| Step | Statement | Explanation |
| 1 | Let S be a subset of the positive integers. | Starting point of the proof. |
| 2 | Suppose S has no smallest element. | Assumption for contradiction. |
| 3 | 1 does not belong to S. | Since 1 is the smallest positive integer, if S had no smallest element, it cannot contain 1. |
| 4 | Let T be the complement of S in the positive integers. | Defining the set of all positive integers not in S. |
| 5 | T must contain 1. | Since 1 cannot be in S, it must be in T. |
| 6 | Suppose every positive integer less than (n) belongs to T. | Inductive assumption for the proof. |
| 7 | Then (n+1) must belong to S. | Based on the assumption that S has no smallest element and the inductive step. |
| 8 | But (n+1) would be the smallest element in S. | Contradiction arises because S is assumed to have no smallest element. |
| 9 | Thus, (n+1) must belong to T. | To avoid contradiction, (n+1) cannot be in S and must be in T. |
| 10 | T contains all positive natural numbers. | By induction, all positive integers are in T, not S. |
| 11 | S is empty. | Since all positive integers are in T, S has no elements. |