# How to prove a pseudo-random function has certain expansion factor

How should I answer the following problem:

Let $F = F0 \times F1$ be a pseudorandom generator with expansion factor $2n$ such that for all $x \in \{0, 1\}^n$ $$F(x) = (F_0 (x) || F_1 (x))$$ and $$|x| = |F_0 (x)| = |F_1(x)|.$$ Prove that $$F'(x) = (F_0 (F_0 (x)) || F_0 (F_1 (x)) || F_1 (F_0 (x)) || F_1 (F_1 (x)))$$ is a pseudorandom generator with expansion factor $4n.$

Here I am not able to understand how expansion factor changes?

I am more interested in general proof structure for solving similar proofs.

I am beginner so more detailed explanation will be helpful.

The concatenation operator $||$ in e definition $F$ ensures an expansion $n \rightarrow 2n$ since both $F_0$ and $F_1$ map $n$ bits to $n$ bits.
The argument for $F'$ is similar since repeated application of two $n$ bit to $n$ bit functions gives an $n$ bit to $n$ bit function, and four of those are concatenated in $F'$.