This is quite tricky to do, because it’s almost a sleight of hand and expressing it in English is like watching the trick from behind the magician - you can obv

This is quite tricky to do, because it’s almost a sleight of hand and expressing it in English is like watching the trick from behind the magician - you can obviously see the wrongness. But: if you skip a pair of horses (which is how the trick is done/the error introduced) it might be: Assume any group of 3 horses are the same colour. Can we show that adding another horse will result in a group of horses that are all the same colour? Well yes: take your new group of 4 horses, and extract 1. (h1). Three horses remain and they must by definition be the same colour. Put h1 back and extract a different horse (h2). Now three remain and must by definition be the same colour as each other. But when h2 was out, h1 was part of a group that was all the same colour. And when h1 was out, h2 was part of a group that was all the same colour. Therefore they must all be the same colour. It doesn’t matter how many different horses you find and bring to your initial group of 3, the rules will mean the resulting group of 4 is always the same colour. And so it is true for a group of 4 horses. By the same reasoning, it must then be true for a group of 5, and so on up to all horses. So all horses are the same colour, so long as the first assumption (that any group of 3 horses must be the same colour) holds. The trick/error is jumping from “a group of 1 horses must all be the same colour” to “a group of 3 horses must be the same colour” which is an easy jump to miss if you go from “1” to “n”, instead of “1” to “3”. (The logic doesn’t work for a pair of horses, because once h1 is taken out there are no other horses left that h2 must be the same colour as.)