diff --git a/notes_hypernova.pdf b/notes_hypernova.pdf
index d52b0c3..f24a0f6 100644
Binary files a/notes_hypernova.pdf and b/notes_hypernova.pdf differ
diff --git a/notes_hypernova.tex b/notes_hypernova.tex
index 188df69..c9dc398 100644
--- a/notes_hypernova.tex
+++ b/notes_hypernova.tex
@@ -349,17 +349,15 @@ $$\theta_{k,j} = \sum_{y \in \{0, 1\}^{s'}} \widetilde{M}_j(r_x', y) \cdot \wide
 
 And in \emph{step 5}, $V$ checks
 
-% TODO check orange gamma^j...
 \begin{align*}
 	c &= \left(\sum_{j \in [t]} \gamma^j \cdot e_1 \cdot \sigma_{1,j}
-~\textcolor{orange}{+ \gamma^{t+j} \cdot e_1 \cdot \sigma_{2,j}}\right)\\
-	  &+ \gamma^{2t+1} \cdot e_2 \cdot \left( \sum_{i=1}^q c_i \cdot \prod_{j \in S_i} \theta_j \right)
-	  + \textcolor{cyan}{\gamma^{2t+2} \cdot e_2 \cdot \left( \sum_{i=1}^q c_i \cdot \prod_{j \in S_i} \theta_j \right)} 
+~\textcolor{orange}{+ \gamma^{t+j} \cdot e_2 \cdot \sigma_{2,j}}\right)\\
+	  &+ \gamma^{2t+1} \cdot e_3 \cdot \left( \sum_{i=1}^q c_i \cdot \prod_{j \in S_i} \theta_j \right)
+	  + \textcolor{cyan}{\gamma^{2t+2} \cdot e_4 \cdot \left( \sum_{i=1}^q c_i \cdot \prod_{j \in S_i} \theta_j \right)} 
 \end{align*}
 
 where
-% TODO check e_4
-$e_1 \leftarrow \widetilde{eq}(r_{1,x}, r_x'),~ e_2 \leftarrow \widetilde{eq}(r_{2,x}, r_x')$, $e_3 \leftarrow \widetilde{eq}(\beta, r_x'),~ e_4 \leftarrow \widetilde{eq}(\beta', r_x')$ (note: wip, pending check for $\beta, \beta'$ used in step 3).
+$e_1 \leftarrow \widetilde{eq}(r_{1,x}, r_x'),~ e_2 \leftarrow \widetilde{eq}(r_{2,x}, r_x')$, $e_3, e_4 \leftarrow \widetilde{eq}(\beta, r_x')$.
 
 \vspace{0.5cm}
 
diff --git a/weil-pairing.pdf b/weil-pairing.pdf
index 1968457..ec20e7a 100644
Binary files a/weil-pairing.pdf and b/weil-pairing.pdf differ
diff --git a/weil-pairing.tex b/weil-pairing.tex
index 6d660b6..cc637f6 100644
--- a/weil-pairing.tex
+++ b/weil-pairing.tex
@@ -151,14 +151,16 @@ Observe that
 \end{enumerate}
 
 Observe that
-$$\sum{P \in E(\Bbbk)} ord_P(r) \cdot P = 0$$
+$$\sum_{P \in E(\Bbbk)} ord_P(r) \cdot P = 0$$
 
-\begin{definition}{Support}
-  $$\sum_P n_P[P], ~\forall P \in E(\Bbbk) \mid n_P \neq 0$$
+\begin{definition}{Support of a divisor}
+  $$\sum_P n_P[P], ~\forall P \in E(\Bbbk) ~\text{s.t.}~ n_P \neq 0$$
 \end{definition}
 
 \begin{definition}{Principal divisor}
-  iff $deg(D)=0$ and $sum(D)=0$
+  iff
+  $$deg(D)=0$$
+  $$sum(D)=0$$
 \end{definition}
 $D \sim D'$ iff $D - D'$ is principal.
 
@@ -169,11 +171,24 @@ $D \sim D'$ iff $D - D'$ is principal.
 
 \section{Weil reciprocity}
 \begin{theorem}{(Weil reciprocity)}
-  Let $E/ \Bbbk$ be an e.c. over an alg. closed field. If $r,~s \in \Bbbk\setminus \{0\}$ are rational functions whose divisors have disjoint support, then
+  Let $E/ \Bbbk$ be an e.c. over an algebraically closed field. If $r,~s \in \Bbbk\setminus \{0\}$ are rational functions whose divisors have disjoint support, then
 $$r(div(s)) = s(div(r))$$
 \end{theorem}
 Proof. (todo)
 
+\paragraph{Example}
+\begin{align*}
+  p(x)=x^2 - 1,&~ q(x)=\frac{x}{x-2}\\
+  div(p)&= 1 \cdot [1] + 1 \cdot [-1] - 2 \cdot [\infty]\\
+  div(q)&= 1 \cdot [0] - 1 \cdot [2]\\
+	&\text{(they have disjoint support)}\\
+  p(div(q)) &= p(0)^1 \cdot p(2)^{-1}= (0^2 - 1)^1 \cdot (2^2 - 1)^{-1} = \frac{-1}{3}\\
+  q(div(p)) &= q(1)^1 \cdot q(-1)^1 - q(\infty)^2\\
+	    &= (\frac{1}{1-2})^1 \cdot (\frac{-1}{-1-2})^1 \cdot (\frac{\infty}{\infty - 2})^2 = \frac{-1}{3}
+\end{align*}
+
+so, $p(div(q))=q(div(p))$.
+
 \section{Generic Weil Pairing}
 Let $E(\Bbbk)$, with $\Bbbk$ of char $p$, $n$ s.t. $p \nmid n$.
 
@@ -236,6 +251,24 @@ with $S \neq \{O, P, -Q, P-Q \}$.
 
 
 \section{Properties}
+\begin{enumerate}[i.]
+  \item $e_n(P, Q)^n = 1 ~\forall P,Q \in E[n]$\\
+    ($\Rightarrow~ e_n(P,Q)$ is a $n^{th}$ root of unity)
+  \item Bilinearity
+    $$e_n(P_1+P_2, Q) = e_n(P_1, Q) \cdot e_n(P_2, Q)$$
+    $$e_n(P, Q_1+Q_2) = e_n(P, Q_1) \cdot e_n(P, Q_2)$$
+    \emph{proof:}
+    recall that $e_n(P,Q)=\frac{g(S+P)}{g(S)}$, then,
+    \begin{align*}
+      e_n(P_1, Q) &\cdot e_n(P_2, Q) = \frac{g(P_1 + S)}{g(S)} \cdot \frac{g(P_2 + P_1 + S)}{g(P_1 + S)}\\
+				    &\text{(replace $S$ by $S+P_1$)}\\
+				    &= \frac{g(P_2 + P_1 + S)}{g(S)} = e_n(P_1+P_2, Q)
+    \end{align*}
+  \item Alternating
+    $$e_n(P, P)=1 ~\forall P\in E[n]$$
+  \item Nondegenerate
+    $$\text{if}~ e_n(P,Q)=1 ~\forall Q\in E[n],~ \text{then}~ P=0$$
+\end{enumerate}
 
 
 \section{Exercises}