This is a complete proof (I hope) that the position below is indeed a complete seki.
As the situation is (almost) symetrical, we have only to check Black moves at a, b, x, and y, followed or not by c. Note first that the sequence of attack after a (White capture, Black plays back at the vital point) is a capturing race (semeai), with Black having 7(BigEyes)+4 liberties, and White 11 BigEyes liberties, and White to play, so a fails. Similarly, capturing the 6 stones at y (followed by White kill), and playing c fails, except that it puts White in shortage of liberties (damezumari) on the left, enabling Black to play a next and hope to win that semeai. This is the exchange variation, but even if White accepts the exchange, Black loses one more stone than White (because of the initial symetry), so c fails in all cases
Now, what happens if Black plays y, White kills, and then Black doesn't play c ? White begins to fill the big eye. If Black plays c at any time later, the situation reverts to the case already studied. If not, after 4 more moves inside, it turns into a temporary seki on the left, but the fight between the two groups on the right is lost by Black (eyes of different size)
If Black plays b, White take the 6 stones, Black plays the vital point, and White plays a. The fight is now 10 liberties for Black, against 11 for White, so b fails too.
Robert Pauli: Wait a moment, couldn't Black mimic White after having played 'b'?
Last, if Black plays x, White kills ; Black has not gained liberties, so his best move is to pass, but then White has only to play one more stone inside to be able to win the semeai.
 I fist thought (because of the similar variation in problem 4 , see StrangeSekis/Solution4) that White had no way to prevent the exchange. But this is not true, as soon as there is more than one mutual liberty between the two middle groups. As it is a little harder to justify, I will let the proof that the sequence below is, indeed, a win for White in all variations as an exercise for the diligent reader :-)