# Can we use a pseudorandom function in both models of a simulation based proof?

Assume in the real model a party blinds a fixed element $b$ as: $v_i=r_i\cdot b$, where $r_i$ is a output of pseudorandom function. So we give $v_i$ to a semi-honest adversary.

Now we want to sketch the proof and we want to show that we can generate the adversary's view in the ideal model that is computationally indistinguishable from its view in the real model.

So we pick a value $b'$ and blind it as $v'_i=r'_i\cdot b'$, where $r'_i$ is a output of the same pseudorandom function but using a random (distinct) key. We give $v'_i$ to the adversary.

We want to say that $v_i$ and $v'_i$ are computationally indistinguishable.

Question: Is the above simulation correct? In general, can we use pseudorandom value in the ideal model (rather than truly random value)?

• Given the ZK and MPC tags, it seems likely you haven't given enough context here... (i.e. with additional context, the following 'answer' would change, most likely...) But -- all else equal.. -- you can swap $PRF(K_0,x)$ -> $\mathcal{U}_{|\mathsf{range}(PRF)|}$ -> $PRF(K_1, x)$ using the security of the PRF twice (one after the other) – Daniel Apon Sep 3 '16 at 17:07
• @DanielApon So the answer is YES? – user153465 Sep 3 '16 at 18:05
• The answer is: "You have not given enough context to decide your question." – Daniel Apon Sep 4 '16 at 6:09
• ..though, if you're trying to use a second PRF key in place of uniform, because uniform does not work, then the answer is a resounding NO. – Daniel Apon Sep 4 '16 at 6:10
• @DanielApon thank you for the answer. But uniform does work, I mean we can use the truly random and it works. My question is what if we replace the truly random value in the ideal world with pseudorandom values. – user153465 Sep 4 '16 at 9:52