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I've been reading the proof in these slides, the last page, and the author is using the lemma:

$Pr[A(c)=1|M=m_0]=Pr[A(c)=1|M=m_1]$

I understand on the intuitive level that it's a consequence of the assumed perfect secrecy, but can someone provide a proof to the aforementioned?

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  • $\begingroup$ The ciphertext $c$ is independent of the message rv $M$, so the conditionals are the same for either message. $\endgroup$ – pg1989 Oct 11 '16 at 19:19

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