Add description for 2.3

report/report.tex

@@ -103,10 +103,34 @@ change from 1 to 0.
 
 \paragraph{Paper proofs}
 
+To prove \texttt{nsteps\_stop}, induction over the amount of executed steps $n$
+is used. The base case starts with $n = 0$. This case is trivial since
+\texttt{nsteps 0 te cs == cs}. The induction hypothesis is \texttt{nsteps (n-1)
+te cs == cs} and it is assumed that $n > 0$. The induction then goes as follows:
+\begin{verbatim}
+nsteps n te cs = nsteps (n-1) te (step_simp te cs)
+               = nsteps (n-1) te cs
+               = cs
+\end{verbatim}
+It is allowed to reduce \texttt{step\_simp te cs} to \texttt{cs}, because
+\texttt{cs} is already final.
 
+Now it is possible to prove the lemma \texttt{progress}: \texttt{cs ==
+step\_simp te cs} $\rightarrow$ \texttt{isFinal cs}. Since, \texttt{step\_simp}
+has the same value as the \texttt{step} function, when the outcome is
+\texttt{Next}, the call stack is different from \texttt{cs}. This means that the
+left side of the lemma does not hold and therefore the right side does not have
+to hold to satisfy the implication. If the outcome is \texttt{Stop}, \texttt{cs}
+can only equal \texttt{Ter ts []} and therefore the left side is satisfied. When
+the left side is satisfied, \texttt{cs} is a final state and therefore it
+follows that \texttt{isFinal cs} is also true.
 
 \paragraph{Explanation on used equalities}
-%TODO
+
+The lemma uses propositional equality, because \texttt{callstack} has the
+keyword \texttt{noeq}, which means that the type does not satisfy decidable
+equality. The usual equality only works with types that satisfy decidable
+equality and therefore propositional equality is needed.
 
 \section{Uniqueness of call stacks}
 
@@ -114,7 +138,6 @@ change from 1 to 0.
 %TODO
 
 \paragraph{Explanation on lemma \texttt{order\_ineq}}
-%TODO
 
 \section{Exception Propagation}