Weird thinking, Robert. Your second sentence is true for (2a) also.

Wait and see


Let's ring the bell.

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

Your turn, Robert
Perhaps you will realise even better now that it is not necessary to have the strings primary points included in local-2 for this proof.



In a position, a string of a player is "two-eye-alive" if the opponent cannot force no intersection of the string with a two-eye-formation on.

...

For the implication two-eye-alive ->J2003-alive, imagine that a string is two-eye-alive. ***The string can either be uncapturable or not uncapturable*** Either the opponent cannot force capture of the string or he can force capture of the string.

(1) ***The string is uncapturable*** The opponent cannot force capture of the string -> It is uncapturable. -> It is J2003-alive

(2) ***It is not uncapturable -> The string is either capturable-1 or not capturable-1*** The opponent can force capture of the string. ->
Because the string is two-eye-alive there is in every hypothetical-strategy of its opponent a hypothetical-sequence in which we reach a two-eye-formation that includes one of its intersections. For every hypothetical-strategy H of the opponent, we choose a hypothetical-sequence S(H) in it where the oponent reaches a two-eye-formation and subsequently only passes. Because the two-eye-formation cannot be captured by only moves of its opponent, it consists of permanent stones.
-> In S(H) the two-eye-formation that is formed on at least one intersection of the captured string has either a stone on local-1 of the string or it does not have a stone on local-1 of the string.

(2a) ***It is capturable-1*** The two-eye-formation has a stone on local-1 of the string -> the string is capturable-1 -> It is J2003-alive

(2b) ***It is not capturable-1*** The two-eye-formation has no stone on local-1 of the string -> ***Because the string is two-eye-alive there is in every hypothetical-strategy of its opponent a hypothetical-sequence in which we reach a two-eye-formation that includes one of its intersections. For every hypothetical-strategy H of the opponent, we choose a hypothetical-sequence S(H) in it where the oponent reaches a two-eye-formation and subsequently only passes. Because the two-eye-formation cannot be capture by only moves of its opponent, it consists of permanent stones. In S(H) the two-eye-formation that is formed on the captured string has either a stone on local-1 of the string or it does not have a stone on local-1 of the string***

***(2b1) If it has a stone on local-1 of the string, it is also on local-2.
(2b2) If it does not have a stone on local-1 of the string, then*** local-1 of the string consists of the one or both of the empty intersections of the two-eye-formation. Actually, it consists of one of the intersections since if it would consist of both, these would have to be adjacent to each other which contradicts the definition of a two-eye-formation. So, local-1 of the string consists of one intersection and during S(H) it becomes one of the the empty points of a two-eye-formation. This implies that this two-eye formation includes strings that occupy the intersections adjacent to local-1. Because local-1 consists of one intersection these adjacent intersections where empty or occupied by opposing stones. Hence, these intersections belong to local-2\1 of the string.

***In both (2b1) and (2b2)*** we see that the two-eye-formation that is formed in S(H) has permanent stones on local-2\1 of the string. Hence, in every hypothetical-strategy of the opponent of the string there is a hypothetical-sequence where a permanent-stone is played on local-2\1. Hence, the opponent cannot force ***both capture of the string and*** ---(superfluous, but does not any harm, repeats first sentence of (2), which is valid as long as we are in (2))--- no local-2\1 permanent stone. Hence, the string is capturable-2\1. Hence it is capturable-2. Hence, it is J2003-alive.

Hence, under the assumption that the string is two-eye-alive, we find that
it is J2003-alive . QED.

----------------------------
BLUE underlined = edited, see Robert's posting below.

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

If one takes into acoount the amount of time, Robert has spent for "rules", and the level of maturity of his texts, it would be impudent to expect more than small steps from the level he has reached.


What Robert refers to as "Basic Rules" does not provide any serious problem.

Basic Rules

- The usual fundamental rules apply.
- The game consists of the phases 1. Competition, 2. Analysis, 3. Scoring.
- The Competition is a sequence of alternating moves.
- A move is either a play or a pass.
- Two successive passes end each sequence of moves.
- Removed stones become prisoners.
- Suicide is prohibited.
- The basic ko rule prohibits a player to make a play just after which the position just before the preceding opposing play would be recreated.

The second sentence is a kind of meta-statement; the following ones refer to "Competition" in the first instance.

"Competition" is named "Play" by me in my "Cassandra's rules" thread.
"Scoring" is named "Score" by me.
"Analysis" is named "Evaluation" by me. Robert likes to have "identification of stones that can be removed from the primary position and their removal" and "identification of territory" as part of "Analysis"; my preference is to have these steps in "Score".

All these aspects are nothing more than a matter of personal taste.


Within territory oriented rules you have to identify strings, which can border territory, and strings, which cannot. In one of the phases after Competition / Play, strings of the second type, which are complety surrounded by opponent's strings of the first type, will be removed without play and become prisoners.


Robert's identifying concept relies on "Independently Alive", following his J2003.

Independently Alive

- The only considered type of life is being independently alive.
- The concept independently alive relies in two other concepts: force and two-eye-formation.
- "force" is a technical term used in the Japanese 2003 Rules. It can be understood also intuitively though: That a player forces something means that he can and does choose his moves well enough to always achieve it - regardless of how the opponent replies.
- A "two-eye-formation" consists of one or several strings of the same player and exactly two empty intersections so that these properties are fulfilled: 1) Each of the strings is adjacent to each of the two intersections. 2) None of the strings is adjacent to another empty intersection. 3) Each of the two intersections is adjacent only to the strings.
- A string is "independently alive" if its player moving second can force to get a two-eye-formation on at least one of its intesections.

The difference to my proposal can be found in the last sentence: "at least one".

This will come to life only, when evaluating strings, which are part of not outplayed positions, named "Seki" colloquially in the world of J1989.
Usually one of two neighboured strings of different colours will become "independently alive", the other one will become "not independently alive". The latter would become prisoners after Competition / Play.
In my proposal, both strings would become "not two-eyed", so none would become prisoners after Play.
Robert's version is disadvantageous for the player, who owns the "not independently alive" string; my version is disadvantageous for this player, too, but to a smaller extent. In both versions this player would be better, had he played out the position during Competition / Play.

As Robert stated correctly, positions of this type will not be "common" after endgame. Despite this fact, it is important that the rule set gives a consistent solution for this type of position.


Robert's Analysis has the same effect on identifying strings, which can border territory, and strings, which cannot, as my proposal.

Analysis

- The Analysis consists of these steps: 1. Determination of the independently alive strings. 2. Removal. 3. Determination of territory.
- Each string is analysed separately. It is determined whether or not it is independently alive.
- For analysis of each string, imagined move-sequences starting by the attacker are considered. The defender shall try to prove "independently alive" while the attacker shall try to prove "not independently alive". Until the truth has been revealed, yet more sequences must be considered.

The content of the following sentences can be found in my proposal within "Score":

- During the step Removal, one considers the connected regions that are adjacent only to one player's independently alive strings and that consist of intersections being empty or having opposing not independently alive strings on, which are removed.
- During the step Determination of territory, a player's territory consists of the intersections of connected empty regions adjacent only to his independently alive strings.

Scoring

- Territory Scoring applies according to the territory determined in the Analysis.
- The score is visualized by means of Japanese Fill-in Counting.


The final result of "Long Cycle Repetition" will be the same, too.

Long Cycle Repetition

- During Competition, a cycle ends the game exceptionally and immediately. It is, however, tolerated that the players notice occurrence of a cycle delayed.
- During Competition, a cycle with an equal number of new prisoners lets the game result be a tie. For traditional reasons, this is also called "No Result" or "Neither Victory Nor Defeat" and the players are perceived to agree on the fact that such a cycle has occurred.
- During Competition, a cycle with an unequal number of new prisoners lets the game result be a win for the player with the smaller number. For traditional reasons, it is tolerated though that the players perform the cycle up to ca., say, a thousand times because the player having collected enough prisoners can at some time depart, let the opponent control the entire board but still win by Scoring.
- During Analysis, every sequence of moves leading to a cycle is treated as if ending upon completion of its first cycle. No stone played during such a cycle is "independently alive" though. This is equivalent to the original rules' implicit assumption of infite recurrence of that cycle.

None of the strings, which will become part of a long cycle repetion (including the strings, which are "neighboured" to a long cycle and cannot become independently alive / two-eyed due to the cycle), will become independently alive / two-eyed.


Within "Ignored Rules and Rules Aspects", you will find the aspect that makes the difference to J1989 (and between Robert's text and mine):

Ignored Rules and Rules Aspects

...
- The hypothetical ko rule is ignored. Instead the basic ko rule applies also during the Analysis. A "triple ko with one eye and one external ko" becomes a precedent: The strings of the player having the "eye" are independently alive while the opposing strings are not independently alive. The combination of a "double ko death" and a "basic ko fight about whether to dissolve it like a teire" elsewhere on the board becomes a precedent: In the double ko death, the strings of the player having the "eye" are independently alive while the opposing strings are not independently alive. The ko stone in the basic ko is not independently alive.

The type of Ko rule during Analysis / Evaluation is part of the "forcing procedure".

The J1989 results for Life&Death-examples 16, 17, and 18, cannot be derived by the J1989 procedure.

Robert's J2003 hypothetical Ko rule is a genious idea to get the J1989-wished results be using a consistent procedure.

If he uses the basic Ko rule borrowed from Competition (what I do also in my proposal), there arises the need to have some precedents to keep the application of the rule set as "user-friendly" as possible, if you want the rule set to provide a wished result for some special positions.

If the number of precedents is limited, there will be some very "exotic" positions (like J1989 Life&Death-examples 16, 17, and 18), which results will differ from those of J1989.



Referring to the discussion of Chris' proof, primary developed in another context, it is very important in the course of the comparison of different rule sets and has a side effect on / from Robert's "user-friendly" rule set.

What Chris has proven in the second part of his proof, is that every string, which is a "two-eye-formation", is either one of Robert's J2003 "uncapturable", "capturable-1", or "capturable-2" strings. Under the precondition that the "forcing procedure" used is the same.

Perhaps this precondition has been underestimated by Robert so far. The "forcing procedure" is immanent part of "two-eye-formation", so one has to pay attention to this dependency. This finding may be one of the small steps I mentioned in the very beginning of this posting.

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

There is really a lot to be done still: rules and formal go theory. Much of it would be big tasks. Most tasks require huge amounts of time though. E.g., although I have educated myself now to solve Ing Ko Rules in principle, I still consider the remainder a huge task. E.g., the step from defining ko to defining ko threat will be similarly tough as the one from basic ko to ko in general, I think.



No. I just had no time to work on this much so far.

LOL.

I hope I will find some time for reading during the following days.

Perhaps you should concentrate on the preliminaries this week.

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

This must be "[...] formed on at least one intersection of the captured string", right?

Yes, indeed.

I just copied the text from below.

At its primary position in Chris' original proof the string consists of only one stone, so the extension "at least one intersection of" is now badly needed at the new position, of course.

Apparently I had been too overcautious with not changing this part of the original text.

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

[now proven, see below]

The second part of Chris Dams's proof for "WAGC-alive equals J2003-alive" is done in an alternative way by Cassandra with help from Jasiek:


For the implication two-eye-alive -> J2003-alive, imagine that a string is two-eye-alive. Either the opponent cannot force capture of the string or he can force capture of the string.

(1) The opponent cannot force capture of the string -> It is uncapturable. -> It is J2003-alive.

(2) The opponent can force capture of the string. -> Because the string is two-eye-alive there is in every hypothetical-strategy of its opponent a hypothetical-sequence in which we reach the player's two-eye-formation that includes one of its intersections. For every hypothetical-strategy H of the opponent, we choose a hypothetical-sequence S(H) in it where the player reaches such a two-eye-formation and subsequently only passes. Because the two-eye-formation cannot be captured by only moves of its opponent, it consists of permanent-stones. -> In S(H) the two-eye-formation that is formed on at least one intersection of the captured string has either a stone on local-1 of the string or it does not have a stone on local-1 of the string.

(2a) The two-eye-formation has a stone on local-1 of the string -> the string is capturable-1 -> It is J2003-alive.

(2b) The two-eye-formation has no stone on local-1 of the string -> Local-1 of the string consists of one or both of the empty intersections of the two-eye-formation. Actually, it consists of one of the intersections since if it would consist of both, these would have to be adjacent to each other which contradicts the definition of a two-eye-formation. So, local-1 of the string consists of one intersection and during S(H) it becomes one of the empty points of a two-eye-formation. This implies that this two-eye-formation includes strings that occupy the intersections adjacent to local-1. Because local-1 consists of one intersection these adjacent intersections were empty or occupied by opposing stones. Hence, these intersections belong to local-2\1 of the string. We see that the two-eye-formation that is formed in S(H) has permanent-stones on local-2\1 of the string. Hence, in every hypothetical-strategy of the opponent of the string there is a hypothetical-sequence where a permanent-stone is played on local-2\1. Hence, the opponent cannot force both capture of the string and no local-2\1 permanent-stone. Hence, the string is capturable-2\1. Hence it is capturable-2. Hence, it is J2003-alive.

Hence, under the assumption that the string is two-eye-alive, we find that it is J2003-alive. QED.

Given a final-position, a string in it, and the definitions of local-2\1, capturable-2\1, J2003-alive-2\1, WAGC-alive-in-seki-2\1, WAGC-alive-2/1.

Proposition:

The string is WAGC-alive-2\1 equals the string is J2003-alive-2\1.

Proof:

Part 1 of the proof is analogue to part 1 of Chris Dams's proof.

Part 2 of the proof is as follows:

For the implication two-eye-alive -> J2003-alive-2\1, imagine that a string is two-eye-alive. Either the opponent cannot force capture of the string or he can force capture of the string.

(1) The opponent cannot force capture of the string -> It is uncapturable. -> It is J2003-alive-2\1.

(2) The opponent can force capture of the string. -> Because the string is two-eye-alive there is in every hypothetical-strategy of its opponent a hypothetical-sequence in which we reach the player's two-eye-formation that includes one of its intersections. For every hypothetical-strategy H of the opponent, we choose a hypothetical-sequence S(H) in it where the player reaches such a two-eye-formation and subsequently only passes. Because the two-eye-formation cannot be captured by only moves of its opponent, it consists of permanent-stones. -> In S(H) the two-eye-formation that is formed on at least one intersection of the captured string has either a stone on local-1 of the string or it does not have a stone on local-1 of the string.

(2a) The two-eye-formation has a stone on local-1 of the string -> the string is capturable-1 -> It is J2003-alive-2\1.

(2b) The two-eye-formation has no stone on local-1 of the string -> Local-1 of the string consists of one or both of the empty intersections of the two-eye-formation. Actually, it consists of one of the intersections since if it would consist of both, these would have to be adjacent to each other which contradicts the definition of a two-eye-formation. So, local-1 of the string consists of one intersection and during S(H) it becomes one of the empty points of a two-eye-formation. This implies that this two-eye-formation includes strings that occupy the intersections adjacent to local-1. Because local-1 consists of one intersection these adjacent intersections were empty or occupied by opposing stones. Hence, these intersections belong to local-2\1 of the string. We see that the two-eye-formation that is formed in S(H) has permanent-stones on local-2\1 of the string. Hence, in every hypothetical-strategy of the opponent of the string there is a hypothetical-sequence where a permanent-stone is played on local-2\1. Hence, the opponent cannot force both capture of the string and no local-2\1 permanent-stone. Hence, the string is capturable-2\1. Hence, it is J2003-alive-2\1.

Hence, under the assumption that the string is two-eye-alive, we find that it is J2003-alive-2\1. QED.

Some minor suggestions:



(2) The opponent can force capture of the string. -> Because the string is two-eye-alive there is in every hypothetical-strategy of its opponent a hypothetical-sequence in which we reach the player's two-eye-formation that includes one of its intersections. For every hypothetical-strategy H of the opponent, we choose a hypothetical-sequence S(H) in it where the player reaches such a two-eye-formation and subsequently only passes. Because the two-eye-formation cannot be captured by only moves of its opponent, it consists of permanent-stones. -> In S(H) the two-eye-formation that is formed on at least one intersection of the captured string has either at least one stone on local-1 of the string or it does not have a stone on local-1 of the string.

(2a) The two-eye-formation has at least one stone stone on local-1 of the string -> the string is capturable-1 -> It is J2003-alive-2\1.

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

It is certainly possible to make lots of further minor improvements in annotation style. E.g., one could add a few "of the player" phrases.

More importantly, I would like to see studies of the differences and relations between capturable-2 and capturable-2\1. Can we classify the set of different positions?

What do you mean with "different positions" ?

Capturable-2 can be seen as extension of capturable-2\1 or capturable-2\1 as part of capturable-2.

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

a) "different positions" in its strict sense.
b) You suggest some classification scheme and count for us the number of classes.



Do you mean this?

Proposition:

The set of capturable-2\1 strings is a subset of the set of capturable-2 strings.

Proof:

"Trivial."

Remark:

Nice. Now where are example positions?:)

Your wish to have "classes" defined for the sake of camparison of capturable-2 and capturable-2\1 is a result of the somewhat ineffective, somewhat overdefined multi-step procedure to identify "life" within your J2003. And what is even worse: the steps cannot be seen independent from each other.



Step 1: Identify uncapturable strings.
Is applied to: all strings.
"Identify" depends on: "force".

A player's final-string is uncapturable if the opponent cannot force capture of its stones.
A permanent-stone is a stone that is played during a hypothetical-sequence and then not removed during the rest of the hypothetical-sequence.
For a final-string, local-1 is all the string's intersections.


Step 2: Identify capturable-1 strings.
Is applied to: all not uncapturable strings.
"Applied to" depends on: "uncapturable"
"Identify" depends on: "force", "local-1".

A player's final-string is capturable-1 if

• it is not uncapturable and
• the opponent cannot - with the same hypothetical-strategy - force both capture of the string's stones and no local-1 permanent-stone of the player.

Step 3: Identify local-2(\1).
Is applied to: all not uncapturable strings.
"Applied to" depends on: "uncapturable"
"Identify" depends on: "force", "uncapturable", "capturable-1".

(3a) For a player's final-string, local-2 is local-1 and, recursively, any adjacent intersection without a stone of a string that is of the player and either uncapturable or capturable-1.
(3b) For a player's final string, local-2\1 is any intersection of the string's local-2, which does not belong to the string's local-1.

Step 3 and step 4 are NOT restricted to those strings without status so far, when using local-2.
There is no need to include strings with the known status "capturable-1", what is done by including local-1 in local-2.


Step 4: Identify capturable-2(\1).
Is applied to: all not uncapturable strings (with local-2\1: which are not capturable-1).
"Applied to" depends on: "uncapturable"
"Identify" depends on: "force", "uncapturable", "capturable-1", "local-1" (only if referring to "local-2"), "local-2\1".

A player's final-string is capturable-2 if

• it is neither uncapturable nor capturable-1 and
• the opponent cannot - with the same hypothetical-strategy - force both capture of the string's stones and no local-2 permanent-stone of the player.

A player's final-string is capturable-2\1 if

• it is neither uncapturable nor capturable-1 and
• the opponent cannot - with the same hypothetical-strategy - force both capture of the string's stones and no local-2\1 permanent-stone of the player.

Your fear is:
There will be a "forced" hypothetical strategy in step 4, which places a permanent stone in local-1 of a string that had not been identified as "capturable-1" in step 2.

But this permanent stone in local-1 fulfils the condition for "capturable-1" in step 2, what apparently both players overlooked when busy with step 2.

So - late, but not too late - the string has been identified as "capturable-1".

Thereafter step 3 und step 4 (both dependent on the result of step 2) must be done again.

Your very special chain will not take part in the second course !



What is needed to identify "capturable-2" is nothing more than local-2\1.

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

Sure. It is necessary for J2003's purposes. For other purposes, one can approach things much more easily. E.g., "alive" is "being on the board".



It is not defined that way. My "fear" persists.



If you can prove that:)



Interesting claim. Can you prove it?

Robert, the need to have local-1 included in local-2 follows from this tiny flaw in your construction, which is not free from feedback loops.

But no further information of the string's status can be gained that is not known already by using local-2\1. This is the decisive point: "no further information".

It is not decisive that the opponent can force "no permanent stone in local 1" when the aim is to answer the question "Is the string capturable-1 ?", and can force either "at least one permanent stone in local-1" or "at least one permanent stone in local-2\1" when the aim is to answer the question "Is the string capturable-2 ?". This is true even for White's single stone in 1989 Nihon Kiin Life & Death example #1. But the "ordinary" use of "force" would end in "local-2\1", an end in "local-1" is just a prestidigitation by the opponent.

When the opponent does not have this choice, there is no possibility to have a permanent stone in local-1 (Nihon Kiin #4, for example).

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)

It is not a flaw in my construction and there is no feedback loop. Regardless, as you have pointed out, it is (for other purposes) also possible to use different constructions like such with fewer levels and greater segregation of the local-1.



Prove it! (BTW, "information" is a very mighty word. You might try to start with much more modest claims. Even each such will be very difficult to prove. E.g., "Capturable-2 equals capturable-2\1." or "Given a hypothetical-strategy for capturable-2\1, the hypothetical-strategy together with certain further left-parts would not [do something new] when used for capturable-2".)



As long as we do not know much better the relation between capturable-2 and capturable-2\1, we should be careful with such statements.

Let me try to explain my current view on your J2003-world, Robert. Perhaps this will give you a better understanding of what lies behind my quoted statements above.

Starting point of our journey will be your fear that the opponent might misuse the definition of "capturable-2\1".

When player and opponent cooperate, they are able to create each of the following results for the status evaluation of a single string:

(A) The string will remain on the board.
(B) The string will not remain on the board.
(B1) A permanent stone of the player becomes established on every point of the primary string.
(B2) A permanent stone of the player becomes established on at least one point of the primary string.
(B3) There is no point of the primary string, where a permanent stone of the player becomes established.
(B31) At least one permanent stone of the player becomes established in a certain area of the board.
(B32) There in no point in a certain area of the board, where a permantent stone of the player becomes established.

But this sort of cooperation is clearly not meant when using "force".

Instead the instructions for the opponent are as follows:

• Use all your capabilities to gain (B32).
• If this is not possible, use all your capabilities to gain (B31).
• If this is not possible, use all your capabilities to gain (B2).
• (B1 is not relevant for J2003.)
• If this is not possible, the default result is (A).

The instrctions for the player read:

• Use all your capabilities to gain (A).
• (B1 is not relevant for J2003.)
• If this is not possible, use all your capabilities to gain (B2).
• If this is not possible, use all your capabilities to gain (B31).
• If this is not possible, the default result is (B32).

The analysis of the set of "possible" evaluation sequences will end well-defined, with only one (1 !) status for every single string.

If the well-defined status is (A), this corresponds to "uncapturable" of J2003.

If the well-defined status is (B2), this corresponds to "capturable-1" of J2003.
All the points of the primary string are refered to as "local-1" in J2003.

When talking about the well-defined status (B31), the aspect comes to life refered to as "feedback loop" by me. This "feedback loop" has to do with what is "a certain area of the board" in my primary list.

But first I should highlight that in (B31) there will be no permanent stone of the player become established in "local-1". It follows, that the definition of a "certain area", which includes "local-1" is equivalent to the definition of the same (rest of the) "certain area" (defined before), with "local-1" excluded.
In J2003 this "certain area" is defined as "local-2", and this equivalent to "local-2\1" for the purpose of analysis.

The definition of "local-2" in J2003 refers to "capturable-1", what is equivalent to (B2) here. This will make it mandatory to repeat the analysis for all what has been identified as (B2) so far, if "later" in the course a status (B2) is found, which has not been visible to player and opponent before. This is because some "certain area" has changed now.
(This is true for the "later" finding of (A), too, with analogue consequences.)

But even if this happens accidentally, all final evaluation statuses will remain well-defined.

The well-defined status (B31) is refered to as "capturable-2" in J2003 and is equivalent to "capturable-2\1".

_________________
The really most difficult Go problem ever: https://igohatsuyoron120.de/index.htm
Igo Hatsuyōron #120 (really solved by KataGo)