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Groups > comp.theory > #49731 > unrolled thread
| Started by | Mr Flibble <flibble@reddwarf.jmc> |
|---|---|
| First post | 2022-05-05 17:50 +0100 |
| Last post | 2022-05-06 16:39 +0300 |
| Articles | 20 on this page of 67 — 7 participants |
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On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-05 17:50 +0100
Re: On recursion and infinite recursion (reprise #3) olcott <polcott2@gmail.com> - 2022-05-05 13:58 -0500
Re: On recursion and infinite recursion (reprise #3) Ben <ben.usenet@bsb.me.uk> - 2022-05-05 20:56 +0100
Re: On recursion and infinite recursion (reprise #3) olcott <polcott2@gmail.com> - 2022-05-05 17:15 -0500
Re: On recursion and infinite recursion (reprise #3) Ben <ben.usenet@bsb.me.uk> - 2022-05-06 02:43 +0100
Re: On recursion and infinite recursion (reprise #3) olcott <polcott2@gmail.com> - 2022-05-05 20:59 -0500
Re: On recursion and infinite recursion (reprise #3) Ben <ben.usenet@bsb.me.uk> - 2022-05-06 04:08 +0100
Re: On recursion and infinite recursion (reprise #3) Ben <ben.usenet@bsb.me.uk> - 2022-05-05 20:51 +0100
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-05 20:52 +0100
Re: On recursion and infinite recursion (reprise #3) Ben <ben.usenet@bsb.me.uk> - 2022-05-05 21:15 +0100
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-05 21:16 +0100
Re: On recursion and infinite recursion (reprise #3) olcott <polcott2@gmail.com> - 2022-05-05 16:35 -0500
Re: On recursion and infinite recursion (reprise #3) olcott <polcott2@gmail.com> - 2022-05-05 16:37 -0500
Re: On recursion and infinite recursion (reprise #3) Ben <ben.usenet@bsb.me.uk> - 2022-05-06 02:38 +0100
Re: On recursion and infinite recursion (reprise #3) olcott <polcott2@gmail.com> - 2022-05-05 20:42 -0500
Re: On recursion and infinite recursion (reprise #3) Ben <ben.usenet@bsb.me.uk> - 2022-05-06 03:59 +0100
Re: On recursion and infinite recursion (reprise #3) Richard Damon <Richard@Damon-Family.org> - 2022-05-05 22:33 -0400
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-06 15:02 +0100
Re: On recursion and infinite recursion (reprise #3) Python <python@example.invalid> - 2022-05-06 16:18 +0200
Re: On recursion and infinite recursion (reprise #3) olcott <polcott2@gmail.com> - 2022-05-06 11:55 -0500
Re: On recursion and infinite recursion (reprise #3) Mikko <mikko.levanto@iki.fi> - 2022-05-07 12:52 +0300
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-07 13:42 +0100
Re: On recursion and infinite recursion (reprise #3) Mikko <mikko.levanto@iki.fi> - 2022-05-07 16:59 +0300
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-07 15:06 +0100
Re: On recursion and infinite recursion (reprise #3) Mikko <mikko.levanto@iki.fi> - 2022-05-07 17:16 +0300
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-07 15:20 +0100
Re: On recursion and infinite recursion (reprise #3) Mikko <mikko.levanto@iki.fi> - 2022-05-07 18:13 +0300
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-07 16:19 +0100
Re: On recursion and infinite recursion (reprise #3) olcott <polcott2@gmail.com> - 2022-05-07 11:03 -0500
Re: On recursion and infinite recursion (reprise #3) Ben <ben.usenet@bsb.me.uk> - 2022-05-07 23:41 +0100
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-07 23:43 +0100
Re: On recursion and infinite recursion (reprise #3) Richard Damon <Richard@Damon-Family.org> - 2022-05-07 19:59 -0400
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-08 01:01 +0100
Re: On recursion and infinite recursion (reprise #3) Richard Damon <Richard@Damon-Family.org> - 2022-05-07 20:18 -0400
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-08 01:55 +0100
Re: On recursion and infinite recursion (reprise #3) Mikko <mikko.levanto@iki.fi> - 2022-05-08 11:31 +0300
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-08 09:49 +0100
Re: On recursion and infinite recursion (reprise #3) Mikko <mikko.levanto@iki.fi> - 2022-05-08 13:00 +0300
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-08 12:57 +0100
Re: On recursion and infinite recursion (reprise #3) Richard Damon <Richard@Damon-Family.org> - 2022-05-08 07:45 -0400
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-08 13:02 +0100
Re: On recursion and infinite recursion (reprise #3) Richard Damon <Richard@Damon-Family.org> - 2022-05-08 08:31 -0400
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-08 13:39 +0100
Re: On recursion and infinite recursion (reprise #3) Richard Damon <Richard@Damon-Family.org> - 2022-05-08 14:10 -0400
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-08 20:06 +0100
Re: On recursion and infinite recursion (reprise #3) Richard Damon <Richard@Damon-Family.org> - 2022-05-08 15:39 -0400
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-08 21:14 +0100
Re: On recursion and infinite recursion (reprise #3) Richard Damon <Richard@Damon-Family.org> - 2022-05-08 17:40 -0400
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-08 23:04 +0100
Re: On recursion and infinite recursion (reprise #3) Ben <ben.usenet@bsb.me.uk> - 2022-05-09 00:26 +0100
Re: On recursion and infinite recursion (reprise #3) Richard Damon <Richard@Damon-Family.org> - 2022-05-08 19:40 -0400
Re: On recursion and infinite recursion (reprise #3) olcott <NoOne@NoWhere.com> - 2022-05-09 10:55 -0500
Re: On recursion and infinite recursion (reprise #3) Richard Damon <Richard@Damon-Family.org> - 2022-05-09 20:13 -0400
Re: On recursion and infinite recursion (reprise #3) olcott <NoOne@NoWhere.com> - 2022-05-09 10:51 -0500
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-09 18:31 +0100
Re: On recursion and infinite recursion (reprise #3) olcott <NoOne@NoWhere.com> - 2022-05-09 12:36 -0500
Re: On recursion and infinite recursion (reprise #3) Richard Damon <Richard@Damon-Family.org> - 2022-05-09 20:15 -0400
Re: On recursion and infinite recursion (reprise #3) Richard Damon <Richard@Damon-Family.org> - 2022-05-09 20:14 -0400
Re: On recursion and infinite recursion (reprise #3) olcott <NoOne@NoWhere.com> - 2022-05-09 10:49 -0500
Re: On recursion and infinite recursion (reprise #3) Mr Flibble <flibble@reddwarf.jmc> - 2022-05-09 18:31 +0100
Re: On recursion and infinite recursion (reprise #3) olcott <NoOne@NoWhere.com> - 2022-05-09 12:37 -0500
Re: On recursion and infinite recursion (reprise #3) olcott <NoOne@NoWhere.com> - 2022-05-09 10:47 -0500
Re: On recursion and infinite recursion (reprise #3) Mikko <mikko.levanto@iki.fi> - 2022-05-09 19:33 +0300
Re: On recursion and infinite recursion (reprise #3) olcott <NoOne@NoWhere.com> - 2022-05-09 11:36 -0500
Re: On recursion and infinite recursion (reprise #3) Richard Damon <Richard@Damon-Family.org> - 2022-05-09 20:17 -0400
Re: On recursion and infinite recursion (reprise #3) olcott <polcott2@gmail.com> - 2022-05-07 11:06 -0500
Re: On recursion and infinite recursion (reprise #3) Mikko <mikko.levanto@iki.fi> - 2022-05-06 16:39 +0300
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| From | Mikko <mikko.levanto@iki.fi> |
|---|---|
| Date | 2022-05-07 12:52 +0300 |
| Message-ID | <t55fhq$u5e$1@dont-email.me> |
| In reply to | #49846 |
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. Mikko
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| From | Mr Flibble <flibble@reddwarf.jmc> |
|---|---|
| Date | 2022-05-07 13:42 +0100 |
| Message-ID | <20220507134250.00007acc@reddwarf.jmc> |
| In reply to | #49931 |
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 does prevent compilation unless the decider is made part of the program that is being decided in which case we get a function call-like infinite recursion instead (as described by Pete Olcott) and we are attempting (and failing) to decide if a procedure halts rather than if a program halts. /Flibble
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| From | Mikko <mikko.levanto@iki.fi> |
|---|---|
| Date | 2022-05-07 16:59 +0300 |
| Message-ID | <t55u0r$7lf$1@dont-email.me> |
| In reply to | #49934 |
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. Mikko
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| From | Mr Flibble <flibble@reddwarf.jmc> |
|---|---|
| Date | 2022-05-07 15:06 +0100 |
| Message-ID | <20220507150612.00003fab@reddwarf.jmc> |
| In reply to | #49935 |
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. /Flibble
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| From | Mikko <mikko.levanto@iki.fi> |
|---|---|
| Date | 2022-05-07 17:16 +0300 |
| Message-ID | <t55uvv$fpf$1@dont-email.me> |
| In reply to | #49936 |
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
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| From | Mr Flibble <flibble@reddwarf.jmc> |
|---|---|
| Date | 2022-05-07 15:20 +0100 |
| Message-ID | <20220507152017.00000990@reddwarf.jmc> |
| In reply to | #49937 |
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
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| From | Mikko <mikko.levanto@iki.fi> |
|---|---|
| Date | 2022-05-07 18:13 +0300 |
| Message-ID | <t562a4$9ao$1@dont-email.me> |
| In reply to | #49938 |
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
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| From | Mr Flibble <flibble@reddwarf.jmc> |
|---|---|
| Date | 2022-05-07 16:19 +0100 |
| Message-ID | <20220507161901.00002e54@reddwarf.jmc> |
| In reply to | #49944 |
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
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| From | olcott <polcott2@gmail.com> |
|---|---|
| Date | 2022-05-07 11:03 -0500 |
| Message-ID | <t56585$uq$1@dont-email.me> |
| In reply to | #49945 |
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
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| From | Ben <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2022-05-07 23:41 +0100 |
| Message-ID | <87pmkpvuce.fsf@bsb.me.uk> |
| In reply to | #49945 |
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.
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| From | Mr Flibble <flibble@reddwarf.jmc> |
|---|---|
| Date | 2022-05-07 23:43 +0100 |
| Message-ID | <20220507234312.00005179@reddwarf.jmc> |
| In reply to | #49979 |
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
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| From | Richard Damon <Richard@Damon-Family.org> |
|---|---|
| Date | 2022-05-07 19:59 -0400 |
| Message-ID | <RJDdK.19199$h6X.16518@fx04.iad> |
| In reply to | #49980 |
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.
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| From | Mr Flibble <flibble@reddwarf.jmc> |
|---|---|
| Date | 2022-05-08 01:01 +0100 |
| Message-ID | <20220508010142.00007bf1@reddwarf.jmc> |
| In reply to | #49992 |
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
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| From | Richard Damon <Richard@Damon-Family.org> |
|---|---|
| Date | 2022-05-07 20:18 -0400 |
| Message-ID | <h%DdK.6916$t72a.5563@fx10.iad> |
| In reply to | #49993 |
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.
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| From | Mr Flibble <flibble@reddwarf.jmc> |
|---|---|
| Date | 2022-05-08 01:55 +0100 |
| Message-ID | <20220508015553.000016a8@reddwarf.jmc> |
| In reply to | #49996 |
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
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| From | Mikko <mikko.levanto@iki.fi> |
|---|---|
| Date | 2022-05-08 11:31 +0300 |
| Message-ID | <t57v4c$7do$1@dont-email.me> |
| In reply to | #49945 |
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
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| From | Mr Flibble <flibble@reddwarf.jmc> |
|---|---|
| Date | 2022-05-08 09:49 +0100 |
| Message-ID | <20220508094913.00002e7c@reddwarf.jmc> |
| In reply to | #50015 |
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
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| From | Mikko <mikko.levanto@iki.fi> |
|---|---|
| Date | 2022-05-08 13:00 +0300 |
| Message-ID | <t584ba$d38$1@dont-email.me> |
| In reply to | #50016 |
On 2022-05-08 08:49:13 +0000, Mr Flibble said: > Strachey is wrong because he neglected to account for the infinite > recursion; this should be obvious to anyone who has actually read and > understood what Strachey wrote: it seems that you haven't. Don't forget that Strachey is not presenting a new proof but merely offering another point of view to an already existing proof. Therefore it is not important that his presentantion be complete. More important is that it covers the main idea in a way that is easy to understand. Mikko
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| From | Mr Flibble <flibble@reddwarf.jmc> |
|---|---|
| Date | 2022-05-08 12:57 +0100 |
| Message-ID | <20220508125704.00001b8f@reddwarf.jmc> |
| In reply to | #50019 |
On Sun, 8 May 2022 13:00:10 +0300 Mikko <mikko.levanto@iki.fi> wrote: > On 2022-05-08 08:49:13 +0000, Mr Flibble said: > > > Strachey is wrong because he neglected to account for the infinite > > recursion; this should be obvious to anyone who has actually read > > and understood what Strachey wrote: it seems that you haven't. > > Don't forget that Strachey is not presenting a new proof but merely > offering another point of view to an already existing proof. Therefore > it is not important that his presentantion be complete. More important > is that it covers the main idea in a way that is easy to understand. It is a pretty stunning omission. No, Occams's Razor suggests it is an oversight. The proof is invalid. /Flibble
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| From | Richard Damon <Richard@Damon-Family.org> |
|---|---|
| Date | 2022-05-08 07:45 -0400 |
| Message-ID | <Q3OdK.15730$Bm21.3438@fx07.iad> |
| In reply to | #50016 |
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.
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