I am reading here: http://senseis.xmp.net/?BigEyeLiberties

The first example for the four-space big eye explains that there are 5 liberties, two for the marked stones in the first diagram, plus three for the resulting 3-space eye.

Why doesn't the atari at a in the third diagram count? Wouldn't it take 6 moves for W to pick up the black group?

I am not sure which diagram you are talking about, but I think I can answer your question. Ataris in big eyes do not count because they are answered. It is the net number of moves to capture that we are interested in.

Click Here To Show Diagram Code

[go]$$W Big eye vs. small eye
$$ . . . . . . . . .
$$ . O O O O O O O .
$$ . O X X X X X O .
$$ . O X O O . X O .
$$ . O X X . X X O .
$$ . O O X X X O O .
$$ . . X O O O X . .
$$ . . X O . O X . .
$$ . . X O . O X . .
$$ . . X O O O X . .
$$ . . X X X X X . .
$$ . . . . . . . . .[/go]

Black's big eye has three liberties to two liberties in White's small eye. Therefore Black just wins the capturing race if White plays first.

Click Here To Show Diagram Code

[go]$$W White first
$$ . . . . . . . . .
$$ . O O O O O O O .
$$ . O X X X X X O .
$$ . O X O W 1 X O .
$$ . O X X 2 X X O .
$$ . O O X X X O O .
$$ . . X O O O X . .
$$ . . X O . O X . .
$$ . . X O . O X . .
$$ . . X O O O X . .
$$ . . X X X X X . .
$$ . . . . . . . . .[/go]

Click Here To Show Diagram Code

[go]$$W Black just wins
$$ . . . . . . . . .
$$ . O O O O O O O .
$$ . O X X X X X O .
$$ . O X . 3 . X O .
$$ . O X X X X X O .
$$ . O O X X X O O .
$$ . . X O O O X . .
$$ . . X O 4 O X . .
$$ . . X O . O X . .
$$ . . X O O O X . .
$$ . . X X X X X . .
$$ . . . . . . . . .[/go]

After the Black eye has only two liberties, just enough to win the capturing race. The sequence, - - took away only one liberty of Black's big eye.

¿Es claro?

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[edit]: there used to be a brain fart here.

Thank you. I think there's some dissonance between these answers, but they both give me some fresh insights.

I don't think that's true. It's five liberties before white plays the first inside move. White needs two moves for the third to be atari, at which point black needs to respond (i.e. it takes white two moves to reduce the 4-space to a 3-space). Then the 3-space eye takes three moves to capture.

Yes, of course. I forgot to look at black having to take and white placing in the middle of 3, which cannot be counted as an extra liberty.

Thanks everyone, I actually understand this now!