Please prove distinguishability given a non-perfectly secure cipher I'm trying to prove that a perfectly secure cipher yields indistinguishability. I already know and can prove that a perfect cipher => indistinguishability by the following proof: 

\begin{align}
\Pr(Enc(m_1) = c) = &amp; \Pr(C = c | P = m_1) \\
 &amp; = \frac{\Pr(C = c \land P = m_1)}{\Pr(P = m_1)} \\
 &amp; = \frac{\Pr(C = c) | \Pr(P = m_1|C = c)}{\Pr(P = m_1)}\\
 &amp; = \frac{\Pr(C = c) | \Pr(P = m_1)}{\Pr(P = m_1)}\\
 &amp; = \Pr(C = c)\\
\end{align}

Please help me prove indistinguishability will yield a perfect/secure cipher (indistinguishability => perfect cipher) which is the opposite direction.
The proof in the question is not about the perfect secrecy $\implies$ indistinguishability. It is about proving that; if perfect secrecy than we have $$ Pr[C=c| M=m] = Pr[C=c]$$ Also, we don't need the $\wedge$ step, just use Bayes' Theorem directly. The reverse (1) of this also true that is if we have $$ Pr[C=c| M=m] = Pr[C=c]$$ that we have perfect secrecy.

Now assuming that we have indistinguishability. That is; for every probability distribution over the message space $\mathcal{M}$ and every $m_0,m_1 \in \mathcal{M}$, and for every $c \in \mathcal{C}$;

$$Pr[C=c| M=m_0] = Pr[C=c | M=m_1] $$ we will show that we have prefect secrecy.

Since, the equation holds for ever $m_0,m_1$ we can say $Pr[C=c| M=m_0] = Pr[C=c | M=m_1] = p$ (2).

Now, write $Pr[C=c]$ as;

\begin{align}
Pr[C=c] &amp; = \sum_{m\in\mathcal{M}} Pr[C=c | M=m] \cdot Pr[M=m] \\
 &amp; = \sum_{m\in\mathcal{M}} p \cdot  Pr[M=m] \\
 &amp; = p \sum_{m\in\mathcal{M}} Pr[M=m] \quad\quad\quad\quad \\
 &amp; = p \\ 
 &amp; = Pr[C=c| M = m_0]  \quad\quad\quad\quad\quad\quad//(\text{use }2)
\end{align}
we have;

$Pr[C=c] = Pr[C=c| M = m_0]$ ,

now use (2);

$$Pr[M=m | C=c] = Pr[M = m_0]$$ therefore, indistinguishability $\implies$ perfect cipher