# Proving two definitions of perfect security are equivalent

I'm trying to prove that the following two definitions are equivalent:

$$\forall m\in M$$ and $$c\in C$$ $$\Pr[C=c \mid M=m]=\Pr[C=c]$$

$$\forall m_1,m_2 \in M$$, $$E_k(m_1)=E_k(m_2)$$, where $$E_k(m_i)$$ stands for the distribution over $$k$$ of the encrypted message $$m_i$$.

First - just to make sure, I am indeed supposed to show two directions, right? (i.e. first $$\Rightarrow$$ second and second $$\Rightarrow$$ first). This is my understanding regarding showing an equivalence for any two definitions.

If so, I'm able to prove the direction first $$\Rightarrow$$ second but I'm unable to do the second direction. How do I use the fact that for every pair of messages I have some conclusion about any single general message $$m$$?

Thanks.

• I don't really understand the second definition. Do you mean something like $Pr[Enc_k(m_1) = c] = Pr[Enc_k(m_2) = c]$ Commented Apr 7, 2022 at 7:14
• The second statement would somehow show that the encryption is not correct? Since you can not decrypt, the ciphertext is the same for all messages. On the bright side, very secure! Commented Apr 7, 2022 at 13:31
• @Titanlord yes, exactly.
– Anon
Commented Apr 7, 2022 at 14:27

Yes, you are right, you need to prove both directions. In Katz & Lindell's textbook (2nd edition) you can find the proof for first $$\Rightarrow$$ second. The other direction is left for exercise. I try my best to give a correct solution for that.

First we have to know that the following holds:

$$Pr[Enc_k(m) = c] = Pr[C = c | M = m]$$

First $$\Rightarrow$$ Second proves, that assuming $$Pr[Enc_k(m_1) = c] = Pr[Enc_k(m_2) = c]$$ is correct, then $$Pr[C = c | M = m] = Pr[M = m]$$ holds.

We want to show that assuming $$Pr[C = c | M = m] = Pr[M = m]$$ is correct, then $$Pr[Enc_k(m_1) = c] = Pr[Enc_k(m_2) = c]$$ holds.

My solution is:

$$Pr[Enc_k(m_1) = c] = Pr[C = c | M = m_1] = \frac{Pr[M = m_1 | C = c] \cdot Pr[C = c]}{Pr[M=m_1]}$$

Because we assumed that $$Pr [ M = m_1 | C = c] = Pr[M=m_1]$$ we get:

$$\Rightarrow Pr[C = c] = Pr[C = c | M = m_2] = Pr[Enc_k(m_2) = c]$$