# What happens if the plaintext $m1$ from the first block of CBC is known to adversary?

The adversary is has unlimited access to the encryption oracle. My thinking is that upon receiving the ciphertext C = $$$$ it will just use $$c_0$$ to XOR it with $$F_k(m_1)$$ to get the first ciphertext, which does nothing. The subsequent blocks remain undecrypted, right?

Argument: any algorithm that could extract some additional knowledge could be turned into one breaking CPA security. For CBC, CPA security can be proven from security of the block cipher, under the assumptions that the IV is random, the number of blocks enciphered remains low enough (like, less than $$2^{b/2-n}$$ where $$b$$ is the block size in bits and $$n$$ is a security parameter), and the decryption side does not leak information about the deciphered plaintext (such as, incorrect padding).