Proof: Let T represents the set of positive integer not in S. Assuming that T is non empty, we pick n to be the smallest integer in T. Then n > 1, by supposition (i)

The minimal nature of n allows us to conclude that none of the integers 1, 2, 3, …, (n -1) lies in T. Or if one prefers a positive assertion 1, 2, 3, …,(n -1) all belongs to T. Property (ii) then puts n = (n – 1) + 1 in S, which is an obvious contradiction. *The result of all this make T is empty.*