Yep that seems to be the case

Probably you can prove it too with reversible options too, one of the discussions for simplifying games to a canoncial form. Though I have a feeling then when you do the reversible options calculation, you end up just proving that {↓|}≥0 and {↓|}≤0 and hence {↓|}=0. So probably one could start with doing that instead either if one wanted.

Apparently that does work XD I think then what I was confused about was that it didn’t fit with the calculation like

because I guess this relative way of calculating score is trying to ignore most stones entirely, but in area scoring they are worth something, exactly as much as an empty space in area scoring.

```
⚫⚪⚪⚪
⚫➕⚫⚪
⚫⚪⚪⚪
Black ↙ ↘ White
⚫⚪⚪⚪ ⚫⚪⚪⚪
⚫⚫⚫⚪ ⚫⚪⬛⚪
⚫⚪⚪⚪ ⚫⚪⚪⚪
-2 -6
```

while in this way I’m factoring in each stone being worth a point. So if in the second way we remove two white stones, it would be like adding +2 onto the position in some sense (2 points to black)

```
⚫⚪⚪➕
⚫➕⚫⚪
⚫⚪⚪➕
Black ↙ ↘ White
⚫⚪⚪➕ ⚫⚪⚪➕
⚫⚫⚫⚪ ⚫⚪⬛⚪
⚫⚪⚪➕ ⚫⚪⚪➕
0 -4
```

assuming the removed stones aren’t in white’s territory, otherwise the score doesn’t change, and two dame would cancel out anyway.

And in fact 2+{-2|-6} = {0|-4}.

I think that feels to me, like it’s going in the right direction, having the properties you want, if you really wanted to do it with area scoring over territory.

I think this is an example of a more general thing called the Number Translation Theorem, (Thm 3.21 in Siegel).

Suppose x is equal to a number and G is not, then

G+x={G^{L}+x|G^{R}+x}

So I think that as long as you score positions in the right way with area scoring (like I’m mentioning above), then this number translation theorem allows you to add or remove any number of stones from a position, and it will simply just translate the score/game by a fixed amount.