Q: can anybody explain why a four-space big eye only has 5 liberties? 3 for the original form, but the result is a three-space eye. And according to SmallEyeLiberties this new form has 3 liberties. this makes 6 in total. Where am i wrong?
A: In the number of stones played by each side.
After Black throws in his third stone, White invests a stone in capturing the three Black stones. So, Black's third stone and White's capturing stone cancel each other out.
This is the useage of liberties in the sense of capturing races. (If it were the other meaning, being the number of empty adjacent points, the count would be four (4). In neither of the cases it would be 6).
The reasoning is the same for big and small eyes:
For comparison: before the 1-2 exchange, White has two liberties (adjacent points). After the exchange, White has 1 liberty (adjacent points). Still, you'll agree for White's initial group to have two liberties (in capturing races) and not 2+1=3.
The difference between big and small eyes is that for small eyes, capturing like in the previous diagram with , or doing nothing, has the same effect. Not so for big eyes: capturing refreshes the number of liberties to 3, while doing nothing leaves 1 liberty.