Problem 0.11

This one is tricky. The simple observation is that this crap statement break down even when there are two horses, so let's expand the statement in that case.

Assuming any set of one horse has the same color.

Given a set H of two horses, name them by their color, let's say, black and white. The argument about H1 has only one color still follows. The argument that H2 has only one color still follow. But the last 'therefore' statement is flawed. It is possible for a set of horse to have two subset H1 and H2 of horses of the same color but H doesn't.

This shows the induction breaks at the least breaking point, even if S(1), S(2) => S(3) => S(4) are all sound. As long as S(1) does not imply S(2), S(3), S(4) are just all as crap as all horse has the same color!