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- $\begingroup$ Adding comments to reflect my understanding, so I do not make the post a moving target. Now I see that the algorithm is basically using two Schnorr signatures in paralell, and for me the math adds up, and also it seems that simplifying the proof environment is okay. But I am just starting to understand it, so I need confirmation. $\endgroup$Árpád Magosányi– Árpád Magosányi2025-05-20 20:46:48 +00:00Commented May 20 at 20:46
- $\begingroup$ ... but if it is Schnorr signature, then s should be mod q, not mod p! $\endgroup$Árpád Magosányi– Árpád Magosányi2025-05-20 20:49:01 +00:00Commented May 20 at 20:49
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