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Aren't $IV_1$ and $IV_2$ public in TLS 1.2 as well? $IV_1$ certainly is (as that's just the ciphertext block in front of the block we're attacking); however the IV that the TLS 1.2 sender will use for the next message ($IV_2$) is not. In fact, the sender might not know it yet, as it might not have not picked it yet. But doesn't this mean that BEAST ...



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