Indistinguishable encryptions in the presence of an eavesdro

Indistinguishable encryptions in the presence of an eavesdropper :

Prove the equivalence of Definition 3.8 and Definition 3.9.

DEFINITION 3.8 A prveile-key ericryptsow scheme = (Gen, Enc. Dec) we TI = :Gen, Eric, Dec) hag indistinguishable encryptions in the presence of an eavesdropper if for a! probabilistic polynomial-wne adverscrics there coasts a wouigblc fuzction negl such thal Pr Privk;i:I(r) = 1]- + negl(11). 1,11 ureere: ,Fvc pruTotihl2ès icthere eeer ke TWT40teril- CAPgi\'is el-Setl og A, (s wel/ (s ,Fv: 3andon? yytit d in the Pr.iessment chol)ng the key, the minori eat b nd aug Tandom coaus uscd 25s the encryption process).

Solution

The point is that, if we assume that Prob[PriveavA,(n)=1]1/2+negl(n)Prob[PrivA,eav(n)=1]1/2+negl(n), then Prob[PriveavA,(n)=0]1/2+negl(n)Prob[PrivA,eav(n)=0]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)1/2+negl(n)).

This allows us to use the same argument I\'ve used for proving the first inequality for getting the inequality with absolute value.

Prob[PriveavA,(n)=1]=Prob[output(PriveavA,(n,0))=0]+Prob[output(PriveavA,(n,1))=1]=(1/2Prob[output(PriveavA,(n,0))=1])+Prob[output(PriveavA,(n,1))=1]1/2+negl(n)Prob[PrivA,eav(n)=1]=Prob[output(PrivA,eav(n,0))=0]+Prob[output(PrivA,eav(n,1))=1]=(1/2Prob[output(PrivA,eav(n,0))=1])+Prob[output(PrivA,eav(n,1))=1]1/2+negl(n)

And we get that:

Prob[output(PriveavA,(n,1))=1]Prob[output(PriveavA,(n,0))=1]negl(n)Prob[output(PrivA,eav(n,1))=1]Prob[output(PrivA,eav(n,0))=1]negl(n)

Now I\'m trying to prove that:

Prob[output(PriveavA,(n,0))=1]Prob[output(PriveavA,(n,1))=1]negl(n)