Bill Spight wrote:
Gérard TAILLE wrote:
Oops the notation I used is not correct because it is not clear when I am reasonning one two boards and when I switch to one board. As a a consequence I see that what I wrote is quite unclear.
Let me take some time to rephrase all my analysis before rediscussing.
Using a second board is not necessary. It is only a convenience to indicate that the two positions are independent.
In the past, discussions about difference games on a single board have run into difficulties because truly independent positions were not constructed. Using two boards makes the construction easy.

In the board I used as illustration, for instance, I did not take the time to construct independently living groups, as I technically should have. I intended the positions to be understood as independent, given the prior discussion between us. Fortunately you understood my intent.

Please do not get sidetracked by the question of two boards or one. Independence and the lack of a ko fight are the criteria.
Here is my updated analysis using only one board:
Let's consider two positions A et B surrounded by an environment E with "good" carateristics (independancy? non-ko? ...)
Assume for example that, black to play, the optimum result of a game beginning from A+E is a win for black by say 10 points
Assume also that, black to play, the optimum result of a game beginning from B+E is a win for black by say 7 points
Let's call B' the mirror position of position B
The amazing result of theory is the following: black to play will wins the game A+B' by
at least 3 points !!!
Let's us try to prove this "theorem":
Let's call E' the mirror position of position E and put on the board the four areas A, B', E, E'
Add on this board a very small area D made of one dame point.
Because that cannot penalise white, you can always impose to white to avoid using the D region except maybe at the very last move of the game.
First of all it is easy for black to win this new game by exactly 3 points by means of the following strategy:
Black begins by playing in A or E exactly as if she where playing a game with only the area A and E.
From this point till the end of the game the strategy of black is the following:
if white plays a move in A or E, black answers also in A or E exactly as if black were playing a game with only the area A and E on the board
if white plays a move in B' or E', black answers also in B' or E' exactly as if black were playing a game with only the area B' and E' on the board
in case white plays the last move in region A+E or in region B'+E', black answers with a move in D.
With this strategy is is clear that black will win the game by exactly 3 points.
Of course it may happen that this winning strategy for black is not the best result for black. That means that, by following the real best strategy, black may win by more than 3 points, say for example 5 points.
At that point comes the assumption that all areas (A, B', E, E', D) have good caracteristics allowing the following simplification:
The above game is made of the five areas A, B', E, E' and D. When you look at these five areas you see in particular the areas E and E' which look like perfect miai areas.
Here, we discover the basic assumption of all the theory:
Because E and E' are perfect miai areas, if black can win the game A+B'+E+E'+D by 5 points then, providing good independance between the five areas, we can completly ignore the presence of the two areas E and E' => black wins the game A+B'+D by again 5 points. In addition you can now ignore also the dame in D => black wins the game A+B' by again 5 points.
As a conclusion we have the equation
minMax(A-B) >= minMax(A+E) - minMax(B+E)
Is it now understandable and correct Bill?