The equivalence transformation says that any sequence of non-zero complex numbers satisfy the general continued fraction in the following manner. Here is the link: https://en.wikipedia.org/wiki/Generalized_continued_fraction#The_equivalence_transformation
Wikipedia says that this can be proved via induction but gives no sources or readings. I can't find any other mention of this equivalence transformation on continued fractions anywhere else on the internet (maybe I haven't looked well enough).
Can someone provide me with a proof? Thanks.
Note: I don't have a clear idea if this proof needs to involve this operation: $[x_1,x_2,...,x_m]$ or not. I am familiar with all the operations mentioned in the wikipedia article but I haven't managed to read too much about that specific operation. Can the proof avoid this (if possible)?
update: https://poset.jp/posts/continued-fractions-attempt-1/continued-fractions-part-2/ Found this link on it, but I'll have to understand [] notation.
$x_n:=b_0+\frac{a_1}{b_1+\frac{a_2}{b_2+\frac{a_3}{b_3+...}}}$ all the way to the $n$th denominator, or to the point where it will be $\frac{a_n}{b_n}$.
${x_n}':=b_0+\frac{c_1a_1}{c_1b_1+\frac{c_1c_2a_2}{c_2b_2+...}}$ to the $n$th version of that, or to the point where it is $\frac{c_{n-1}c_na_n}{c_nb_n}$
Assuming $x_n={x_n}'$ holds true for some values of $n$,
I know there are also some other relationships that I understand like
$x_n=\frac{A_n}{B_n}=\frac{b_nA_{n-1}+a_nA_{n-2}}{b_nB_{n-1}+a_nB_{n-2}}$ which can be proved with induction.
Additionally, I know that $A_{n-1}B_n-A_nB_{n-1}=(-1)^na_1a_2a_3...a_n$ which can also be proved with induction.
I'm struggling with the inductive step of showing that $x_{n+1}={x_{n+1}}'$
Ok, it was simpler than I thought because the recursive relationship of $A_n$ and $B_n$ was defined.
$c_n{b_n}' = c_n(b_n+\frac{a_{n+1}}{b_{n+1}})$ and I already have proven that recurssive relationship before.