Proof by Will Dyer

Given: 0 < 1 < 2 < …

Proof: Let X be a subset of N that has no least value, and XC be the complement of X in N. Now suppose all natural numbers less than or equal to k are in XC. Therefore, if k+1 is in X, k+1 is the least value of X. This cannot be, since X has no least value, so k+1 is in XC. By strong induction, XC = N and X is the empty set.

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