +0000, Mr Flibble said:
>>>>>>>>
>>>>>>>>> On Sat, 7 May 2022 18:13:08 +0300
>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>>>
>>>>>>>>>> On 2022-05-07 14:20:17 +0000, Mr Flibble said:
>>>>>>>>>>
>>>>>>>>>>> On Sat, 7 May 2022 17:16:31 +0300
>>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
>>>>>>>>>>>>
>>>>>>>>>>>>> On Sat, 7 May 2022 16:59:55 +0300
>>>>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On Sat, 7 May 2022 12:52:58 +0300
>>>>>>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> The decider could never be compiled and run in the
>>>>>>>>>>>>>>>>> first place due to the category error in the
>>>>>>>>>>>>>>>>> definition of the proof.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> An error in the definition of the proof does not
>>>>>>>>>>>>>>>> prevent compilation and execution of the program.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> In this case the error in the definition of the proof
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> No part of the proof is identified as errorneous, so the
>>>>>>>>>>>>>> rest is irrelevant. Anyway,
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> does prevent compilation unless
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> There is no option here, so the "unless" is void.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> the decider is made part of the program that is being
>>>>>>>>>>>>>>> decided
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The decider is made a part of the program discussed in
>>>>>>>>>>>>>> the proof.
>>>>>>>>>>>>>>> in which case we get a function call-like infinite
>>>>>>>>>>>>>>> recursion instead
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> We get or we don't get, depending on how the halt decider
>>>>>>>>>>>>>> candidate attempts to decide.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> (as described by Pete Olcott) and we are attempting (and
>>>>>>>>>>>>>>> failing) to decide if a procedure halts rather than if a
>>>>>>>>>>>>>>> program halts.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> In any case, the decider candidate fails to give the
>>>>>>>>>>>>>> correct answer, and therefore is not a halt decider. Note
>>>>>>>>>>>>>> that this is correctly inferred in the proof. Therefore
>>>>>>>>>>>>>> either the conclusion of the proof is correct or you are
>>>>>>>>>>>>>> wrong.
>>>>>>>>>>>>>
>>>>>>>>>>>>> No answer is ever given due to the infinite recursion.
>>>>>>>>>>>>
>>>>>>>>>>>> True about some candidates, false about others. For
>>>>>>>>>>>> example, a candidate that always says "no" is not
>>>>>>>>>>>> infinitely recursive but is wrong in the particular case
>>>>>>>>>>>> discussed in the proof. But anyway, you have confirmed the
>>>>>>>>>>>> conclusion of the proof.
>>>>>>>>>>>
>>>>>>>>>>> False. The proof is not cognisant of the presence of the
>>>>>>>>>>> infinite recursion. The decider can never return an answer
>>>>>>>>>>> of halts/doesn't halt due to the infinite recursion.
>>>>>>>>>>
>>>>>>>>>> The correct term is decider candidate. It is not a decider
>>>>>>>>>> because the infinite recursion prevents it from halting in
>>>>>>>>>> finite time.
>>>>>>>>>>
>>>>>>>>>> You cannot call the proof incorrect merely because it agrees
>>>>>>>>>> with you.
>>>>>>>>>
>>>>>>>>> Try actually reading what Strachey wrote: a contradiction
>>>>>>>>> arises based on the result of evaluating T[P] however T[P] is
>>>>>>>>> never evaluated due to the infinite recursion: a fact ignored
>>>>>>>>> by the proof.
>>>>>>>>
>>>>>>>> If you can prove that the program does give the correct result
>>>>>>>> in that case you have proven Strachey wrong. Otherwise you
>>>>>>>> haven't.
>>>>>>>
>>>>>>> Strachey is wrong because he neglected to account for the
>>>>>>> infinite recursion; this should be obvious to anyone who has
>>>>>>> actually read and understood what Strachey wrote: it seems that
>>>>>>> you haven't.
>>>>>>>
>>>>>>> /Flibble
>>>>>>>
>>>>>>
>>>>>> No, because if T gets stuck in an infinite recursion, it is
>>>>>> wrong, because it failed to answer in finite time.
>>>>>>
>>>>>> If T does answer in finite time, then there never was an infinite
>>>>>> recursion.
>>>>>>
>>>>>> All you have shown is that there exists a WRONG method to build
>>>>>> a T that gets stuck, Like a T that needs to run its input to
>>>>>> completion to answer about it.
>>>>>
>>>>> I have shown that [Strachey, 1965] contains an infinite recursion
>>>>> and is thus invalid.
>>>>>
>>>>> /Flibble
>>>>>
>>>>
>>>> No, you haven't. All you have shown is that one way to attempt to
>>>> make the program that Strachev says doesn't exist, fails due to
>>>> getting caught in infinite recursion.
>>>>
>>>> That just helps confirm Strachev, not refute it.
>>>
>>> Nope. Strachey's proof is based on a contradiction relating to
>>> evaluating the result of T[P] however T[P] can never be evaluated if
>>> there is an infinite recursion.
>>>
>>> /Flibble
>>>
>>
>> Nope, because if T actually meets the requirement to be able to take
>> ANY program and answer, then it need to be able to take a program
>> that uses a copy of T in it, as that is a valid program.
>>
>> If T gets into an infinite recursion on such a program, it is that
>> *T* fails to meet the requirements, not that the such a program is
>> invalid to give to T.
>>
>> This means that the program T is Strachey's proof can't just execute
>> its input to get the needed answer, as that makes it (T) not meet its
>> requirements.
>>
>> Thus, the "recursion" error isn't in the proof of impossibility, but
>> in the candidate T that is attempting to be the counter example for
>> the proof.
>>
>> Recursion is allowed in programming and in math. If you try to just
>> ban it, you will find you logic can't handle what it needs to.
>
> You are simply wrong, and fractally so. Try reading what Strachey
> actually wrote.
>
> /Flibble
>


Click here to read the complete article
Re: On recursion and infinite recursion (reprise #3)

<20220508211421.000031a0@reddwarf.jmc>

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https://www.novabbs.com/devel/article-flat.php?id=31955&group=comp.theory#31955

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From: flib...@reddwarf.jmc (Mr Flibble)
Newsgroups: comp.theory
Subject: Re: On recursion and infinite recursion (reprise #3)
Message-ID: <20220508211421.000031a0@reddwarf.jmc>
References: <20220505175034.0000253c@reddwarf.jmc>
<5O%cK.17$wYy9.0@fx11.iad>
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 by: Mr Flibble - Sun, 8 May 2022 20:14 UTC

On Sun, 8 May 2022 15:39:32 -0400
Richard Damon <Richard@Damon-Family.org> wrote:

> On 5/8/22 3:06 PM, Mr Flibble wrote:
> > On Sun, 8 May 2022 14:10:13 -0400
> > Richard Damon <Richard@Damon-Family.org> wrote:
> >
> >> On 5/8/22 8:39 AM, Mr Flibble wrote:
> >>> On Sun, 8 May 2022 08:31:10 -0400
> >>> Richard Damon <Richard@Damon-Family.org> wrote:
> >>>
> >>>> On 5/8/22 8:02 AM, Mr Flibble wrote:
> >>>>> On Sun, 8 May 2022 07:45:52 -0400
> >>>>> Richard Damon <Richard@Damon-Family.org> wrote:
> >>>>>
> >>>>>> On 5/8/22 4:49 AM, Mr Flibble wrote:
> >>>>>>> On Sun, 8 May 2022 11:31:08 +0300
> >>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>>>
> >>>>>>>> On 2022-05-07 15:19:01 +0000, Mr Flibble said:
> >>>>>>>>
> >>>>>>>>> On Sat, 7 May 2022 18:13:08 +0300
> >>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>>>>>
> >>>>>>>>>> On 2022-05-07 14:20:17 +0000, Mr Flibble said:
> >>>>>>>>>>
> >>>>>>>>>>> On Sat, 7 May 2022 17:16:31 +0300
> >>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>>>>>>>
> >>>>>>>>>>>> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
> >>>>>>>>>>>>
> >>>>>>>>>>>>> On Sat, 7 May 2022 16:59:55 +0300
> >>>>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>>>>>>>>>
> >>>>>>>>>>>>>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>>> On Sat, 7 May 2022 12:52:58 +0300
> >>>>>>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
> >>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>> The decider could never be compiled and run in the
> >>>>>>>>>>>>>>>>> first place due to the category error in the
> >>>>>>>>>>>>>>>>> definition of the proof.
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>> An error in the definition of the proof does not
> >>>>>>>>>>>>>>>> prevent compilation and execution of the program.
> >>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>> In this case the error in the definition of the proof
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> No part of the proof is identified as errorneous, so
> >>>>>>>>>>>>>> the rest is irrelevant. Anyway,
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>>> does prevent compilation unless
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> There is no option here, so the "unless" is void.
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>>> the decider is made part of the program that is being
> >>>>>>>>>>>>>>> decided
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> The decider is made a part of the program discussed in
> >>>>>>>>>>>>>> the proof.
> >>>>>>>>>>>>>>> in which case we get a function call-like infinite
> >>>>>>>>>>>>>>> recursion instead
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> We get or we don't get, depending on how the halt
> >>>>>>>>>>>>>> decider candidate attempts to decide.
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>>> (as described by Pete Olcott) and we are attempting
> >>>>>>>>>>>>>>> (and failing) to decide if a procedure halts rather
> >>>>>>>>>>>>>>> than if a program halts.
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> In any case, the decider candidate fails to give the
> >>>>>>>>>>>>>> correct answer, and therefore is not a halt decider.
> >>>>>>>>>>>>>> Note that this is correctly inferred in the proof.
> >>>>>>>>>>>>>> Therefore either the conclusion of the proof is
> >>>>>>>>>>>>>> correct or you are wrong.
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> No answer is ever given due to the infinite recursion.
> >>>>>>>>>>>>
> >>>>>>>>>>>> True about some candidates, false about others. For
> >>>>>>>>>>>> example, a candidate that always says "no" is not
> >>>>>>>>>>>> infinitely recursive but is wrong in the particular case
> >>>>>>>>>>>> discussed in the proof. But anyway, you have confirmed
> >>>>>>>>>>>> the conclusion of the proof.
> >>>>>>>>>>>
> >>>>>>>>>>> False. The proof is not cognisant of the presence of the
> >>>>>>>>>>> infinite recursion. The decider can never return an
> >>>>>>>>>>> answer of halts/doesn't halt due to the infinite
> >>>>>>>>>>> recursion.
> >>>>>>>>>>
> >>>>>>>>>> The correct term is decider candidate. It is not a decider
> >>>>>>>>>> because the infinite recursion prevents it from halting in
> >>>>>>>>>> finite time.
> >>>>>>>>>>
> >>>>>>>>>> You cannot call the proof incorrect merely because it
> >>>>>>>>>> agrees with you.
> >>>>>>>>>
> >>>>>>>>> Try actually reading what Strachey wrote: a contradiction
> >>>>>>>>> arises based on the result of evaluating T[P] however T[P]
> >>>>>>>>> is never evaluated due to the infinite recursion: a fact
> >>>>>>>>> ignored by the proof.
> >>>>>>>>
> >>>>>>>> If you can prove that the program does give the correct
> >>>>>>>> result in that case you have proven Strachey wrong.
> >>>>>>>> Otherwise you haven't.
> >>>>>>>
> >>>>>>> Strachey is wrong because he neglected to account for the
> >>>>>>> infinite recursion; this should be obvious to anyone who has
> >>>>>>> actually read and understood what Strachey wrote: it seems
> >>>>>>> that you haven't.
> >>>>>>>
> >>>>>>> /Flibble
> >>>>>>>
> >>>>>>
> >>>>>> No, because if T gets stuck in an infinite recursion, it is
> >>>>>> wrong, because it failed to answer in finite time.
> >>>>>>
> >>>>>> If T does answer in finite time, then there never was an
> >>>>>> infinite recursion.
> >>>>>>
> >>>>>> All you have shown is that there exists a WRONG method to build
> >>>>>> a T that gets stuck, Like a T that needs to run its input to
> >>>>>> completion to answer about it.
> >>>>>
> >>>>> I have shown that [Strachey, 1965] contains an infinite
> >>>>> recursion and is thus invalid.
> >>>>>
> >>>>> /Flibble
> >>>>>
> >>>>
> >>>> No, you haven't. All you have shown is that one way to attempt to
> >>>> make the program that Strachev says doesn't exist, fails due to
> >>>> getting caught in infinite recursion.
> >>>>
> >>>> That just helps confirm Strachev, not refute it.
> >>>
> >>> Nope. Strachey's proof is based on a contradiction relating to
> >>> evaluating the result of T[P] however T[P] can never be evaluated
> >>> if there is an infinite recursion.
> >>>
> >>> /Flibble
> >>>
> >>
> >> Nope, because if T actually meets the requirement to be able to
> >> take ANY program and answer, then it need to be able to take a
> >> program that uses a copy of T in it, as that is a valid program.
> >>
> >> If T gets into an infinite recursion on such a program, it is that
> >> *T* fails to meet the requirements, not that the such a program is
> >> invalid to give to T.
> >>
> >> This means that the program T is Strachey's proof can't just
> >> execute its input to get the needed answer, as that makes it (T)
> >> not meet its requirements.
> >>
> >> Thus, the "recursion" error isn't in the proof of impossibility,
> >> but in the candidate T that is attempting to be the counter
> >> example for the proof.
> >>
> >> Recursion is allowed in programming and in math. If you try to just
> >> ban it, you will find you logic can't handle what it needs to.
> >
> > You are simply wrong, and fractally so. Try reading what Strachey
> > actually wrote.
> >
> > /Flibble
> >
>
> I have.
>
> T[R] is defined to be a boolean function that returns True if routine
> R (taking no parameters) Halts, and False if R never halts.
>
> Thus, for ANY input program, T must itself Halt, that requirement
> INCLUDES if R calls T[R], so if T gerts into infinite recursion and
> thus not answer, T has failed its specification.
>
> If you want to define that T can't take the R he defines, then why is
> that NOT a program? What "rule of programs" has it violated, note R
> getting stuck in an infinite loop is one of the purposes that T is
> supposed to handle.
>
> T has no requirement that the input R halts or in fact, any
> requirement other than it be a program. Thus the fact that R gets
> "hung up" in an infinite recursion isn't an error in building R, but
> it IS a error in the design of T, as T is REQUIRED, to meet its
> definition to always answer.
>
> Thus, your claim about T not answering because it got stuck in an
> infinite recursion is just a statement that a T built that way jus
> fails to meet its requriement.
>
> Note, there is NO requirement in the problem that T[R] actually runs
> R, and in fact, you are proving that it CAN'T (at least not without
> some way to stop it) as that leads to T inherently failing for this
> sort of program.
>
> Note, any "rule" you try to invent that prohibits this "impossible
> program" must also take into account the "compliant" program C that
> should be allowed that just calls T[C] and doesn't what it says, and
> the two fixed behavior ones that call T with themselfs and then
> unconditionally Halt or Loop. All three of these should be "legal" as
> they do have definite answers that work, so arguements on
> contrariness don't apply.


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Re: On recursion and infinite recursion (reprise #3)

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From: Rich...@Damon-Family.org (Richard Damon)
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Date: Sun, 8 May 2022 17:40:52 -0400
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 by: Richard Damon - Sun, 8 May 2022 21:40 UTC

On 5/8/22 4:14 PM, Mr Flibble wrote:
> On Sun, 8 May 2022 15:39:32 -0400
> Richard Damon <Richard@Damon-Family.org> wrote:
>
>> On 5/8/22 3:06 PM, Mr Flibble wrote:
>>> On Sun, 8 May 2022 14:10:13 -0400
>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>
>>>> On 5/8/22 8:39 AM, Mr Flibble wrote:
>>>>> On Sun, 8 May 2022 08:31:10 -0400
>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>
>>>>>> On 5/8/22 8:02 AM, Mr Flibble wrote:
>>>>>>> On Sun, 8 May 2022 07:45:52 -0400
>>>>>>> Richard Damon <Richard@Damon-Family.org> wrote:
>>>>>>>
>>>>>>>> On 5/8/22 4:49 AM, Mr Flibble wrote:
>>>>>>>>> On Sun, 8 May 2022 11:31:08 +0300
>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>>>
>>>>>>>>>> On 2022-05-07 15:19:01 +0000, Mr Flibble said:
>>>>>>>>>>
>>>>>>>>>>> On Sat, 7 May 2022 18:13:08 +0300
>>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> On 2022-05-07 14:20:17 +0000, Mr Flibble said:
>>>>>>>>>>>>
>>>>>>>>>>>>> On Sat, 7 May 2022 17:16:31 +0300
>>>>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> On 2022-05-07 14:06:12 +0000, Mr Flibble said:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On Sat, 7 May 2022 16:59:55 +0300
>>>>>>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> On 2022-05-07 12:42:50 +0000, Mr Flibble said:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> On Sat, 7 May 2022 12:52:58 +0300
>>>>>>>>>>>>>>>>> Mikko <mikko.levanto@iki.fi> wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> On 2022-05-06 14:02:53 +0000, Mr Flibble said:
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> The decider could never be compiled and run in the
>>>>>>>>>>>>>>>>>>> first place due to the category error in the
>>>>>>>>>>>>>>>>>>> definition of the proof.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> An error in the definition of the proof does not
>>>>>>>>>>>>>>>>>> prevent compilation and execution of the program.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> In this case the error in the definition of the proof
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> No part of the proof is identified as errorneous, so
>>>>>>>>>>>>>>>> the rest is irrelevant. Anyway,
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> does prevent compilation unless
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> There is no option here, so the "unless" is void.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> the decider is made part of the program that is being
>>>>>>>>>>>>>>>>> decided
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> The decider is made a part of the program discussed in
>>>>>>>>>>>>>>>> the proof.
>>>>>>>>>>>>>>>>> in which case we get a function call-like infinite
>>>>>>>>>>>>>>>>> recursion instead
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> We get or we don't get, depending on how the halt
>>>>>>>>>>>>>>>> decider candidate attempts to decide.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> (as described by Pete Olcott) and we are attempting
>>>>>>>>>>>>>>>>> (and failing) to decide if a procedure halts rather
>>>>>>>>>>>>>>>>> than if a program halts.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> In any case, the decider candidate fails to give the
>>>>>>>>>>>>>>>> correct answer, and therefore is not a halt decider.
>>>>>>>>>>>>>>>> Note that this is correctly inferred in the proof.
>>>>>>>>>>>>>>>> Therefore either the conclusion of the proof is
>>>>>>>>>>>>>>>> correct or you are wrong.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> No answer is ever given due to the infinite recursion.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> True about some candidates, false about others. For
>>>>>>>>>>>>>> example, a candidate that always says "no" is not
>>>>>>>>>>>>>> infinitely recursive but is wrong in the particular case
>>>>>>>>>>>>>> discussed in the proof. But anyway, you have confirmed
>>>>>>>>>>>>>> the conclusion of the proof.
>>>>>>>>>>>>>
>>>>>>>>>>>>> False. The proof is not cognisant of the presence of the
>>>>>>>>>>>>> infinite recursion. The decider can never return an
>>>>>>>>>>>>> answer of halts/doesn't halt due to the infinite
>>>>>>>>>>>>> recursion.
>>>>>>>>>>>>
>>>>>>>>>>>> The correct term is decider candidate. It is not a decider
>>>>>>>>>>>> because the infinite recursion prevents it from halting in
>>>>>>>>>>>> finite time.
>>>>>>>>>>>>
>>>>>>>>>>>> You cannot call the proof incorrect merely because it
>>>>>>>>>>>> agrees with you.
>>>>>>>>>>>
>>>>>>>>>>> Try actually reading what Strachey wrote: a contradiction
>>>>>>>>>>> arises based on the result of evaluating T[P] however T[P]
>>>>>>>>>>> is never evaluated due to the infinite recursion: a fact
>>>>>>>>>>> ignored by the proof.
>>>>>>>>>>
>>>>>>>>>> If you can prove that the program does give the correct
>>>>>>>>>> result in that case you have proven Strachey wrong.
>>>>>>>>>> Otherwise you haven't.
>>>>>>>>>
>>>>>>>>> Strachey is wrong because he neglected to account for the
>>>>>>>>> infinite recursion; this should be obvious to anyone who has
>>>>>>>>> actually read and understood what Strachey wrote: it seems
>>>>>>>>> that you haven't.
>>>>>>>>>
>>>>>>>>> /Flibble
>>>>>>>>>
>>>>>>>>
>>>>>>>> No, because if T gets stuck in an infinite recursion, it is
>>>>>>>> wrong, because it failed to answer in finite time.
>>>>>>>>
>>>>>>>> If T does answer in finite time, then there never was an
>>>>>>>> infinite recursion.
>>>>>>>>
>>>>>>>> All you have shown is that there exists a WRONG method to build
>>>>>>>> a T that gets stuck, Like a T that needs to run its input to
>>>>>>>> completion to answer about it.
>>>>>>>
>>>>>>> I have shown that [Strachey, 1965] contains an infinite
>>>>>>> recursion and is thus invalid.
>>>>>>>
>>>>>>> /Flibble
>>>>>>>
>>>>>>
>>>>>> No, you haven't. All you have shown is that one way to attempt to
>>>>>> make the program that Strachev says doesn't exist, fails due to
>>>>>> getting caught in infinite recursion.
>>>>>>
>>>>>> That just helps confirm Strachev, not refute it.
>>>>>
>>>>> Nope. Strachey's proof is based on a contradiction relating to
>>>>> evaluating the result of T[P] however T[P] can never be evaluated
>>>>> if there is an infinite recursion.
>>>>>
>>>>> /Flibble
>>>>>
>>>>
>>>> Nope, because if T actually meets the requirement to be able to
>>>> take ANY program and answer, then it need to be able to take a
>>>> program that uses a copy of T in it, as that is a valid program.
>>>>
>>>> If T gets into an infinite recursion on such a program, it is that
>>>> *T* fails to meet the requirements, not that the such a program is
>>>> invalid to give to T.
>>>>
>>>> This means that the program T is Strachey's proof can't just
>>>> execute its input to get the needed answer, as that makes it (T)
>>>> not meet its requirements.
>>>>
>>>> Thus, the "recursion" error isn't in the proof of impossibility,
>>>> but in the candidate T that is attempting to be the counter
>>>> example for the proof.
>>>>
>>>> Recursion is allowed in programming and in math. If you try to just
>>>> ban it, you will find you logic can't handle what it needs to.
>>>
>>> You are simply wrong, and fractally so. Try reading what Strachey
>>> actually wrote.
>>>
>>> /Flibble
>>>
>>
>> I have.
>>
>> T[R] is defined to be a boolean function that returns True if routine
>> R (taking no parameters) Halts, and False if R never halts.
>>
>> Thus, for ANY input program, T must itself Halt, that requirement
>> INCLUDES if R calls T[R], so if T gerts into infinite recursion and
>> thus not answer, T has failed its specification.
>>
>> If you want to define that T can't take the R he defines, then why is
>> that NOT a program? What "rule of programs" has it violated, note R
>> getting stuck in an infinite loop is one of the purposes that T is
>> supposed to handle.
>>
>> T has no requirement that the input R halts or in fact, any
>> requirement other than it be a program. Thus the fact that R gets
>> "hung up" in an infinite recursion isn't an error in building R, but
>> it IS a error in the design of T, as T is REQUIRED, to meet its
>> definition to always answer.
>>
>> Thus, your claim about T not answering because it got stuck in an
>> infinite recursion is just a statement that a T built that way jus
>> fails to meet its requriement.
>>
>> Note, there is NO requirement in the problem that T[R] actually runs
>> R, and in fact, you are proving that it CAN'T (at least not without
>> some way to stop it) as that leads to T inherently failing for this
>> sort of program.
>>
>> Note, any "rule" you try to invent that prohibits this "impossible
>> program" must also take into account the "compliant" program C that
>> should be allowed that just calls T[C] and doesn't what it says, and
>> the two fixed behavior ones that call T with themselfs and then
>> unconditionally Halt or Loop. All three of these should be "legal" as
>> they do have definite answers that work, so arguements on
>> contrariness don't apply.
>
> You continue to be wrong and fractally so. The infinite recursion will
> ALWAYS happen according to the definition of the proof. Again: read AND
> UNDERSTAND what Strachey actually wrote.
>
> /Flibble
>


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