diff --git a/thesis/chapters/stabilizer.tex b/thesis/chapters/stabilizer.tex
index e6c4315..1213ccd 100644
--- a/thesis/chapters/stabilizer.tex
+++ b/thesis/chapters/stabilizer.tex
@@ -382,7 +382,7 @@ is done by using the symmetric set difference:
 \end{definition}
 
 Toggling an edge $\{i, j\}$ updates $E' = E \Delta \left\{\{i,j\}\right\}$.
-Another transformation on the VOP-free graph states is for a vertex $a \in V$:
+Another transformation on the VOP-free graph states for a vertex $a \in V$ is:
 
 \begin{equation}
     M_a := \sqrt{-iX_a} \prod\limits_{j\in n_a} \sqrt{iZ_j}
@@ -599,15 +599,15 @@ As stated in \ref{ref:g_states_vops} $C_L$ is also generated by $\sqrt{-iX}$ and
 This yields an algorithm to reduce the vertex operator of a non-isolated qbit $j$ to the identity.
 The combined operation of toggling the neighbourhood of $j$ and right-multiplying
 $M_j^\dagger$ is called $L_j$ transformation and it transforms $(V, E, O)$ into a so-called
-local Clifford equivalent graphical representation.
+local Clifford equivalent graphical representation. The algorithm is given by the following steps:
 
 \begin{enumerate}
     \item{Check whether $n_a \setminus \{b\} \neq \{\}$. If this condition does not hold
-            the vertex operator on $a$ cannot be cleared }
+            the vertex operator on $a$ cannot be cleared. }
     \item{Express the vertex operator $o_a$ as a product of $\sqrt{-iX}$ and $\sqrt{iZ}$.}
-    \item{Move from right to left through the product
+    \item{Move from right to left through the product:
         \begin{enumerate}
-            \item{If the current matrix is $\sqrt{-iX}$ apply $L_a$}
+            \item{If the current matrix is $\sqrt{-iX}$ apply $L_a$.}
             \item{If the current matrix is $\sqrt{iZ}$ select a vertex
                 $j \in n_a \setminus \{b\}$ (which is possible as it has been checked before)
                 and apply $L_j$.}
@@ -625,9 +625,9 @@ If $o_a$ could not be cleared $o_b$ can be cleared using the same procedure and
 clearing $o_b$ one can retry to clear $o_a$. 
 
 In any case at least one vertex operator has been cleared. If both vertex operators have been
-cleared Case 1 will be applied, if there is just one cleared vertex operator it is sure that it 
+cleared Case 1 will be applied. If there is just one cleared vertex operator it 
 is the vertex with non-operand neighbours. Using this one can still apply a $CZ$: Without loss of generality
-assume that $a$ has non-operand neighbours and $b$ does not. Now the state $\ket{G}$ has the form \cite{andersbriegel2005}
+assume that $a$ has non-operand neighbours and $b$ does not. Now the state $\ket{G}$ has the form \cite{andersbriegel2005}:
 
 \begin{equation}
     \begin{aligned}
@@ -637,7 +637,7 @@ assume that $a$ has non-operand neighbours and $b$ does not. Now the state $\ket
     \end{aligned}
 \end{equation}
 
-as $o_b$ commutes with all operators but $CZ_{a,b}$ and $s \in \{0, 1\}$ indicates whether there is an edge between $a$ and $b$.
+As $o_b$ commutes with all operators but $CZ_{a,b}$ and $s \in \{0, 1\}$ indicates whether there is an edge between $a$ and $b$.
 
 \begin{equation}
     \begin{aligned}
@@ -646,25 +646,26 @@ as $o_b$ commutes with all operators but $CZ_{a,b}$ and $s \in \{0, 1\}$ indicat
     \end{aligned}
 \end{equation}
 
-Using this one can re-use the method used in Case 2 to apply the $CZ$ while keeping the $o_a = I$.
+This allows to re-use the method in Case 2 to apply the $CZ$ while keeping the $o_a = I$.
 
-As both $CZ$ and $C_L$ can be applied to a graph state $\ket{G}$ this proofs constructively that
+As it has been shown how both $CZ$ and $C_L$  act on a graph state $\ket{G}$ and the resulting state 
+is a graph state as well this proofs constructively that
 the graphical representation of a stabilizer state is indeed able to represent any stabilizer state.
 If one wants to do computations using this formalism it is however also necessary to perform measurements.
 
 \subsubsection{Measurements on Graph States}
 
 Recalling \ref{ref:meas_stab} it is clear that one has to compute the commutator of
-the observable $g_a = Z_a$ with the stabilizers to get the probability amplitudes.
-This is a quite expensive computation in theory however it is possible to simplify
+the observable $g_a = Z_a$ with the stabilizers to get the probability amplitudes
+which is a quite expensive computation in theory. It is possible to simplify
 the problem by pulling the observable behind the vertex operators. For this consider
 the projector $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$:
 
 \begin{equation}
     \begin{aligned}
         P_{a,s} \ket{\psi} &= P_{a,s} \left(\prod\limits_{o_i \in O} o_i \right) \ket{\bar{G}} \\
-                        &= \left(\prod\limits_{o_i \in O \setminus o_a}o_i \right)P_a  o_a \ket{\bar{G}} \\
-                        &= \left(\prod\limits_{o_i \in O \setminus o_a}o_i \right) o_a o_a^\dagger P_a o_a \ket{\bar{G}} \\
+                        &= \left(\prod\limits_{o_i \in O \setminus \{o_a\}}o_i \right)P_a  o_a \ket{\bar{G}} \\
+                        &= \left(\prod\limits_{o_i \in O \setminus \{o_a\}}o_i \right) o_a o_a^\dagger P_a o_a \ket{\bar{G}} \\
                         &= \left(\prod\limits_{o_i \in O} o_i \right) o_a^\dagger P_a o_a \ket{\bar{G}} \\
                         &= \left(\prod\limits_{o_i \in O} o_i \right) \tilde{P}_{a,s} \ket{\bar{G}} \\
     \end{aligned}
@@ -682,10 +683,10 @@ as $o_a$ is a Clifford operator:
     \end{aligned}
 \end{equation}
 
-Where $\tilde{g}_a \in \{+1, -1\}\cdot\{X_a, Y_a, Z_a\}$. It is therefore enough to study the measurements
+Where $\tilde{g}_a \in \{+1, -1\}\cdot\{X_a, Y_a, Z_a\}$. Therefore, it is enough to study the measurements
 of any Pauli operator on the vertex operator free graph states. The commutators of the observable
-with the $K_G^{(i)}$ are quite easy to compute, note that Pauli matrices wither commute or anticommute 
-so it is easier to list the operators that anticommute:
+with the $K_G^{(i)}$ are quite easy to compute. Note that Pauli matrices either commute or anticommute 
+and it is easier to list the operators that anticommute:
 
 \begin{equation}
     \begin{aligned}
@@ -697,8 +698,8 @@ so it is easier to list the operators that anticommute:
 
 This gives one immediate result: if a qbit $a$ is isolated and the operator $\tilde{g}_a = X_a (-X_a)$ 
 is measured the result $s=0(1)$ is obtained with probability $1$ and $(V, E, O)$ is unchanged.
-In any other case the results $s=1$ and $s=0$ both have probability $\frac{1}{2}$ and both 
-graph and vertex operators have to be updated. Further it is clear that measurement of $-\tilde{g}_a$
+In any other case the results $s=1$ and $s=0$ have probability $\frac{1}{2}$ and both 
+graph and vertex operators have to be updated. Further it is clear that measurements of $-\tilde{g}_a$
 and $\tilde{g}_a$ are related by just inverting the result $s$.
 
 The calculations to obtain the transformation on graph and vertex operators are lengthy and follow
@@ -712,10 +713,10 @@ the steps required to obtain the following results:
     \end{aligned}
 \end{equation}
 
-These transformations split it two parts: the first is a result of Lemma \ref{lemma:stab_measurement}
-and the second part makes sure that the qbit $a$ is diagonal in the correct state of the measured state.
+These transformations split it two parts: the first is a result of Lemma \ref{lemma:stab_measurement}.
+The second part makes sure that the qbit $a$ is diagonal in measured observable and has the correct eigenvalue.
 When comparing with Lemma \ref{lemma:stab_measurement} in both cases $Y,Z$ the anticommuting stabilizer 
-$K_G^{(a)}$ is chosen. The graph is changed according to 
+$K_G^{(a)}$ is chosen. The graph is changed according to:
 
 \begin{equation}
     \begin{aligned}
@@ -725,7 +726,7 @@ $K_G^{(a)}$ is chosen. The graph is changed according to
 \end{equation}
 
 
-For $g_a = X_a$ one has to choose a $b \in n_a$ and the transformations are
+For $g_a = X_a$ one has to choose a $b \in n_a$ and the transformations are:
 
 \begin{equation}
     \begin{aligned}