The argument is proof by induction. First, we establish a base case for one horse (n=1). We then prove that if n horses have the same color, then n+1 horses must also have the same color.
Base case: One horse
The case with just one horse is trivial. If there is only one horse in the "group", then clearly all horses in that group have the same color.
Inductive step
Assume that n horses always are the same color. Consider a group consisting of n+1 horses.
First, exclude one horse and look only at the other n horses; all these are the same color, since n horses always are the same color. Likewise, exclude some other horse (not identical to the one first removed) and look only at the other n horses. By the same reasoning, these, too, must also be of the same color. Therefore, the first horse that was excluded is of the same color as the non-excluded horses, who in turn are of the same color as the other excluded horse. Hence, the first horse excluded, the non-excluded horses, and the last horse excluded are all of the same color, and we have proven that:
- If n horses have the same color, then n+1 horses will also have the same color.
We already saw in the base case that the rule ("all horses have the same color") was valid for n=1. The inductive step proved here implies that since the rule is valid for n=1, it must also be valid for n=2, which in turn implies that the rule is valid for n=3 and so on.
Thus, in any group of horses, all horses must be the same color
Good Luck