Hybrid Sequence:
The starting point is $\lib{prf-real}^{\mathbb F}$.
We can add a cache $\prftable[\cdot]$ so that each distinct output is computed only once.
We can replace each PRF instance $F(\key_i,\cdot)$ with a separate lazy random dictionary $\prftable_i[\cdot]$. The standard three-hop maneuvers are not shown.
\FORMATTINGHACK{\needspace{4\baselineskip}} We expect $X_2$ and $X_3$ to never repeat. So we can use the usual PRF Golden Rule proof strategy to make the assignments to $\prftable_2[X_2]$ and $\prftable_3[X_3]$ unconditional, triggering a bad event if these values were already defined. Later, we must show that the bad event's probability is negligible.
Instead of sampling $\prftable_2[X_2]$ uniformly and then computing $X_3$, we can sample $X_3$ uniformly and compute $\prftable_2[X_2]$. The same reasoning applies to $\prftable_3[X_3]$ and $X_4$. The standard three-hop maneuver involving \claimref{provsec.clm.generalized-otp} is not shown.
Now $X_3$ and $X_4$ are sampled uniformly and concatenated into $\prftable[X_0\|X_1]$. These steps can happen as early as possible.
\FORMATTINGHACK{\needspace{3.5\baselineskip}}Nothing after the assignment to $\prftable[\cdot]$ affects what the adversary sees; it exists only to determine whether to trigger a bad event. We can therefore move all bad-event logic to the end of time, without changing the bad event's overall probability.\FORMATTINGHACK{\pagebreak}
$\lib{prf-real}^{\mathbb F}$
$\key_1 \| \key_2 \| \key_3 \gets \bits^{3\secpar}$
$\prfquery_{\mathbb{F}}$($X_0\|X_1$):
$X_2 := X_0 \oplus {}$
$F(\key_1,X_1)$
$\prftable_1[X_1]$
$X_3 := X_1 \oplus {}$
$F(\key_2,X_2)$
$\prftable_2[X_2]$
$X_4 := X_2 \oplus {}$
$F(\key_3,X_3)$
$\prftable_3[X_3]$
return
$X_3 \| X_4$
$\prftable[X_0\|X_1]$