Back to the game tree for E-15
Code:
A
/ \ \
/ \ \_____ -1
/ \
B C
/ \ / \
BIG D 0 -BIG
/ \
1 -1
Let's simplify it by removing White's sente option.
Code:
A'
/ \
B -1
/ \
BIG D
/ \
1 -1
What is this? Another ambiguous sente/gote?
Not really, even though Black could play A'-> B -> D as a ko threat. Kos aside, A' = D. Maybe that is obvious to you, but let me show that.
Now, with no ko fight, games add and subtract. So A' = D means A' - D = 0. How do you subtract D? You form the negative of D and add it. So how do you form the negative of D? In go it's easy, you just reverse the color of the stones. For a game you flip the tree and reverse the signs of the scores.
Let's do that for D.
First, flip the tree:
Code:
D'
/ \
-1 1
Second, reverse the signs:
Code:
D''
/ \
1 -1
So the negative of D is D! How about that?
Now to find out if A' - D = 0. That is so if the result with best play, after each player has made the same number of moves, is 0, no matter who plays first. It is also the case with a seki, or standoff, where neither player wants to play or is able to play. For instance, D' above = 0. Black prefers a score of 0 to -1, and White prefers 0 to 1 (for Black), so neither play wants to play.
Let's look at A' - D = A' + D.
Code:
A' + D
/ \ / \
B -1 1 -1
/ \
BIG D
/ \
1 -1
First, suppose that Black plays first.
1) Let Black play from A' to B; then White will reply from B to D. As we now know, D + D = 0.
2) Let Black play from D to 1. Then White will play from A' to -1. 1 - 1 = 0.
Next, suppose that White plays first.
1) Let White play from A' to -1. Then Black will play from D to 1. 1 - 1 = 0.
2) Let White play from D to -1. Then Black will play A' -> B -> D -> 1. 1 - 1 = 0.
So A' = D. It's just a gote, except in ko fights.

In summary, in A, D is necessary to retain the option, C, by providing a possible reverse sente for it; and C is necessary to keep A from reducing to D.
