# Prove that $F(x)=f(h(x))$ is a secure PRF if $f$ is a secure PRF and $h$ is a CRHF

From my lecture notes, it says that if we have some PRF $$f = \{f_k: \{0,1\}^{n} \rightarrow \{0,1\}^{n}\}$$ and CRHF $$h = \{h_t: \{0,1\}^{2n} \rightarrow \{0,1\}^{n}\}$$, then $$F = \{F_{k,t} = f_k(h_t()): \{0,1\}^{2n} \rightarrow \{0,1\}^{n}\}$$ is also a secure PRF. However, the proof is omitted. How would you go about proving this?

My thought is that we can use a hybrid argument, to show that $$H_1 = f_k(h_t(\cdot))$$ is indistinguishable from $$H_2 = U_1(h_t(\cdot))$$ (where $$U_1$$ is a rand fct) is indistinguishable from $$H_3 = U_2(\cdot)$$ (where $$U_2$$ is also a rand fct). You can easily claim that $$H_1$$ is indist from $$H_2$$ by definition of a PRF because you're simply replacing the PRF with a random function. However I'm not sure how to show that $$H_2$$ is indist from $$H_3$$. Any ideas on how to prove this? Is this even the right approach to this proof? Thanks in advance.