diff --git a/metrics.tex b/metrics.tex
index 2c44df4..9c78212 100644
--- a/metrics.tex
+++ b/metrics.tex
@@ -63,13 +63,13 @@ \subsubsection{Weighted Jensen-Shannon Divergence}
 \subsection{NP-Hardness of the Max Quality Pipeline Instantiation Process}\label{sec:nphard}
 
  \begin{definition}[Max Quality Pipeline Instantiation Process]\label{def:MaXQualityInstance}
-Given \textit{dtloss}$_i$ the value of the quality metrics computed after applying the transformation of the policy matching the service selected to instantiate vertex  \vi{i}$\in$$\V_S$, the Max quality \problem is the case in which the \emph{pipeline instantiation} function returns a \pipelineInstance where the \textit{dtloss}$_i$ sum is maximized.
+Given \textit{dtloss}$_i$ the value of the quality metric computed after applying the transformation of the policy matching the service selected to instantiate vertex  \vi{i}$\in$$\V_S$, the Max quality \problem is the case in which the \emph{pipeline instantiation} function returns a \pipelineInstance where the \textit{dtloss}$_i$ sum is maximized.
 \end{definition}
  
-The Max Quality \problem is a combinatorial selection problem and is NP-hard, as stated by the following theorem.
+The Max Quality \problem is a combinatorial selection problem and is NP-hard, as stated by theorem \ref{theorem:NP}.
 However, while the overall problem is NP-hard, there is a component of the problem that is solvable in polynomial time: matching the profile of each service with the node policy. This can be done by iterating over each node and each service, checking if the service matches the node’s policy. This process would take $O(|N|*|S|)$ time. This is polynomial time complexity.
 
-\begin{theorem}
+\begin{theorem}\label{theorem:NP}
   The Max Quality  \problem is NP-Hard.
 \end{theorem}
 \emph{Proof: }