### Update signing.md to use operatorname

parent 04690658
 ... ... @@ -49,9 +49,9 @@ compromised keys, and sends a pre-key message using a shared secret $S$, where: math S = ECDH\left(I_A,E_E\right)\;\parallel\; ECDH\left(E_A,I_B\right)\;\parallel\; ECDH\left(E_A,E_E\right) S = \operatorname{ECDH}\left(I_A,E_E\right)\;\parallel\; \operatorname{ECDH}\left(E_A,I_B\right)\;\parallel\; \operatorname{ECDH}\left(E_A,E_E\right)  Eve cannot decrypt the message because she does not have the private parts of ... ... @@ -67,9 +67,9 @@ On the other hand, signing the one-time keys leads to a reduction in deniability. Recall that the shared secret is calculated as follows: math S = ECDH\left(I_A,E_B\right)\;\parallel\; ECDH\left(E_A,I_B\right)\;\parallel\; ECDH\left(E_A,E_B\right) S = \operatorname{ECDH}\left(I_A,E_B\right)\;\parallel\; \operatorname{ECDH}\left(E_A,I_B\right)\;\parallel\; \operatorname{ECDH}\left(E_A,E_B\right)  If keys are unsigned, a forger can make up values of $E_A$ and ... ... @@ -82,7 +82,7 @@ a conversation between the two of them, rather than constructed by a forger. If $E_B$ is signed, it is no longer possible to construct arbitrary transcripts. Given a transcript and Alice and Bob's identity keys, we can now show that at least one of Alice or Bob was involved in the conversation, because the ability to calculate $ECDH\left(I_A,\,E_B\right)$ requires because the ability to calculate $\operatorname{ECDH}\left(I_A,E_B\right)$ requires knowledge of the private parts of either $I_A$ (proving Alice's involvement) or $E_B$ (proving Bob's involvement, via the signature). Note that it remains impossible to show that *both* Alice and Bob ... ...
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