The sequence from to yields a ko for life. Black cannot connect at a without removing the white stones because of shortage of liberties.
Note that the order of and can be interchanged.
To avoid the ko is impossible - here, White falsifies an eye so Black is left a cyclops. ( at is a self-atari.)
JoelR: This is one of a class of problems where it looks like the defender has a sure thing because of a set of miai plays, but the task is to find the one combination that makes the pairings false. In this case, Black wants an eye at , and can get it if he holds either a or b, and either c or d. But wait, if White takes both b and d.... I don't see this line of reasoning presented as often as it occurs to me. A good example where it is on the page is six die but eight live.