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-rw-r--r--docs/signing.md12
1 files changed, 7 insertions, 5 deletions
diff --git a/docs/signing.md b/docs/signing.md
index fcc5342..58a2b5e 100644
--- a/docs/signing.md
+++ b/docs/signing.md
@@ -49,13 +49,14 @@ 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 = ECDH\left(I_A,E_E\right)\;\parallel\;
+ ECDH\left(E_A,I_B\right)\;\parallel\;
+ ECDH\left(E_A,E_E\right)
```
Eve cannot decrypt the message because she does not have the private parts of
either $`E_A`$ nor $`I_B`$, so cannot calculate
-$`ECDH\left(E_A,\,I_B\right)`$. However, suppose she later compromises
+$`ECDH\left(E_A,I_B\right)`$. However, suppose she later compromises
Bob's identity key $`I_B`$. This would give her the ability to decrypt any
pre-key messages sent to Bob using the compromised one-time keys, and is thus a
problematic loss of forward secrecy. If Bob signs his keys with his Ed25519
@@ -66,8 +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 = ECDH\left(I_A,E_B\right)\;\parallel\;
+ ECDH\left(E_A,I_B\right)\;\parallel\;
+ ECDH\left(E_A,E_B\right)
```
If keys are unsigned, a forger can make up values of $`E_A`$ and