# All horses are the same color

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The horse paradox is a falsidical paradox that arises from flawed demonstrations, which purport to use mathematical induction, of the statement All horses are the same color. There is no actual contradiction, as these arguments have a crucial flaw that makes them incorrect. This example was originally raised by George Pólya.[1] The paradox was also used by Joel E. Cohen as an example of the subtle errors that can occur in attempts to prove statements by induction.[2]

## The argument

The argument is proof by induction. First we establish a base case for one horse (${\displaystyle n=1}$). We then prove that if ${\displaystyle n}$ horses have the same color, then ${\displaystyle n+1}$ 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 ${\displaystyle n}$ horses always are the same color. Let us consider a group consisting of ${\displaystyle n+1}$ horses.

First, exclude the last horse and look only at the first ${\displaystyle n}$ horses; all these are the same color since ${\displaystyle n}$ horses always are the same color. Likewise, exclude the first horse and look only at the last ${\displaystyle n}$ horses. These too, must also be of the same color. Therefore, the first horse in the group is of the same color as the horses in the middle, who in turn are of the same color as the last horse. Hence the first horse, middle horses, and last horse are all of the same color, and we have proven that:

• If ${\displaystyle n}$ horses have the same color, then ${\displaystyle n+1}$ 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 ${\displaystyle n=1}$. The inductive step showed that since the rule is valid for ${\displaystyle n=1}$, it must also be valid for ${\displaystyle n=2}$, which in turn implies that the rule is valid for ${\displaystyle n=3}$ and so on.

Thus in any group of horses, all horses must be the same color.[3]

## Explanation

The argument above makes the implicit assumption that the two subsets of horses to which the induction assumption is applied have a common element. This is not true when the original set (prior to either removal) only contains two horses.

Let the two horses be horse A and horse B. When horse A is removed, it is true that the remaining horses in the set are the same color (only horse B remains). The same is true when horse B is removed. However the statement "the first horse in the group is of the same color as the horses in the middle" is meaningless, because there are no "horses in the middle" (common elements (horses) in the two sets). Therefore the above proof has a logical link broken. The proof forms a falsidical paradox; it seems to show by valid reasoning something that is manifestly false, but in fact the reasoning is flawed.