  == not mathematical induction? ==  I'm not a logician but it seems to me there is another problem with the line of reasoning that hasn't been mentioned. Even ignoring that the case for 2 horses fails, this is not mathematical induction anyway.   Compare with this: If we had a proof (P) that all horses are white we could prove that one horse is white. My horse is white. Therefore the proof P is valid. Therefore all horses are white.   This is obviously flawed because my horse being white can occur even if not all horses are white. Similarly, we could argue that in the OP's 4 horse proof, having all horses in a group of 1 horse be the same colour can occur for reasons other than all horses being the same colour (as, indeed, it can).   For this to be mathematical induction we would have an argument along the lines of:   I have a proof (P) that shows for any size group of horses all of one colour, a group of one more horse will also be all the same colour. A group of one horse is a group of all one colour, therefore all groups are the same colour.   This is a correct form of argument but of course, since we don't have a proof P, it isn't true.   Thus, as well as the explanation due to 'special cases', there is also an explanation due to a flawed form of argument that only resembles mathematical induction.