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.

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