0
$\begingroup$

In game-based security proof for key-exchange protocols, there is a Test query. The Test(U) query typically is only available to the adversary if the attacked instance U is fresh. (U represents either a participant or an oracle)

Fresh: Before the session expires, there is no SSReveal(U), SKReveal(U) or Corrupt(U) query that has been asked by the adversaries. Both U and its matching session are not locally exposed. Such session is called fresh.

SSReveal(U): this query allows the adversary to learn the session-specific state information held by U.

SKReveal(U): this query allows the adversary to learn the session key held by U.

Corrupt(U): this query allows the adversay to learn the long-term private key of U.

My questions is: if Test(U) can only be issued to fresh session, does that mean "the impersonation of the protocol participants (e.g, U) does not happen" is the premise of asking this query? If so, do we need to proof the mutual authenticity before we ask the Test(U) query?

Thank you.

$\endgroup$

2 Answers 2

0
$\begingroup$

It is important to remember the purpose of «fresh», namely to identify the sessions that have an expectation of confidentiality. The non-fresh sessions are exactly the sessions for which the adversary knows the session keys (because it has used its queries to reveal too much information). A test query for a non-fresh session will then trivially reveal the challenge bit.

If the adversary has engineered a situation where mutual authentication has failed, that would not necessarily mean that the session is non-fresh.

(Security models deal with this either by forbidding such queries, or by ignoring the adversary’s guess after a disallowed test query.)

$\endgroup$
0
$\begingroup$

if Test(U) can only be issued to fresh session, does that mean "the impersonation of the protocol participants (e.g, U) does not happen" is the premise of asking this query?

I don't quite understand the wording of your question, but $\mathsf{Test}(U)$ does not make sense if $U$ is a user. In authenticated key-exchange (AKE) security, you call $\mathsf{Test}$ against a session not a user, i.e., a specific instance of the key exchange protocol between two users. Neither user involved with that session which is being tested can be corrupted, because that would contradict the freshness of the session. If the adversary does not know the long-term keys, then they should not be able to impersonate the users (if the scheme is secure).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.