I'm having some difficulty understanding some material in a book I'm reading, Introduction to Set Theory 3rd ed. by Hrbacek and Jech.

The question is about a proof of a theorem given on page 118 in the book.

The theorem that I'm having difficult with is Theorem 3.6 (given on page 113):

Let G be an operation.For any setathere is a unique infinite sequence ⟨a_{n}|n∈N⟩ such that

(a)a_{0}=a

(b)a_{n+1}=G(a_{n},n)for alln∈N

The proof provided in the book on page 118 is as follows:

Let G be an operation. We want to find, for every seta, a sequence⟨a_{n}|n∈N⟩ such thata_{0 }=aanda_{n+1}=G(a_{n},n) for alln∈N.

By the parametric version of the Transfinite Recursion Theorem 4.11, there is an operationFsuch thatF(0) =aandF(n+1)=G(F(n),n) for alln∈N.

Now we apply the Axiom of Replacement: There exists a sequence ⟨a_{n}|n∈N⟩that is equal toF↾ωand the Theorem follows.

Note:

The parametric version of the Transfinite Recursion Theorem 4.11 is as follows:

Given binary operationsG_{1},G_{2}, andG_{3 }there is an operationFsuch that for all z

F(z,0) =G_{1}(z,0),

F(z,α+1) =G_{2}(z,F_{z}(α)) for all ordinalsα, and(

Fz,α) =G_{3}(z,F_{z}↾α)for all limit ordinalsα.

Now, I understand everything in the proof of Theorem 3.6 except how the parametric version of Theorem 4.11 is used to derive the operationFin the proof. Can can someone please help me fill in blanks?