# proving indistinguishability of joint distribution using hybrid argument

Using Hybrid Argument, I want to prove following equivalence: $$\{X,Y,Z\}\equiv \{X',Y',Z'\}$$

by proving following equivalences:
$\{X,Y,Z\}\equiv \{X,Y,Z'\}$(1)
$\{X,Y,Z'\}\equiv \{X,Y',Z'\}$(2)
$\{X,Y',Z'\}\equiv \{X',Y',Z'\}$ (3)
I know how to prove $\{Z\}\equiv\{Z'\}$, under what condition $\{Z\}\equiv\{Z'\}$ implies (1) ?

(I've changed my post since my problem has nothing to do with SMC (maybe it has nothing to do with cryptography either) and the way I presented the problem was confusing. )

• In the third line what is $c(k.y)$? – pg1989 Jul 16 '16 at 1:11
• If X and Y are independent from Z you can make a simple reduction showing that distinguishing between {X,Y,Z} and {X,Y,Z′} implies distinguishing between Z and Z' (i.e. a proof by contradiction). If not, it is still possible that your claim holds, but you cannot use such a generic argument. – antosecret Jul 17 '16 at 5:40
• For example, here is a case where your claim does not hold: if you choose z' at random (independently of x and y) and z as x + y where x comes from X and y comes from Y, then the two distributions in (1) are distinguishable while Z and Z' aren't. – antosecret Jul 17 '16 at 5:42
• @antosecret Thanks, more specifically my question is, should Z be independent from {X} and {Y}, or be independent from {X,Y}. I guess (considering your example) the second one is correct – Mhy Jul 17 '16 at 13:13
• @Mhy You are right. – antosecret Jul 17 '16 at 19:36