[FMFP] Mostly finish exercises section

2026-07-27 21:29:09 +02:00 · 2026-07-14 14:44:07 +02:00
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+\subsection{Induction on Proof Trees}
+\label{sec:induction-on-proof-trees}
+This type of induction is used to prove that a statement exhibits given behaviour.
+Since often, we are not restricted to just simple statements, such where we know that we are applying the $\textsc{Ass}_{\text{NS}}$ rule,
+we need to perform case distinction on all possible last rules applied in the derivation tree $T$.
+If all are to be proven, this will yield $7$, one for each rule of the big-step semantics.
+
+\inlinedefinition[Subderivation] We define $T' \sqsubset T$, where $T$ is a derivation tree. $T'$ is called a \textit{subderivation} of $T$,
+or more simply, a \textit{subtree} of $T$. Definition is analogous to the subterm relation.
+
+
+\shade{ForestGreen}{Rundown} Below a rundown of how such proofs are expected to be laid out.
+
+We define $P(T) \equiv \forall \sigma, \sigma', s . (\texttt{root}(T) \equiv \text{statement})$.
+The quantifier-bound variables / states / statements can of course differ, so adjust accordingly.
+This step is the crucial step to setting up this type of induction.
+
+We then prove $\forall T . P(T)$ by strong induction on the shape of the derivation tree $T$.
+Thus, for some arbitrary $T$, the induction hypothesis is $\forall T' \sqsubset T . P(T')$ and we prove $P(T)$.
+
+Let $\sigma, \sigma', s$ (and more, if applicable) be arbitrary.
+
+Then, for rules, such as $\textsc{WhT}_{\text{NS}}$, we draw up a derivation tree, like so (from solutions of Exercise Session 11, FS2026):
+\begin{center}
+    \includegraphics[width=0.6\linewidth]{assets/proof-on-shape-of-tree.png}
+\end{center}
+After it, we need to put final constraints and bind free variables in some way, e.g. here:
+
+For some $b, s', \sigma'', T_4, T_5$, such that $s \equiv \texttt{while}\; b \; \texttt{do} \; s' \; \texttt{end}$ and $\cB \llbracket b \rrbracket \sigma = \texttt{tt}$.
+
+Then use the induction hypothesis on the subderivations $T_i$ and finish the proof by showing the original statement holds under the induction hypothesis.
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+\newpage
+\subsection{Induction on Derivation Sequence}
+A finite derivation sequence has a length $n \in \N$.
+We typically reason about finite derivation sequences using \bi{strong induction on the length of a derivation sequence},
+i.e. we reason about a multi-step execution $\gamma \rightarrow_1^k \gamma'$ using strong induction on the number of steps $k$.
+
+We define $P(k) \equiv$ ``for all executions of length $k$, our property holds''
+
+We then prove $P(k)$ for arbitrary $k$, with induction hypothesis $\forall k' < k. P(k')$, as is usually the case with induction on natural numbers.
+
+After the setup, the proof \textit{typically} proceeds by case distinction on:
+\begin{itemize}
+    \item $k = 0$ step execution
+    \item $k > 0$ step execution, by splitting off the first execution (i.e. an execution $\langle s, \sigma \rangle \rightarrow_1^k \sigma''$
+        can be split up as $\langle s, \sigma \rangle \rightarrow_1^1 \gamma \rightarrow_1^{k - 1} \sigma''$.)
+\end{itemize}
+In the $k > 0$ case, we can then apply the induction hypothesis. Sometimes a further case distinction can become necessary, depending on the rules that are possible.
+For instance, if $s = s_1; s_2$, two rules are possible, \textsc{Seq1} and \textsc{Seq2} from the SOS.