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If an adversary $M$ observes a session of the protocol between $A$ and $B$ and learns the session key $k_{AB}$.How Can the adversary mount a successful impersonation attack and communicate with $B$ on behalf of $A$?And this session are still used. Can adversary $M$ compute the key pair of $A$?

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  • $\begingroup$ That is a type problem。Correct adversary A to M $\endgroup$ – belle Jan 4 '18 at 21:00
  • $\begingroup$ Is the session that uses key $k_{AB}$ still active? $\endgroup$ – poncho Jan 4 '18 at 21:33
  • $\begingroup$ If you know the session key, and presuming that session key is the only key, then $M$ can do anything that $A$ can do, cryptographically speaking; decryption of messages, forging of messages, change of messages you name it. Re-authentication of the entity or renewed establishment of the secured channel is generally not possible with just a session key. What exactly is the question? $\endgroup$ – Maarten Bodewes Jan 4 '18 at 21:33
  • $\begingroup$ yes the Kab still active $\endgroup$ – belle Jan 4 '18 at 23:03
  • $\begingroup$ Thanks for your help .If the adversary M can do these because they can compute the private key of A $\endgroup$ – belle Jan 4 '18 at 23:08
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To summarize from the comments:

How Can the adversary mount a successful impersonation attack and communicate with B on behalf of A?

Answer:

If you know the session key, and presuming that session key is the only key, then M can do anything that A can do, cryptographically speaking; decryption of messages, forging of messages, change of messages you name it. Re-authentication of the entity or renewed establishment of the secured channel is generally not possible with just a session key. ~ Maarten Bodewes

Second Question:

Can adversary M compute the key pair of A?

No. If you are given $(g,g^a,g^b,g^{ab})$ you cannot infer $a$ nor $b$. If you could, then you would have shown hardness-equivalence of the Computational Diffie-Hellman Problem (CDH) and the Discrete Logarithm Problem (DLOG). Because we don't know whether these two are actually equally hard, we also don't know a way to do what you asked above.

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