hoshi911: solution counts one liberty less for white as it in fact has.
In particular, white's T-shape at the top is worth 2 liberties, not
just one:
white can force J13, black needs to answer J15, and when white invests
another two stones at J14 and J15, the T-shape with additionally 4
liberties is connected. So 4 liberties - 2 moves invested = 2 liberties.
d4rkm4tter: Wow. That makes it even more tricky. I modified the problem.
solution counts one liberty less for white as it in fact has. In particular, white's T-shape at the top is worth 2 liberties, not just one: white can force J13, black needs to answer J15, and when white invests another two stones at J14 and J15, the T-shape with additionally 4 liberties is connected. So 4 liberties - 2 moves invested = 2 liberties.