Update 2023-01-24-Kate.md

Author: RisenCrypto

_posts/2023-01-24-Kate.md

@@ -193,10 +193,10 @@ $e((a-b)\cdot C_Q,G_2) \stackrel {?}{=} e(C_F - G_1\cdot c, G_2)$
 
 Using the bilinear property,
 
-$e(C_Q,(a-b)\cdot G_2) \stackrel {?}{=} e(C_F - G_1\cdot c, G_2)$
+$e(C_Q,(a-b)\cdot G_2) \stackrel {?}{=} e(C_F - c \cdot G_1, G_2)$
 
 
-$e(C_Q,a\cdot G_2 - b\cdot G_2) \stackrel {?}{=} e(C_F - G_1\cdot c, G_2)$
+$e(C_Q,a\cdot G_2 - b\cdot G_2) \stackrel {?}{=} e(C_F - c \cdot G_1, G_2)$
 
 Everything except $a\cdot G_2$ is either calculated or found in the SRS for $G_1$. And $a\cdot G_2$ will be part of the SRS for $G_2$ & that is also known and thus the equality can be verified.