diff --git a/hardness-app.tex b/hardness-app.tex
index e217a65..8f5c038 100644
--- a/hardness-app.tex
+++ b/hardness-app.tex
@@ -702,19 +702,19 @@ If \circuit.\type $= \circplus$, then $\degree(\circuit) = \max\left(\degree(\ci
 If \circuit.\type $= \circmult$, then, 
 substituing values, the following should hold,  
 \begin{align}
-&2\left(\degree(\circuit_\linput) + \degree(\circuit_\rinput)\right) \cdot \left(\max(\depth(\circuit_\linput), \depth(\circuit_\rinput)) + 1\right) + 1 \label{eq:times-lhs}\\
+&2\left(\degree(\circuit_\linput) + \degree(\circuit_\rinput)\right) \cdot \left(\max(\depth(\circuit_\linput), \depth(\circuit_\rinput)) + 1\right) + 1 \nonumber\\%\label{eq:times-lhs}\\
 &\qquad\geq 2\degree(\circuit_\linput) \cdot \depth(\circuit_\linput) + 2 \degree(\circuit_\rinput) \cdot \depth(\circuit_\rinput) + 3\label{eq:times-middle} \\
 &\qquad\geq 1 + \cost(\circuit_\linput) + \cost(\circuit_\rinput) = \cost(\circuit) (\ref{eq:cost-sampmon})\label{eq:times-rhs}.
 \end{align}
 
-To prove the inequality $\Cref{eq:times-lhs} \geq \Cref{eq:times-middle}$, first, let us expand \Cref{eq:times-lhs},  
-\begin{align}
-(\ref{eq:times-lhs}) &= 2\degree(\circuit_\linput)\depth_{\max} + 2\degree(\circuit_\rinput)\depth_{\max} + 2\degree(\circuit_\linput) +\nonumber \\
-&  2\degree(\circuit_\rinput) + 1\label{eq:times-lhs-expanded}
-\end{align}
+To prove (\ref{eq:times-middle}), first, the LHS expands to, %\Cref{eq:times-lhs},  
+\begin{equation}
+%(\ref{eq:times-lhs}) 
+2\degree(\circuit_\linput)\depth_{\max} + 2\degree(\circuit_\rinput)\depth_{\max} + 2\degree(\circuit_\linput) +  2\degree(\circuit_\rinput) + 1\label{eq:times-lhs-expanded}
+\end{equation}
 where $\depth_{\max}$ is used to denote the maximum depth of the two input subcircuits.
 
-Let us simplify the inequality $(\ref{eq:times-lhs}) \geq (\ref{eq:times-middle})$. 
+Let us now simplify the inequality (\ref{eq:times-middle}). 
 \begin{align}
 &2\degree(\circuit_\linput)\depth_{\max} + 2\degree(\circuit_\rinput)\depth_{\max} + 2\degree(\circuit_\linput) + 2\degree(\circuit_\rinput) + 1 \nonumber\\
 &\qquad \geq 2\degree(\circuit_\linput) \cdot \depth(\circuit_\linput) + 2 \degree(\circuit_\rinput) \cdot \depth(\circuit_\rinput) + 3\nonumber\\
@@ -722,31 +722,33 @@ Let us simplify the inequality $(\ref{eq:times-lhs}) \geq (\ref{eq:times-middle}
 \end{align}
 Note that by the \emph{reduced} invariant of \reduce, a circuit \circuit with $\depth(\circuit) \geq 1$ will always have at least one input with $\degree(\circuit_i) \geq 1$.  Thus, \Cref{eq:times-lhs-middle-step1} follows, and the inequality is upheld.
 
-Now to justify the inequality (\ref{eq:times-middle}) $\geq$ (\ref{eq:times-rhs}) which holds for the following reasons.  First, \Cref{eq:times-rhs} is the result of \Cref{eq:cost-sampmon} when $\circuit.\type = \circmult$.  \Cref{eq:times-middle} is then produced by substituting the upperbound of (\ref{eq:ih-bound-cost}) for each $\cost(\circuit_i)$, trivially establishing an upper bound to (\ref{eq:times-rhs}).  This proves \Cref{eq:strict-upper-bound} for the $\circmult$ case.
+Now to justify (\ref{eq:times-rhs}) which holds for the following reasons.  First, the RHS%\Cref{eq:times-rhs} 
+is the result of \Cref{eq:cost-sampmon} when $\circuit.\type = \circmult$.  The LHS %\Cref{eq:times-middle}
+is then produced by substituting the upperbound of (\ref{eq:ih-bound-cost}) for each $\cost(\circuit_i)$, trivially establishing the upper bound of (\ref{eq:times-rhs}).  This proves \Cref{eq:strict-upper-bound} for the $\circmult$ case.
 
 For the case when \circuit.\type $= \circplus$, substituting values yields
 \begin{align}
-&2\max(\degree(\circuit_\linput), \degree(\circuit_\rinput)) \cdot \left(\max(\depth(\circuit_\linput), \depth(\circuit_\rinput)) + 1\right) +1\label{eq:plus-lhs-inequality}\\
+&2\max(\degree(\circuit_\linput), \degree(\circuit_\rinput)) \cdot \left(\max(\depth(\circuit_\linput), \depth(\circuit_\rinput)) + 1\right) +1\nonumber\\%\label{eq:plus-lhs-inequality}\\
 &\qquad \geq \max\left(2\degree(\circuit_\linput) \cdot \depth(\circuit_\linput) + 1, 2\degree(\circuit_\rinput) \cdot \depth(\circuit_\rinput) +1\right) + 1\label{eq:plus-middle}\\
 &\qquad \geq 1 + \max(\cost(\circuit_\linput), \cost(\circuit_\rinput)) = \cost(\circuit)\label{eq:plus-rhs}
 \end{align}
 
-To prove that $(\ref{eq:plus-lhs-inequality}) \geq (\ref{eq:plus-middle})$, we can rewrite (\ref{eq:plus-lhs-inequality}) as
+To prove  (\ref{eq:plus-middle}), we can rewrite the LHS as %(\ref{eq:plus-lhs-inequality}) as
 \begin{equation}
 2\degree_{\max}\depth_{\max} + 2\degree_{\max} + 1.\label{eq:plus-lhs-expanded}
 \end{equation}
-Since $\degree_{\max} \cdot \depth_{\max} \geq \degree(\circuit_i)\cdot \depth(\circuit_i),$ the following upper bound holds for \Cref{eq:plus-middle}:
+Since $\degree_{\max} \cdot \depth_{\max} \geq \degree(\circuit_i)\cdot \depth(\circuit_i),$ the following upper bound holds for the RHS of (\ref{eq:plus-middle}):
 \begin{equation}
 2\degree_{\max}\depth_{\max} + 2 \geq  \max\left(2\degree(\circuit_\linput) \cdot \depth(\circuit_\linput) + 1, 2\degree(\circuit_\rinput) \cdot \depth(\circuit_\rinput) +1\right) + 1.\label{eq:plus-middle-expanded}
 \end{equation}
-Substituting the upperbound of (\ref{eq:plus-middle-expanded}) in for (\ref{eq:plus-middle}) we obtain the following for the inequality (\ref{eq:plus-lhs-inequality}) $\geq$ (\ref{eq:plus-middle}):
+Substituting the upperbound (LHS) of (\ref{eq:plus-middle-expanded}) in for the RHS of (\ref{eq:plus-middle}) we obtain the following for (\ref{eq:plus-middle}):
 \begin{align}
 &2\degree_{\max}\depth_{\max} + 2\degree_{\max} + 1 \geq 2\degree_{\max}\depth_{\max} + 2\nonumber\\
 &\implies 2\degree_{\max} + 1 \geq 2\label{eq:plus-upper-bound-final}.
 \end{align}
-As in the $\circmult$ case the \emph{reduced} invariant of \reduce implies that $\degree_{\max} \geq 1$, and (\ref{eq:plus-upper-bound-final}) follows.  This proves (\ref{eq:plus-lhs-inequality}) $\geq$ (\ref{eq:plus-middle}).
+As in the $\circmult$ case the \emph{reduced} invariant of \reduce implies that $\degree_{\max} \geq 1$, and (\ref{eq:plus-upper-bound-final}) follows.  This proves (\ref{eq:plus-middle}).
 
-Similar to the case of $\circuit.\type = \circmult$, the inequality $(\ref{eq:plus-middle}) \geq (\ref{eq:plus-rhs})$ follows by equations $(\ref{eq:cost-sampmon})$ and $(\ref{eq:ih-bound-cost})$.
+Similar to the case of $\circuit.\type = \circmult$, (\ref{eq:plus-rhs}) follows by equations $(\ref{eq:cost-sampmon})$ and $(\ref{eq:ih-bound-cost})$.
 
 This proves (\ref{eq:strict-upper-bound}) for the $\circplus$ case and thus the claimed $O(k\log{k}\cdot\depth(\circuit))$ runtime for $k = \degree(\circuit)$ follows.