Mauriès Robert wrote:The correction, François, would be for you to address to me exactly what you addressed to Denis.

Thank you

Robert

Of course, but rather tomorrow morning.

48 posts
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Mauriès Robert wrote:The correction, François, would be for you to address to me exactly what you addressed to Denis.

Thank you

Robert

Of course, but rather tomorrow morning.

- DEFISE
**Posts:**258**Joined:**16 April 2020**Location:**France

Hi Robert,

Finally I was able to do it tonight, but without translating:

1ere étape :

je démontre qu’une piste issue d’un ensemble ne contient pas forcément un candidat de cet ensemble.

Soit une grille ayant au moins 2 solutions S1 et S2 et une entité contenant 4 candidats A1, A2, A3, A4 tels que A1 ∈ S1 et A2 ∈ S2.

Soit la paire d’ensembles E1 = {A1,A2} et E2 = {A3,A4}

Par définition P(E1) = P(A1) ∩ P(A2).

Si P(E1) contenait A1, alors A1 appartiendrait à la fois à P(A1) et à P(A2),

donc P(A2) contiendrait A1 et A2 (candidats d’une même entité) et serait donc contradictoire.

Or P(A2) ne peut pas présenter de contradiction puisque A2 appartient à une solution et que tous ses autres candidats se déduisent logiquement de A2.

Donc P(E1) ne peut pas contenir A1.

De façon symétrique P(E1) ne peut pas contenir A2.

Donc P(E1) ne contient ni A1 ni A2 donc aucun candidat de E1.

CQFD

N.B: la case r2c3 de la grille 630 de assistant-sudoku.com vérifie le schéma ci-dessus.

2eme étape

Au début de la démonstration du Th 2-1 on trouve la phrase suivante:

« Comme un des candidats Ak de E1 doit être placé pour appliquer R, on peut dire que un des Ak de E1 est un candidat de P ’(E2) »

Je retiens de cela uniquement la proposition PR :

« un des Ak de E1 est un candidat de P ’(E2) »

Si le Th 2-1 est vrai, alors P ’(E2) = P(E1) et la proposition PR devient :

« un des Ak de E1 est un candidat de P(E1) » ce qui peut être impossible comme l’a montré la 1ère étape.

Donc, en résumé : si le théorème 2-1 est vrai alors sa démonstration est fausse.

N.B : J’ajoute qu’il est précisé au début de votre document que la théorie est censée s’appliquer aux grilles de tous types (à solution unique ou non).

N.B: tomorrow I will be busy all day.

Finally I was able to do it tonight, but without translating:

1ere étape :

je démontre qu’une piste issue d’un ensemble ne contient pas forcément un candidat de cet ensemble.

Soit une grille ayant au moins 2 solutions S1 et S2 et une entité contenant 4 candidats A1, A2, A3, A4 tels que A1 ∈ S1 et A2 ∈ S2.

Soit la paire d’ensembles E1 = {A1,A2} et E2 = {A3,A4}

Par définition P(E1) = P(A1) ∩ P(A2).

Si P(E1) contenait A1, alors A1 appartiendrait à la fois à P(A1) et à P(A2),

donc P(A2) contiendrait A1 et A2 (candidats d’une même entité) et serait donc contradictoire.

Or P(A2) ne peut pas présenter de contradiction puisque A2 appartient à une solution et que tous ses autres candidats se déduisent logiquement de A2.

Donc P(E1) ne peut pas contenir A1.

De façon symétrique P(E1) ne peut pas contenir A2.

Donc P(E1) ne contient ni A1 ni A2 donc aucun candidat de E1.

CQFD

N.B: la case r2c3 de la grille 630 de assistant-sudoku.com vérifie le schéma ci-dessus.

2eme étape

Au début de la démonstration du Th 2-1 on trouve la phrase suivante:

« Comme un des candidats Ak de E1 doit être placé pour appliquer R, on peut dire que un des Ak de E1 est un candidat de P ’(E2) »

Je retiens de cela uniquement la proposition PR :

« un des Ak de E1 est un candidat de P ’(E2) »

Si le Th 2-1 est vrai, alors P ’(E2) = P(E1) et la proposition PR devient :

« un des Ak de E1 est un candidat de P(E1) » ce qui peut être impossible comme l’a montré la 1ère étape.

Donc, en résumé : si le théorème 2-1 est vrai alors sa démonstration est fausse.

N.B : J’ajoute qu’il est précisé au début de votre document que la théorie est censée s’appliquer aux grilles de tous types (à solution unique ou non).

N.B: tomorrow I will be busy all day.

- DEFISE
**Posts:**258**Joined:**16 April 2020**Location:**France

.

As François says he will be busy today (Paris time, maybe still yesterday in the West of US), let me do the translation work for our non French speaking readers.

If the whole theory had to depend on the assumption of uniqueness, it wouldn't have much value.

As François says he will be busy today (Paris time, maybe still yesterday in the West of US), let me do the translation work for our non French speaking readers.

François, translated by Denis wrote:First step : I show that a track based on a set does not necessarily contain a candidate from this set.

Consider a puzzle with at least two solutions S1 and S2, and an entity containing 4 candidates A1, A2, A3, A4 such that A1 ∈ S1 and A2 ∈ S2.

Let the pair of sets be E1 = {A1, A2} and E2 = {A3, A4}

By definition, P(E1) = P(A1) ∩ P(A2).

If P(E1) contained A1, then A1 would belong to both P(A1) and P(A2).

Therefore P(A2) would contain A1 and A2, and it would be contradictory (A1 and A2 are candidates of the same entity).

But P(A2) cannot be contradictory, because A2 belongs to a solution and all the other candidates in P(A2) are logical consequences of A2.

Therefore P(E1) cannot contain A1.

By symmetry, P(E1) cannot contain A2.

Finally, P(E1) contains neither A1 nor A2, i.e. no candidate from E1.

qed

N.B.: cell r2c3 from puzzle #630 at assistant-sudoku.com satisfies the conditions at the start of the above proof.

Second step

At the start of the proof of Th 2-1, there is the following sentence:

« As one of the candidates Ak in E1 must be true for applying R, one can say that one of the Ak's in E1 is a candidate of P’(E2) »

From this, I keep only the proposition PR :

« one of the candidates Ak in E1 is in P'(E2) »

If Th 2-1 is true, then P ’(E2) = P(E1) and proposition PR becomes:

« one of the candidates Ak in E1 is in P(E1); but the 1st step has shown that this is not necessarily true.

To sum up : if theorem 2-1 is true, then its proof is false.

François, translated by Denis wrote:N.B : let me add that, at the start of the document, it is stated that the theory is supposed to apply all types of puzzles (with unique solution or not).

If the whole theory had to depend on the assumption of uniqueness, it wouldn't have much value.

- denis_berthier
- 2010 Supporter
**Posts:**3676**Joined:**19 June 2007**Location:**Paris

DEFISE wrote:Hi Robert,

Finally I was able to do it tonight, but without translating ....

Bonjour François,

Votre démonstration est sans appel effectivement : ou le théorème et faux ou ma démonstration est fausse. J'ai retrouvé votre mail de 2019 où vous me la donniez. Pourquoi à l'époque je l'ai omise, laissant mon texte en l'état... je ne sais pas ? Sans doute avais-je l'intention de donner une autre démonstration de ce théorème (je n'ai jamais publié mes réflexions à ce sujet, il faut que je m'y replonge).

Ce théorème que je crois vrai reste donc une conjecture pour moi, et je vais modifier mon texte en conséquence.

En tout cas je vous adresse mes excuses, ainsi qu'à mes lecteurs, pour ne pas vous avoir suivi sur ce point, vous qui aviez pris soin de discuter de tout avec moi et de relire intégralement mon document.

Amicalement

Hello François,

Your demonstration is indeed without appeal: either the theorem is false or my demonstration is false. I found your email of 2019 where you gave it to me. Why at the time I omitted it, leaving my text as it was... I don't know? No doubt I intended to give another demonstration of this theorem (I never published my thoughts on this subject, I must go back to it).

This theorem, which I believe to be true, remains a conjecture for me, and I will modify my text accordingly.

In any case, I apologise to you and to my readers for not having followed you on this point, as you had taken care to discuss everything with me and to reread my document in full.

Sincerely

Robert

- Mauriès Robert
**Posts:**585**Joined:**07 November 2019**Location:**France

Mauriès Robert wrote:Your (DEFISE) demonstration is indeed without appeal

At last, my delirium had a result.

Mauriès Robert wrote:either the theorem is false or my demonstration is false.

As I've shown, it's not either/or. It's merely the proof is false (which, of course, doesn't prevent the theorem from being false also).

Mauriès Robert wrote:This theorem, which I believe to be true,

You may have hard work to prove it. See my forthcoming post (when I have time to do it).

- denis_berthier
- 2010 Supporter
**Posts:**3676**Joined:**19 June 2007**Location:**Paris

Robert, thank you for acknowledging your mistake.

Denis, thank you for your translation.

Denis, thank you for your translation.

- DEFISE
**Posts:**258**Joined:**16 April 2020**Location:**France

.

Let me propose a different proof of François's proposition that a track based on a subset E of an entity does not necessarily contain some candidate from this set.

It will not suppose the existence of several solutions.

It will require only trivial boolean logic.

1) I shall first prove something much stronger:

Theorem: a track based on an arbitrary set E of candidates contains a candidate A0 iff for each candidate Ak≠A0 in E, A0 is a consequence of Ak in TR (TB).

(Note that:

- E doesn't have to be a subset of any entity;

- A0 doesn't have to be in E;

- the result is true even in the trivial cases where E contains only 1 or 2 elements

- no hypothesis is made on the number of solutions of the puzzle.)

If you think of it in pure logic terms, cleaned of all the useless definitions, you will realise that it is totally obvious.

"iff" means "if and only if".

Let A, B, C, D... be any candidates.

I'll write "A=>B" to mean that B can be derived from A within TR (i.e. within TB, in the only interpretation of TR that makes sense).

By the definition of a track P(A), B ∊ P(A) merely means A=>B.

Proof of the theorem:

By definition, P(A B C...) = P(A) ∩ P(B) ∩ P(C)...,

i.e. A0 ∊ P(A B C...) iff A0 ∊ P(A) and A0 ∊ P(B) and A0 ∊ P(C)...

i.e. A0 ∊ P(A B C...) iff A=>A0 and B=>A0 and C=>A0...

qed

2) In particular:

A ∊ P(A B C...) iff A=>A and B=>A and C=>A iff B=>A and C=>A ...

B ∊ P(A B C...) iff A=>B and B=>B and C=>B iff A=>B and C=>B ...

C ∊ P(A B C...) iff A=>C and B=>C and C=>C iff A=>C and B=>C ...

...

Therefore, one of the candidates A, B, C... is in P(A B C...) iff this candidate is a consequence in TR (TB) of each of the other ones separately.

This is obviously impossible (unless the puzzle is contradictory) when E is a subset of an undecided 2D-cell ("entity" in Robert's vocabulary).

3) Corollary: whether theorem 2.1 is true or not, its proof is false.

But we can go further. Let E1 = A, B, C, ... and let E2 be its complement in some 2D-cell ("entity"), where none of E1 or E2 is empty.

4) By definition, P'(E2) is defined by X ∊ P'(E2) iff eliminating all the elements of E2 allows to prove X in TR (TB)

i.e. X ∊ P'(E2) iff the disjunction of all the elements in E1 implies X in TR (TB)

i.e. X ∊ P'(E2) iff (A or B or C...) => X

i.e. X ∊ P'(E2) iff A=>X or B=>X or C=>X...

i.e. X ∊ P'(E2) iff X ∊ P(A) or X ∊ P(B) or X ∊ P(C) ...

Said otherwise, P'(E2) is the union of all the P(Ak), Ak in E1 (while P(E1) is their intersection).

This will be the last nail in this rotten coffin:

5) Théorème 2.1 is grossly false, except for 1-candidate sets E1, E2 (in which case it is trivial).

P(E1) is the intersection of all the P(Ak), Ak in E1

P'(E2) is the union of all the P(Ak), Ak in E1

One always has P(E1) ⊂ P'(E2).

The first part of the theorem is true iff P(Ak) is included in P(E1) for any Ak in E1. This requires that each Ak be an element of P(E1). But we have seen that this is impossible for any Ak if E1 has more than one candidate.

And the "conversely" part is false if E2 has more than one candidate.

Let me propose a different proof of François's proposition that a track based on a subset E of an entity does not necessarily contain some candidate from this set.

It will not suppose the existence of several solutions.

It will require only trivial boolean logic.

1) I shall first prove something much stronger:

Theorem: a track based on an arbitrary set E of candidates contains a candidate A0 iff for each candidate Ak≠A0 in E, A0 is a consequence of Ak in TR (TB).

(Note that:

- E doesn't have to be a subset of any entity;

- A0 doesn't have to be in E;

- the result is true even in the trivial cases where E contains only 1 or 2 elements

- no hypothesis is made on the number of solutions of the puzzle.)

If you think of it in pure logic terms, cleaned of all the useless definitions, you will realise that it is totally obvious.

"iff" means "if and only if".

Let A, B, C, D... be any candidates.

I'll write "A=>B" to mean that B can be derived from A within TR (i.e. within TB, in the only interpretation of TR that makes sense).

By the definition of a track P(A), B ∊ P(A) merely means A=>B.

Proof of the theorem:

By definition, P(A B C...) = P(A) ∩ P(B) ∩ P(C)...,

i.e. A0 ∊ P(A B C...) iff A0 ∊ P(A) and A0 ∊ P(B) and A0 ∊ P(C)...

i.e. A0 ∊ P(A B C...) iff A=>A0 and B=>A0 and C=>A0...

qed

2) In particular:

A ∊ P(A B C...) iff A=>A and B=>A and C=>A iff B=>A and C=>A ...

B ∊ P(A B C...) iff A=>B and B=>B and C=>B iff A=>B and C=>B ...

C ∊ P(A B C...) iff A=>C and B=>C and C=>C iff A=>C and B=>C ...

...

Therefore, one of the candidates A, B, C... is in P(A B C...) iff this candidate is a consequence in TR (TB) of each of the other ones separately.

This is obviously impossible (unless the puzzle is contradictory) when E is a subset of an undecided 2D-cell ("entity" in Robert's vocabulary).

3) Corollary: whether theorem 2.1 is true or not, its proof is false.

But we can go further. Let E1 = A, B, C, ... and let E2 be its complement in some 2D-cell ("entity"), where none of E1 or E2 is empty.

4) By definition, P'(E2) is defined by X ∊ P'(E2) iff eliminating all the elements of E2 allows to prove X in TR (TB)

i.e. X ∊ P'(E2) iff the disjunction of all the elements in E1 implies X in TR (TB)

i.e. X ∊ P'(E2) iff (A or B or C...) => X

i.e. X ∊ P'(E2) iff A=>X or B=>X or C=>X...

i.e. X ∊ P'(E2) iff X ∊ P(A) or X ∊ P(B) or X ∊ P(C) ...

Said otherwise, P'(E2) is the union of all the P(Ak), Ak in E1 (while P(E1) is their intersection).

This will be the last nail in this rotten coffin:

Robert_Mauriès wrote:"Theorem 2.1: If E1 and E2 form a pair of sets, the track P(E1) based on E1 is identical to the antitrack P’(E2) based on E2, and conversely."

5) Théorème 2.1 is grossly false, except for 1-candidate sets E1, E2 (in which case it is trivial).

P(E1) is the intersection of all the P(Ak), Ak in E1

P'(E2) is the union of all the P(Ak), Ak in E1

One always has P(E1) ⊂ P'(E2).

The first part of the theorem is true iff P(Ak) is included in P(E1) for any Ak in E1. This requires that each Ak be an element of P(E1). But we have seen that this is impossible for any Ak if E1 has more than one candidate.

And the "conversely" part is false if E2 has more than one candidate.

- denis_berthier
- 2010 Supporter
**Posts:**3676**Joined:**19 June 2007**Location:**Paris

Hi Denis,

I had decided not to "enter" into your delirium, because it was not a question of a cordial confrontation on technical points of the TDP, but of denigrating it.

But with so much fervour on your part to continue this denigration, I will respond, and this will take several comments.

Robert

I notice that you only translate parts of the document, so you miss some parts and this leads to misinterpretation (or even translation).

The document you quote is intended for those who are new to the "Technique" of tracks. It is therefore simplified and this is written in the preamble where it says: This document is a simplified presentation of the Track Technique (TDP) for the attention of those who are discovering this technique or are not yet well familiar with it. This you have not read, let alone translated.

I had decided not to "enter" into your delirium, because it was not a question of a cordial confrontation on technical points of the TDP, but of denigrating it.

But with so much fervour on your part to continue this denigration, I will respond, and this will take several comments.

Robert

denis_berthier wrote:.

[Edit 2]: the reference for my forthcoming posts: the december 2017 original French version of the document "Théorie de la technique des pistes en sudoku. Par Robert Mauriès (*)". I'll cite the original and do the translations myself.

I notice that you only translate parts of the document, so you miss some parts and this leads to misinterpretation (or even translation).

denis_berthier wrote:[Edit 3]: On the website's left part, there are indeed two buttons, leading to 2 different versions of the paper. So, the previous version has not disappeared. It remains that any reference to tracks and anti-tracks based on arbitrary sets of candidates have disappeared from the new paper.

The document you quote is intended for those who are new to the "Technique" of tracks. It is therefore simplified and this is written in the preamble where it says: This document is a simplified presentation of the Track Technique (TDP) for the attention of those who are discovering this technique or are not yet well familiar with it. This you have not read, let alone translated.

Last edited by Mauriès Robert on Fri Jan 28, 2022 3:04 pm, edited 2 times in total.

- Mauriès Robert
**Posts:**585**Joined:**07 November 2019**Location:**France

.

Robert,

We've gone past my first post. We've reached a point where I've proven that theorem 2.1 is false.

My translations are not missing any part related to theorem 2.1. and they are totally faithful. My proof uses only what I've translated. If you claimed the contrary, you would have to say where precisely.

The only questions are:

- after François told you the proof of Theorem 2.1 has an error, why did you not investigate it seriously but keep it online for 2 or 3 years as if it was true?

- after my proof, do you finally admit that theorem 2.1 of the 2017 paper is false or will you continue trying to gaslight us and the readers of your forum?

Robert,

We've gone past my first post. We've reached a point where I've proven that theorem 2.1 is false.

My translations are not missing any part related to theorem 2.1. and they are totally faithful. My proof uses only what I've translated. If you claimed the contrary, you would have to say where precisely.

The only questions are:

- after François told you the proof of Theorem 2.1 has an error, why did you not investigate it seriously but keep it online for 2 or 3 years as if it was true?

- after my proof, do you finally admit that theorem 2.1 of the 2017 paper is false or will you continue trying to gaslight us and the readers of your forum?

- denis_berthier
- 2010 Supporter
**Posts:**3676**Joined:**19 June 2007**Location:**Paris

denis_berthier wrote:.

Robert,

We've gone past my first post. We've reached a point where I've proven that theorem 2.1 is false.

My translations are not missing any part related to theorem 2.1. and they are totally faithful. My proof uses only what I've translated. If you claimed the contrary, you would have to say where precisely.

The only questions are:

- after François told you the proof of Theorem 2.1 has an error, why did you not investigate it seriously but keep it online for 2 or 3 years as if it was true?

- after my proof, do you finally admit that theorem 2.1 of the 2017 paper is false or will you continue trying to gaslight us and the readers of your forum?

I'm getting there, so give me time, because I have a lot to write.

- Mauriès Robert
**Posts:**585**Joined:**07 November 2019**Location:**France

I continue to comment on your writings in the order in which they were published...

You want to translate, then translate my definitions exactly.

Here is the translation of what I wrote about R: R stands for the rule of the game, which stipulates that one candidate and only one candidate per entity must be placed in all the cells of the grid.

Here is the full text on this subject:

A solving technique is a logical approach (reasoning) allowing to place or eliminate candidates on a grid by applying R. We designate by TR the set of solving techniques allowing to solve a grid, by TE (elementary techniques) a part of these techniques including the simplest ones which are: the search for unique candidates by sweeping the zones, the search for alignments and the search for closed sets (doublet, triplet, quadruplet, etc). These three techniques together constitute the basic techniques designated by TB.

I agree that it should be more detailed, and it was planned in a new edition that I cannot find the time to finalise.

So I specify here that TRs include TBs, Wings and other fish, etc... but also reasoning by absurdity, i.e. T&E, so everything that can be done by logical reasoning.

denis_berthier wrote:Note: R is the 4 rules of Sudoku (see p.1).

Honestly, I'm re-assured. Robert will not speak of sacrificing virgins, drinking potions or eating brains of alive monkeys. Phew!!!!

You want to translate, then translate my definitions exactly.

Here is the translation of what I wrote about R: R stands for the rule of the game, which stipulates that one candidate and only one candidate per entity must be placed in all the cells of the grid.

denis_berthier wrote:But do I know what a resolution technique (RT) is? Of course not. Does this intend to restrict RTs to pattern-based solutions - which would exclude T&E...? We'll never know.

denis_berthier wrote:As we still don't know what a resolution technique is, TR is not defined. Note that this "definition" of TR is the basis for all the rest of the paper. This seems to be very shaky grounds for a theory.

Here is the full text on this subject:

A solving technique is a logical approach (reasoning) allowing to place or eliminate candidates on a grid by applying R. We designate by TR the set of solving techniques allowing to solve a grid, by TE (elementary techniques) a part of these techniques including the simplest ones which are: the search for unique candidates by sweeping the zones, the search for alignments and the search for closed sets (doublet, triplet, quadruplet, etc). These three techniques together constitute the basic techniques designated by TB.

I agree that it should be more detailed, and it was planned in a new edition that I cannot find the time to finalise.

So I specify here that TRs include TBs, Wings and other fish, etc... but also reasoning by absurdity, i.e. T&E, so everything that can be done by logical reasoning.

Last edited by Mauriès Robert on Fri Jan 28, 2022 4:38 pm, edited 3 times in total.

- Mauriès Robert
**Posts:**585**Joined:**07 November 2019**Location:**France

Mauriès Robert wrote:I continue to comment on your writings in the order in which they were published...

Here is the full text on this subject:

A solving technique is a logical approach (reasoning) allowing to place or eliminate candidates on a grid by applying R. We designate by TR the set of solving techniques allowing to solve a grid, by TE (elementary techniques) a part of these techniques including the simplest ones which are: the search for unique candidates by sweeping the zones, the search for alignments and the search for closed sets (doublet, triplet, quadruplet, etc). These three techniques together constitute the basic techniques designated by TB.

I agree that it should be more detailed, and it was planned in a new edition that I cannot find the time to finalise.

TR remains undefined. The broadest interpretation is the first I gave, i.e. applying the full power of logic to R.

No need to talk of the rest.

Mauriès Robert wrote:You have translated "either" which here means consequently or therefore, by "equivalent".

I haven't translated "either" but "i.e.". I don't know in which dictionary you found that "either" means "therefore".

In any case, all this is now totally beside the point, because my proofs apply for any resolution theory, including applying the full power of FOL to the 4 rules of Sudoku; and your theorem 2.1 his false for any of them.

I see you are again trying to drown the fish, as you have done for 2+ years with François; that's not a very good idea.

Again, the only question now is: is there anything wrong with my proofs (according to you)?

And take this as a stern warning: stop insulting me (delirium...) or you will be reported.

Last edited by denis_berthier on Fri Jan 28, 2022 4:36 pm, edited 1 time in total.

- denis_berthier
- 2010 Supporter
**Posts:**3676**Joined:**19 June 2007**Location:**Paris

denis_berthier wrote:TR remains undefined. The broadest interpretation is the first I gave, i.e. applying the full power of logic to R.

No need to talk of the rest.

In any case, all this is now totally beside the point, because my proofs apply for any resolution theory, including applying the full power of FOL to the 4 rules of Sudoku; and your theorem 2.1 his false for any of them.

I see you are again trying to drown the fish, as you have done for 2+ years with François; that's not a very good idea.

Again, the only question now is: is there anything wrong with my proofs (according to you)?

And take this as a stern warning: stop insulting me (delirium...) or you will be reported.

I asked you to let me answer all your comments, I will answer all your questions, but I need time (writing, translating, editing...). By tonight it will be done.

As for the delirium, it's not an insult, but the fact that you're spending all this time to dismantle my TDP.

- Mauriès Robert
**Posts:**585**Joined:**07 November 2019**Location:**France

I continue my comments on what you write...

It is not false, but it is just that you did not understand that I make a distinction between the theory necessary to prove certain properties (or theorems) and the practical use intended for the sudokists. Now my goal is to propose to sudokists a way (Technique of tracks) to solve which avoids them (if they do not master them) the use of the techniques known as advanced (Wing fish, etc...). So to use only the TB, which does not prevent that the properties and theorems are established even for those who would like to use within the TDP Wings fishs etc...

Indeed, I used a more popular vocabulary which "speaks" to sudoku lovers, who for their great majority, I guarantee you, do not know what is a CSP variable and even less a 2D cell.

You have translated "either" which means here by consequently or therefore, by i.e to conclude: P(E) = ∩E P(Ak ) is equivalent to P(E) ⊆ P(Ak ) ∀k.

This is a falsification of the meaning of my sentence to make me look like a moron, it is unacceptable and unworthy of you!

Having said all this, I now come to the questions you are asking me:

I answered François one earlier in this thread. I put this question aside until I could prove that the Theorem is right, because what François showed was only that my reasoning is wrong in the case of a multiple solution puzzle.

But the time was taken up by many other things. And I forgot this problem in the document. But no one, not even François, drew my attention to the fact that it was still on the site without correction, proof without doubt that it was not read by anyone, not to say that it interested no one. I myself did not consult this document for 2 years before you became interested in it.

It also did not escape you that this theorem is not in the TDP document that I published on this forum and which is later than the one we are dealing with. It is thus an oversight !

Yes, I maintain that this theorem, let us say this conjecture since my demonstration is false, is right, in any case as long as a counter-example is not opposed, I say well a counter-example and not a demonstration.

I come back to this below with your demonstration.

In your demonstration you write :

By definition, P'(E2) is defined by X ∊ P'(E2) iff eliminating all the elements of E2 allows to prove X in TR (TB)

i.e. X ∊ P'(E2) iff the disjunction of all the elements in E1 implies X in TR (TB)

i.e. X ∊ P'(E2) iff (A or B or C...) => X

i.e. X ∊ P'(E2) iff A=>X or B=>X or C=>X...

i.e. X ∊ P'(E2) iff X ∊ P(A) or X ∊ P(B) or X ∊ P(C) ...

Said otherwise, P'(E2) is the union of all the P(Ak), Ak in E1 (while P(E1) is their intersection).

I disagree with the conclusion. What you show is: P'(E2) ⊂ in the union of P(Ak).

So for the moment, and unless I have not understood your demonstration, I maintain that the theorem is not false.

I hope I have answered all your questions.

Robert

denis_berthier wrote:Resolution technique: first interpretation

In the absence of an answer, here is my first interpretation of what a TR is. This is the most obvious one, the one that any mathematician would consider as the most likely.

A TR is any way of proving logical consequences of the 4 axioms.

For a puzzle with a single solution, this has the unfortunate consequence that any track is:

- either the set of all the candidates (inconsistent puzzle);

- or the set of all the candidates that are true in the solution.

(The basis for this is the well-known theorem that you can find in the first pages of any Logic book: a formula is provable in a theory T if and only if it is true in all the models of T).

In both cases, a track is totally independent of its starting point and the notion is devoid of any use.

It is not false, but it is just that you did not understand that I make a distinction between the theory necessary to prove certain properties (or theorems) and the practical use intended for the sudokists. Now my goal is to propose to sudokists a way (Technique of tracks) to solve which avoids them (if they do not master them) the use of the techniques known as advanced (Wing fish, etc...). So to use only the TB, which does not prevent that the properties and theorems are established even for those who would like to use within the TDP Wings fishs etc...

denis_berthier wrote:Translator = me a écrit:

An entity is the set of candidates for the same cell or the set of candidates with the same number in the same unit"

denis_berthier wrote:(I've translated to more usual Sudoku vocabulary.)

In my vocabulary, an "entity" is merely the content of a 2D-cell or CSP-Variable.

Indeed, I used a more popular vocabulary which "speaks" to sudoku lovers, who for their great majority, I guarantee you, do not know what is a CSP variable and even less a 2D cell.

denis_berthier wrote:Ah, yes, something more; just a remark on the fly: the end of definition 3 - "P(E) = ∩E P(Ak ), i.e. P(E) ⊆ P(Ak ) ∀k" - must be a new fundamental theorem in Set Theory, making Robert a potential candidate for a Fields Medal.

Starting from today, P(E) = ∩E P(Ak ) is equivalent to P(E) ⊆ P(Ak ) ∀k; you don't need to prove the reverse inclusion.

That'll make a lot of students happy; they can replace half of their home work by playing with their Nintendo.

You have translated "either" which means here by consequently or therefore, by i.e to conclude: P(E) = ∩E P(Ak ) is equivalent to P(E) ⊆ P(Ak ) ∀k.

This is a falsification of the meaning of my sentence to make me look like a moron, it is unacceptable and unworthy of you!

Having said all this, I now come to the questions you are asking me:

denis_berthier wrote:The only questions are:

- after François told you the proof of Theorem 2.1 has an error, why did you not investigate it seriously but keep it online for 2 or 3 years as if it was true?

- after my proof, do you finally admit that theorem 2.1 of the 2017 paper is false or will you continue trying to gaslight us and the readers of your forum?

I answered François one earlier in this thread. I put this question aside until I could prove that the Theorem is right, because what François showed was only that my reasoning is wrong in the case of a multiple solution puzzle.

But the time was taken up by many other things. And I forgot this problem in the document. But no one, not even François, drew my attention to the fact that it was still on the site without correction, proof without doubt that it was not read by anyone, not to say that it interested no one. I myself did not consult this document for 2 years before you became interested in it.

It also did not escape you that this theorem is not in the TDP document that I published on this forum and which is later than the one we are dealing with. It is thus an oversight !

Yes, I maintain that this theorem, let us say this conjecture since my demonstration is false, is right, in any case as long as a counter-example is not opposed, I say well a counter-example and not a demonstration.

I come back to this below with your demonstration.

denis_berthier wrote:We've gone past my first post. We've reached a point where I've proven that theorem 2.1 is false.

My translations are not missing any part related to theorem 2.1. and they are totally faithful. My proof uses only what I've translated. If you claimed the contrary, you would have to say where precisely.

In your demonstration you write :

By definition, P'(E2) is defined by X ∊ P'(E2) iff eliminating all the elements of E2 allows to prove X in TR (TB)

i.e. X ∊ P'(E2) iff the disjunction of all the elements in E1 implies X in TR (TB)

i.e. X ∊ P'(E2) iff (A or B or C...) => X

i.e. X ∊ P'(E2) iff A=>X or B=>X or C=>X...

i.e. X ∊ P'(E2) iff X ∊ P(A) or X ∊ P(B) or X ∊ P(C) ...

Said otherwise, P'(E2) is the union of all the P(Ak), Ak in E1 (while P(E1) is their intersection).

I disagree with the conclusion. What you show is: P'(E2) ⊂ in the union of P(Ak).

So for the moment, and unless I have not understood your demonstration, I maintain that the theorem is not false.

I hope I have answered all your questions.

Robert

- Mauriès Robert
**Posts:**585**Joined:**07 November 2019**Location:**France

Mauriès Robert wrote:In your demonstration you write :

By definition, P'(E2) is defined by X ∊ P'(E2) iff eliminating all the elements of E2 allows to prove X in TR (TB)

i.e. X ∊ P'(E2) iff the disjunction of all the elements in E1 implies X in TR (TB)

i.e. X ∊ P'(E2) iff (A or B or C...) => X

i.e. X ∊ P'(E2) iff A=>X or B=>X or C=>X...

i.e. X ∊ P'(E2) iff X ∊ P(A) or X ∊ P(B) or X ∊ P(C) ...

Said otherwise, P'(E2) is the union of all the P(Ak), Ak in E1 (while P(E1) is their intersection).

I disagree with the conclusion. What you show is: P'(E2) ⊂ in the union of P(Ak).

So for the moment, and unless I have not understood your demonstration, I maintain that the theorem is not false.

I hope I have answered all your questions.

No, what I show is exactly what I've written; i.e. the equality: P'(E2) is the union of all the P(Ak), Ak in E1. It's amazing that you're unable to understand such an elementary proof!

- denis_berthier
- 2010 Supporter
**Posts:**3676**Joined:**19 June 2007**Location:**Paris

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