diff --git a/whitepaper/Dissertation.pdf b/whitepaper/Dissertation.pdf
index 6b9ac39..15d5a4f 100644
Binary files a/whitepaper/Dissertation.pdf and b/whitepaper/Dissertation.pdf differ
diff --git a/whitepaper/Dissertation.tex b/whitepaper/Dissertation.tex
index 3f93dde..4c139a4 100644
--- a/whitepaper/Dissertation.tex
+++ b/whitepaper/Dissertation.tex
@@ -721,27 +721,23 @@ Consider point (1). One option is to prove that the sum of the committed values
     \begin{tikzpicture}
         \tikzstyle{style}=[circle,minimum size=15mm,draw=black,fill=white]
 
-        \node (im1) {\begin{tikzpicture}
-                \node[style] (A) at (-1.5, 3) {$n_1$};
-                \node[style] (B) at (-3, 0) {$n_2$};
-                \node[style] (C) at (0, 0) {$n_3$};
-                \node[style] (D) at (1.5, 3) {$n_4$};
+        \node[style] (A) at (-1.5 - 4, 3) {$n_1$};
+        \node[style] (B) at (-3   - 4, 0) {$n_2$};
+        \node[style] (C) at (0    - 4, 0) {$n_3$};
+        \node[style] (D) at (1.5  - 4, 3) {$n_4$};
 
-                \path[draw] (A) -- (C) -- (B) -- (A);
-                \path[draw] (A) -- (D);
-        \end{tikzpicture}};
+        \path[draw] (A) -- (C) -- (B) -- (A);
+        \path[draw] (A) -- (D);
 
-        \node (im2) at (0.5\textwidth, 0) {\begin{tikzpicture}
-                \node[style,label=center:$n_1 + 0$] (A) at (-1.5, 3) {};
-                \node[style,label=center:$n_2 + 0$] (B) at (-3, 0) {};
-                \node[style,label=center:$n_3 + 1$] (C) at (0, 0) {};
-                \node[style,label=center:$n_4 + 0$] (D) at (1.5, 3) {};
+        \node[style,label=center:$n_1 + 0$] (A2) at (-1.5 + 4, 3) {};
+        \node[style,label=center:$n_2 + 0$] (B2) at (-3   + 4, 0) {};
+        \node[style,label=center:$n_3 + 1$] (C2) at (0    + 4, 0) {};
+        \node[style,label=center:$n_4 + 0$] (D2) at (1.5  + 4, 3) {};
 
-                \path[draw] (A) -- (C) -- (B) -- (A);
-                \path[draw] (A) -- (D);
-        \end{tikzpicture}};
+        \path[draw] (A2) -- (C2) -- (B2) -- (A2);
+        \path[draw] (A2) -- (D2);
 
-        \path[draw,->,very thick] (im1) -- (im2);
+		\path[draw,->,very thick] (-1.5, 1.5) -- (0, 1.5);
     \end{tikzpicture}
     \caption{Example state change from reinforce action.}
 \end{figure}
@@ -896,27 +892,23 @@ Point (5) still remains, as the range proof alone only works to prevent negative
 \begin{tikzpicture}
     \tikzstyle{style}=[circle,minimum size=15mm,draw=black,fill=white]
 
-    \node (im1) {\begin{tikzpicture}
-            \node[style] (A) at (-1.5, 3) {$n_1$};
-            \node[style] (B) at (-3, 0) {$n_2$};
-            \node[style] (C) at (0, 0) {$n_3$};
-            \node[style] (D) at (1.5, 3) {$n_4$};
+    \node[style] (A) at (-1.5 - 4, 3) {$n_1$};
+    \node[style] (B) at (-3   - 4, 0) {$n_2$};
+    \node[style] (C) at (0    - 4, 0) {$n_3$};
+    \node[style] (D) at (1.5  - 4, 3) {$n_4$};
 
-            \path[draw] (A) -- (C) -- (B) -- (A);
-            \path[draw] (A) -- (D);
-        \end{tikzpicture}};
+    \path[draw] (A) -- (C) -- (B) -- (A);
+    \path[draw] (A) -- (D);
 
-    \node (im2) at (0.5\textwidth, 0) {\begin{tikzpicture}
-            \node[style,label=center:$n_1 + k$] (A) at (-1.5, 3) {};
-            \node[style,label=center:$n_2 + 0$] (B) at (-3, 0) {};
-            \node[style,label=center:$n_3 - k$] (C) at (0, 0) {};
-            \node[style,label=center:$n_4 + 0$] (D) at (1.5, 3) {};
+    \node[style,label=center:$n_1 + k$] (A2) at (-1.5 + 4, 3) {};
+    \node[style,label=center:$n_2 + 0$] (B2) at (-3   + 4, 0) {};
+    \node[style,label=center:$n_3 - k$] (C2) at (0    + 4, 0) {};
+    \node[style,label=center:$n_4 + 0$] (D2) at (1.5  + 4, 3) {};
 
-            \path[draw] (A) -- (C) -- (B) -- (A);
-            \path[draw] (A) -- (D);
-    \end{tikzpicture}};
+    \path[draw] (A2) -- (C2) -- (B2) -- (A2);
+    \path[draw] (A2) -- (D2);
 
-    \path[draw,->,very thick] (im1) -- (im2);
+    \path[draw,->,very thick] (-1.5, 1.5) -- (0, 1.5);
 \end{tikzpicture}
 \caption{Example state change from fortify action.}
 \end{figure}