Error in 14:00. Can be fixed like this:
f(f(r)*f(s)) = -1 IMPLIES f(f(r)*f(s) + 1) = 0.... (E1)
This is so because we have proven it at 10:00.
Earlier, we established that the root of f is 1. So E1 implies f(r)*f(s) + 1 = 1 and so f(r)*f(s) = 0. The rest is the same as in the video.
I would like to thank Khaled Karaman for pointing out the error and Mike Lyons for providing an easy patch.