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2

Yes, as @DrLecter said in the comments, that equation holds from the bilinear property. Here is a step-by-step proof. Let $e : \mathbb G_1 \times \mathbb G_2 \to \mathbb G_T$ be a bilinear pairing. The bilinear property states that: \begin{align}e(g_1 ^ a, g_2 ^b) = e(g_1,g_2)^{ab}\end{align} Since you don't seem to distinguish between $\mathbb G_1$ and ...

0

The only way i can see to evaluate $e(C_x,D)$ is to recover $D$ from a previous known pairing $e(C_i,D)$. However due to the Fixed Argument Pairing Inversion 1 assumption this is computationally hard.

1

If you define $u$ as element of $G_1$, then you can't just use it in $G_2$ (in the first equation), like you did in your equation. Besides, you missed to state, if $u$ and $g$ are generators of those groups. The following holds: $$e(u^a,g^{\frac{b}{a}}) = e(u,g)^b = e(u^b,g)$$ Anyway, you didn't do that in the actual question, where I don't really see the ...

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