diff --git a/tex-stuff/math.tex b/tex-stuff/math.tex
index 54a0fc8..04cc1dc 100644
--- a/tex-stuff/math.tex
+++ b/tex-stuff/math.tex
@@ -127,7 +127,7 @@ is $5nk*32 = 160nk$ bytes large.
 
 $\forall i,j:$ Compute and publish \\[2.0ex]
 $\gamma_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(\sum_{h=1}^n\sum_{d=j+1}^k\alpha_{hd}\right)+\left(\sum_{d=1}^{j-1}\alpha_{id}\right)+\left(\sum_{h=1}^{i-1}\alpha_{hj}\right)\right)$ and \\[2.0ex]
-$\delta_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(\sum_{h=1}^n\sum_{d=j+1}^k\beta_{hd}\right)+\left(\sum_{d=1}^{j-1}\beta_{id}\right)+\left(\sum_{h=1}^{i-1}\beta_{hj}\right)\right)$ \\[2.0ex]
+$\delta_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(\sum_{h=1}^n\sum_{d=j+1}^k \beta_{hd}\right)+\left(\sum_{d=1}^{j-1} \beta_{id}\right)+\left(\sum_{h=1}^{i-1} \beta_{hj}\right)\right)$ \\[2.0ex]
 with a corresponding Proof 2 for $ECDL(\gamma_{ij}^{\times a}) = ECDL(\delta_{ij}^{\times a})$.
 
 \subsubsection{Round 3: Decrypt outcome}
@@ -164,8 +164,8 @@ The message has $nk$ parts, each consisting of $5$ Points. Therefore the message
 is $5nk*32 = 160nk$ bytes large.
 
 $\forall i,j:$ Compute and publish \\[2.0ex]
-$\gamma_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(2M+2\right)\left(\sum_{h=1}^n\left(\sum_{d=j}^k\alpha_{hd}+\sum_{d=j+1}^k\alpha_{hd}\right)+\sum_{d=1}^{j}\alpha_{id}\right) - \left(2M+1\right)Y \right)$ and \\[2.0ex]
-$\delta_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(2M+2\right)\left(\sum_{h=1}^n\left(\sum_{d=j}^k\beta_{hd}+\sum_{d=j+1}^k\beta_{hd}\right)+\sum_{d=1}^{j}\beta_{id}\right)\right)$ \\[2.0ex]
+$\gamma_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\sum_{h=1}^n\left(\sum_{d=j}^k\alpha_{hd}+\sum_{d=j+1}^k\alpha_{hd}\right)+\left(2M+2\right)\sum_{d=1}^{j}\alpha_{id} - \left(2M+1\right)Y \right)$ and \\[2.0ex]
+$\delta_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\sum_{h=1}^n\left(\sum_{d=j}^k \beta_{hd}+\sum_{d=j+1}^k \beta_{hd}\right)+\left(2M+2\right)\sum_{d=1}^{j} \beta_{id}\right)$ \\[2.0ex]
 with a corresponding Proof 2 for $ECDL(\gamma_{ij}^{\times a}) = ECDL(\delta_{ij}^{\times a})$.
 
 \subsubsection{Round 3: Decrypt outcome}