[FMFP] Restructure summary

semester4/fmfp/parts/01_induction-proofs/00_intro.tex

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+To prove recursive formulas, or more precisely formulated, a formula $P$ (with free variable $n$) for all $n \in \N$,
+we have can use weak or strong induction.
+Weak induction may be a \textit{slightly} misleading term, because it isn't necessarily weaker than strong induction.
+
+This section has been moved to the very start of the theory part, even though many of the topics mentioned have not been covered in the summary yet,
+such that all the induction proofs can be covered in the same place.

semester4/fmfp/parts/01_induction-proofs/01_mathematical-induction.tex

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+\subsection{Mathematical Induction}
+{\small NOTE: These types of induction were (primarily) mentioned in the Formal Methods part of the course, but made most sense to be put here}
+
+\subsubsection{Weak Mathematical Induction}
+To prove $\forall n \in \N. P$ (with $n$ free in $P$), we do the following:
+
+\shade{blue}{Base case} We show that $P[n \mapsto 0]$ is correct
+
+\shade{green}{Step case} For an arbitrary $m$ not free in $P$, we show that $P[n \mapsto m + 1]$ is correct under the assumption that $P[n \mapsto m]$ is correct.
+
+For \bi{well-founded} domains, we have to adjust the induction hypothesis slightly: We assume $\forall l \in \N. l < m \rightarrow P[n \mapsto l]$
+and then prove $P[n \mapsto m]$ under our assumption.
+
+The same, but expressed as a Natural Deduction rule:
+\[
+    \begin{prooftree}
+        \hypo{\Gamma \vdash P(0)}
+        \hypo{\Gamma, P(n) \vdash P(n + 1)}
+        \infer2{\Gamma \vdash \forall n. P(n)}
+    \end{prooftree}
+\]
+
+
+\subsubsection{Strong Mathematical Induction}

semester4/fmfp/parts/01_induction-proofs/02_structural-induction.tex

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+\subsection{Structural Induction}
+\subsubsection{Weak Structural Induction}
+Induction is based on the structure of terms
+\mint{haskell}+data T t = Leaf t | Node1 (T t) | Node2 t (T t) (T t)+
+
+\shade{blue}{Base Case} $T_0 = \{ \texttt{Leaf}\ a \divider a \in t \}$
+
+\shade{green}{Step Case} $T_i = T_{i - 1} \cup \{ \texttt{Node1}\ s \divider s \in T_{i - 1} \} \cup \{ \texttt{Node2}\ a\ l\ r \divider a \in t \text{ and } l, r \in T_{i - 1} \}$
+
+A natural deduction rule structural induction is:
+\[
+    \begin{prooftree}
+        \hypo{\Gamma \vdash P[x \mapsto \texttt{Leaf}\ a]}
+        \hypo{\Gamma, P[x \mapsto s] \vdash P[xs \mapsto \texttt{Node1}\ s]}
+        \hypo{\Gamma, P[x \mapsto l], P[x \mapsto r] \vdash P[ \mapsto \texttt{Node2}\ a\ l\ r]}
+        \infer3[$(*)$]{\Gamma \vdash \forall x \in \texttt{T}\ t. P}
+    \end{prooftree}
+\]
+(*) $a$, $l$, $r$, $s$ not free in $\Gamma, P$

semester4/fmfp/parts/01_induction-proofs/04_other-induction.tex

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+\subsection{Other types of induction}
+\subsubsection{Induction over Lists}
+To prove $P$ for all $xs$ in \texttt{[T]}, we do the following:
+
+\shade{blue}{Base case} We prove that $P[xs \mapsto []]$ is correct
+
+\shade{green}{Step case} We prove that $\forall y :: T, ys :: [T]. P[xs \mapsto ys] \rightarrow P[xs \mapsto y : ys]$, or in other words:
+We fix arbitrary $y :: T$ and $ys :: [T]$, which both are not free in $P$.
+We then apply our induction hypothesis $P[xs \mapsto ys]$ to prove $P[xs \mapsto y : ys]$
+
+