Trying to prove the above implication I thought about the following proof. However, I think the reasoning is flawed. What is wrong here?
Assume a non-empty set of natural numbers S. Let /S/ be the number of elements in S.
Base step: If /S/=1, S contains one element, which is obviously the least element.
Induction step: The well ordering principle holds for sets with /S/=n. Take a set with /S/=n and add any element x not equal to an element in S. The new set is S’ with /S’/=n+1. If x is smaller than the least element of the set S, then it is the least element of S’. If x is greater than the least element of S, then the least element of the set S is also the least element of S’. In either case S’ has a least element, that is the well ordering principle holds for set with /S/=n+1.
Hence the well ordering principle holds for all n.
