# Perfect Secrecy, two Definitions

I'm reading the proof of the implication "Def 2.1 $\Rightarrow$ Def 2.4" in these slides about Adversarial Indistinguishability and Perfectly-Secret Encryption. I have a doubt in the slide 10. Here it says:

Construct an adversary $A$ who outputs $m_0$, $m_1$ in the ﬁrst step of $\mathsf{PRIV_{EAV}}(\Pi,A,n)$. Then if $A$ receives any ciphertext $c' \neq \tilde{c}$ it outputs a random bit $b_0$; if it receives the ciphertext $\tilde{c}$, it outputs $b_0 = 0$.

How is $A$ able to "receive any ciphertext $c'$". Perhaps the adversary not receives only the ciphertext $Enc(m_b)$?

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Why I have vote -1? –  juaninf May 20 '13 at 13:52

The adversary $\mathcal A$ is an entity (think of a computer program) designed to participate in the experiment $\operatorname{PrivK}^{\text{eav}}_{A,\Pi}$.

So the adversary produces two messages, then is given the encryption of one of them, and has to guess which one it was.

Of course, you can give the adversary other "ciphertexts" too, but this wouldn't be the same experiment, and the results of this thus don't matter for the experiment (they might be interesting in other experiments, though).

The proof does only have to consider the behavior of the adversary in the experiment, not in other cases.

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understand, I found here cs.stevens.edu/~nicolosi/classes/13sp-cs579/scribing-11sp/… (in the last page) other proof for the same implication I cann't understand Why $$Pr[\mathsf{PRIV_{EAV}}(\Pi,A,n)=1] = \dfrac{1}{2}$Pr[A(c)=B|B=0] + Pr[A(c)=B|B=1]$ = \dfrac{1}{2}$Pr[A(c)=0|B=0] + \dfrac{1}{2}Pr[A(c)=1|B=1]$$$ –  juaninf May 20 '13 at 13:52