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I have two statements which I think are correct:

  1. $\beta, s$ are unknown. Only $g^\beta$ is known to the prover but not any of $g^{\beta s^i}$ for $i \ge 1$. If the prover is able to find group elements $A$ and $B$ such that $e(A, g^\beta) = e(B, g)$ then the prover knows a field element $a$ such that $A = g^a$

  2. And a more general statement: $\alpha, s$ are unknown to the prover. $g^\alpha, g^{\alpha s}, g^{\alpha s^2}, \dots ,g^{\alpha s^q}$ are known to the prover. If the prover is able to provide group elements $A$ and $B$ such that $e(A, g^\alpha) = e(B, g)$ then the prover knows coefficients $a_i$ such that $A = g^{\sum^q_{i=0}{a_i}\alpha s^i}$

Both of these statements intuitively seem correct to me but what are the commonly used cryptographic assumptions that I should reduce them to and how should I do that?

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  • $\begingroup$ It does not invalidate it as in your example the prover knows a = 1 and that's what the statement is about. So you have to know some a $\endgroup$ Commented Feb 28 at 6:26
  • $\begingroup$ Sorry, I misread. $\endgroup$
    – Maeher
    Commented Feb 28 at 6:28
  • $\begingroup$ Actually, for a type 1 pairing, finding $A, B$ is easy: $A = g, B = g^\beta$ $\endgroup$
    – poncho
    Commented Mar 29 at 18:07

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I think that both statements are untrue.

If the prover is furnished by a third party (who does not need to know $\alpha$) with $R=g^r$ and $S=g^{\alpha r}$) but not $\alpha$, then they can pick any $x$ and set $A=R^x$, $B=S^x$ without knowing $a=rx$.

In particular a malicious prover could obtain such values from a pair provided by a legitimate prover.

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  • $\begingroup$ however this is exactly what is stated. Prover knows x. So the statement is that this is the only way to obtain the pair $\endgroup$ Commented Feb 28 at 8:16
  • $\begingroup$ @NikolayZakirov: No it is not. The prover does know $x$ but this does not furnish them with the knowledge of $a$ such that $g^a=A$. $\endgroup$
    – Daniel S
    Commented Feb 28 at 8:29
  • $\begingroup$ You are right, but I think the problem is that you allow a 3d party into the universe which is not really allowed. I guess I should rewrite my statements in terms of "there exists a polynomial time extractor that given same inputs as prover has which are $g$ and $g^{\beta}$ produces $a$ such that $A=g^a$. I think this is called knowledge of exponent assumption? $\endgroup$ Commented Feb 28 at 9:43

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