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.

So why is the proverb four is five, and not four is six? Because you only start counting the liberties after white plays the vital point of one. Otherwise, black is just alive. Why do we only start counting it after white has killed? That is because the question of liberties for big eyes mostly comes up in regards to capturing races.

So why is the proverb four is five, and not four is six? Because you only start counting the liberties after white plays the vital point of one. Otherwise, black is just alive. Why do we only start counting it after white has killed? That is because the question of liberties for big eyes mostly comes up in regards to capturing races.

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.