# All horses are the same color?

Recall that mathematical induction has two requirements: a base case and an inductive step. Show that a statement is true for x=1 and show that if it is true for x=n, then it is true for x=n+1. The induction follows as a falling series of dominoes.

Evolutionists try to do something similar. They show that evolution is true in some cases (microevolution) and they show that if it is true up to a certain level of adaptation or complexity, then they can show a plausible scenario in which it is true of the next higher level.

But consider a ‘proof’ that all horses are the same color as proposed by Joel E. Cohen: The base case is n=1 which is trivially true: a single horse has the same color as itself. Then if we assume that all groups of n horses have the same color, we can show that all groups of n+1 horses have the same color. First remove the last horse so only n horses are left; they must have the same color. Then remove the first horse so only n horses are left; they must have the same color too. Thus the (n+1)th horse has the same color as the others.

The flaw in this ‘proof’ is that the base case does not match the inductive step. If the base case were n=2, it would be valid but n=1 does not have a valid comparative.

The problem with evolutionary induction is similar: the base case does not match the inductive step. Variation within a kind is not variation between kinds. If it could be shown that variation between kinds exists, then they might have a case but they have only shown that variation within a kind exists. By focusing on ‘species’ they have used variation within and between ‘species’ to promote their case but that begs the question of whether species are proper kinds or not.

December 2014