Cassandra wrote:
***RED*** = deleted
BLUE = inserted
---(BLUE ITALIC)--- = comment
[Conjecture:]
If a string is two-eye-alive, it follows that it is J2003-alive.
That's the topic of part II of Chris' 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*** The opponent can either force capture of the string or not. ---(This is what I mean with "primary")---
(1) ***The string is uncapturable*** The opponent cannot force capture of the string -> It is uncapturable. ---(This is what I mean with "secondary", the J2003-term follows from a "neutral" clause)--- -> 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. -> The opponent can force either no permanent stone on local-1 or not.
(2a) ***It is capturable-1*** The opponent cannot force no permanent stone on local-1 -> the string is capturable-1 -> It is J2003-alive
(2b) ***It is not capturable-1*** The opponent can force no permanent stone on local-1 -> 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 ---(this follows from the first sentence of (2b)-new)---
***(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*** ---(superfluous now)---
I do not read the rest of your proof change attempt yet. Let me comment on the cited part.
First I need to say something about your English. In maths, careful English is essential. So I correct samples (not as precise as J2003 but precise enough for your purpose):
"The opponent can either force capture of the string or not."
->
"Either the opponent can force capture of the player's string or the opponent cannot force capture of the player's string."
"The opponent can force no permanent stone on local-1"
->
"The opponent can - with the same hypothetical-strategy - force both capture of the string's stones and no local-1 permanent-stone of the player."
Your other inserted texts need the same precision.
Now to the contents:
You write:
Quote:
(2b) ***It is not capturable-1*** The opponent can force no permanent stone on local-1 -> 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 ---(this follows from the first sentence of (2b)-new)---
With precise English, this should be:
Code:
(2b) ***It is not capturable-1*** The opponent can - with the same hypothetical-strategy - force both capture of the string's stones and no local-1 permanent-stone of the player. -> 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 ---(this follows from the first sentence of (2b)-new)---
Here you make an oversight that leads to a mistake.
The hypothetical-strategy used by the opponent for "can force both capture of the string's stones and no local-1 permanent-stone of the player" is not necessarily the same hypothetical-strategy H we assume for "the string is two-eye-alive".
Therefore your further deletions in (2b), (2b1), (2b2) for the intention of simplification may not be made.
Your proof rewriting goes to the next round.