$ x-1\notin\mathbb{N} $ olduğunu varsayalım ve $x\in\mathbb{N}\setminus \{0\}$ olsun.
$\begin{array}{rcl} x\in\mathbb{N}\setminus \{0\} & \Rightarrow& (x\in\mathbb{N} \ \wedge \ x\notin \{0\}) \\ & \Rightarrow & (0\leq x \ \wedge \ x\neq 0) \\ & \Rightarrow & 0 < x \\ & \Rightarrow & 0+1 \leq x \\ & \Rightarrow & 1\leq x \\ & \Rightarrow & 1+(-1) \leq x+(-1) \\ & \Rightarrow & \begin{array}{c} \\ \left.\begin{array}{rr} 0\leq x-1 \\ x-1\notin\mathbb{N} \Rightarrow x-1< 0 \end{array}\right\} \Rightarrow \text{ Çelişki.} \end{array} \end{array}$
O halde $x-1\in\mathbb{N}$ olur.