# Indistinguishable encryptions in the presence of an eavesdropper equivalence

I'm trying to prove that definition 5 and definition 6 in this document are equivalent. This is what I've done at the moment: Asume that the scheme has Indistinguishable encryptions in the presence of an eavesdropper (definition 5), then:

$$\mbox{Prob}[\mbox{Priv}_{\mathcal{A},\Pi}^{\mbox{eav}}(n)=1]=\mbox{Prob}[\texttt{output}(\mbox{Priv}_{\mathcal{A},\Pi}^{\mbox{eav}}(n,0))=0]+\mbox{Prob}[\texttt{output}(\mbox{Priv}_{\mathcal{A},\Pi}^{\mbox{eav}}(n,1))=1]=(1/2-\mbox{Prob}[\texttt{output}(\mbox{Priv}_{\mathcal{A},\Pi}^{\mbox{eav}}(n,0))=1])+\mbox{Prob}[\texttt{output}(\mbox{Priv}_{\mathcal{A},\Pi}^{\mbox{eav}}(n,1))=1]\leq 1/2+negl(n)$$

And we get that:

$$\mbox{Prob}[\texttt{output}(\mbox{Priv}_{\mathcal{A},\Pi}^{\mbox{eav}}(n,1))=1]-\mbox{Prob}[\texttt{output}(\mbox{Priv}_{\mathcal{A},\Pi}^{\mbox{eav}}(n,0))=1]\leq negl(n)$$

Now I'm trying to prove that:

$$\mbox{Prob}[\texttt{output}(\mbox{Priv}_{\mathcal{A},\Pi}^{\mbox{eav}}(n,0))=1]-\mbox{Prob}[\texttt{output}(\mbox{Priv}_{\mathcal{A},\Pi}^{\mbox{eav}}(n,1))=1]\leq negl(n)$$

For finishing the inequality with the absolute value. Is there someone who can help me with this? Thanks!

PS: I can understand the last line in the definition 6 in the document I've given: where the probability is taken over...

Can someone explain me that?

The point is that, if we assume that $\mbox{Prob}[\mbox{Priv}_{\mathcal{A},\Pi}^{\mbox{eav}}(n)=1]\leq 1/2+negl(n)$, then $\mbox{Prob}[\mbox{Priv}_{\mathcal{A},\Pi}^{\mbox{eav}}(n)=0]\leq 1/2+negl(n)$ too (if this were not to happen, then we could create an adversary that solves the experiment with a better probability than $1/2+negl(n)$).