Loading docs/signing.md +7 −7 Original line number Diff line number Diff line Loading @@ -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 Loading @@ -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 Loading @@ -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 Loading Loading
docs/signing.md +7 −7 Original line number Diff line number Diff line Loading @@ -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 Loading @@ -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 Loading @@ -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 Loading