# Does truncating a CBC ciphertext result in a truncated plaintext?

I want to know why in the CBC cipher, if we truncate the first block in the ciphertext then the corresponding plaintext block is truncated in the same way. I have an equation for the plaintext $$m_i$$ in terms of the ciphertexts $$c_i$$, $$c_{i+1}$$ so that the ciphertext corresponding to $$m_i$$ is $$c_{i+1}$$.

$$m_i = c_i \oplus D_k(c_{i+1})$$

Here I am skeptical about the random IV and its effect and that it may have.

• Are you truncating the last block as stated, or truncating the message from it's whole last block? If removing 1 byte from the last block removes 1 byte from the deciphered plaintext (and blocks are more than 1 byte), the simplest explanation is that this is not CBC.
– fgrieu
Commented May 2 at 18:32
• @fgrieu Ok i realised that this is actually a flawed question the way i wrote it. I need to actually see what happens when the first block is truncated instead of the last block. I have edited it now Commented May 2 at 19:07

This can be seen from the decryption equation $$p_i = c_{i-1} \oplus D_k(c_i)$$; for the first block of the truncated decryption, the decryptor won't have the correct value for $$c_{i-1}$$ (that's the block you truncated), and so the first $$p_i$$ will be gibberish. For the rest of the blocks, you do have valid $$p_{i-1}, c_i$$ pairs, and so the decryption works on those blocks.