The first step of the proof is to verify that ''<math>f''(γ\gamma) ≥ γ\ge\gamma</math> for all ordinals γ<math>\gamma</math> and that ''<math>f''</math> commutes with suprema. Given these results, inductively define an increasing sequence <α<submath>''\langle\alpha_n\rangle_{n''<\omega}</submath>> (''n'' < ω) by setting α<sub>0</submath>\alpha_0 = α\alpha</math>, and α<submath>''\alpha_{n''+1</sub>} = ''f''(α<sub>''n''\alpha_n)</submath>) for ''<math>n'' ∈ ω\in\omega</math>. Let β<math>\beta = sup \sup_{α<sub>''n''<\omega} \alpha_n</submath> : ''n'' ∈ ω}, so β ≥ α<math>\beta\ge\alpha</math>. Moreover, because ''<math>f''</math> commutes with suprema,
The last equality follows from the fact that the sequence <α<submath>''n''\langle\alpha_n\rangle_n</submath>> increases. <math> \square </math>
As an aside, it can be demonstrated that the β<math>\beta</math> found in this way is the smallest fixed point greater than or equal to α<math>\alpha</math>.
