[FMFP] Restructure summary

2026-07-27 21:29:09 +02:00 · 2026-07-06 09:49:27 +02:00
parent 67500a14cb
commit 294363f3c8
66 changed files with 127 additions and 101 deletions
@@ -1,25 +0,0 @@
-\subsubsection{Induction}
-To prove recursive formulas, or more precisely formulated, a formula $P$ (with free variable $n$) for all $n \in \N$,
-we can't really do a proof by cases, as there are infinitely many cases (one for each input).
-Thus: We can use induction to prove recursive formulas or functions.
-
-
-\paragraph{The schema}
-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.
-
-
-\paragraph{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 $P[xs \mapsto y : ys]$
@@ -0,0 +1,6 @@
+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.
@@ -0,0 +1,24 @@
+\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}
@@ -1,4 +1,5 @@
-\subsubsection{Structural Induction}
+\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)+

@@ -0,0 +1,11 @@
+\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 $P[xs \mapsto y : ys]$
+
+
@@ -41,4 +41,4 @@ We then show that the function is correct for both cases (i.e. LHS and RHS of OR
 \end{enumerate}
 In this proof we used the \textbf{TND} and \textbf{$\lor$-E} (here also called \bi{Case Split}) rules.

-So what we have to show, given $Q \lor R$ for any proposition $P$ with case split is that \bi{(1)} $P$ follows from $Q$ and \bi{(2)} $P$ follows from $R$
+What we have to show, given $Q \lor R$ for any proposition $P$ with case split, is that \bi{(1)} $P$ follows from $Q$ and \bi{(2)} $P$ follows from $R$
@@ -1,3 +1,4 @@
+\newpage
 \subsubsection{The semantics}
 \paragraph{Numerals}
 The semantic function $\cN : \texttt{Numeral} \rightarrow \texttt{Val}$ maps a numeral $n$ to an integer value $\cN\llbracket n \rrbracket$,
@@ -27,7 +28,7 @@ $\cA : \texttt{Aexp} \rightarrow \texttt{State} \rightarrow \texttt{Val}$ maps a
 given by:
 \begin{align*}
    \cA\llbracket x \rrbracket \sigma                         & = \sigma(x)                                                                                            \\
-    \cA\llbracket n \rrbracket \sigma                         & = \cN\llbracket x \rrbracket                                                                           \\
+    \cA\llbracket n \rrbracket \sigma                         & = \cN\llbracket n \rrbracket                                                                           \\
    \cA\llbracket e_1 \; \texttt{op} \; e_2 \rrbracket \sigma & = \cA\llbracket e_1 \rrbracket \sigma \; \overline{\texttt{op}} \; \cA\llbracket e_2 \rrbracket \sigma
 \end{align*}
 For $\texttt{op} \in \texttt{Op}$, $\overline{\texttt{op}}$ is the corresponding operation $\texttt{Val} \times \texttt{Val} \rightarrow \texttt{Val}$
@@ -24,7 +24,6 @@ If on the other hand we define $\cA\llbracket -e \rrbracket \sigma = \cA\llbrack
 it is \textit{not} an inductive definition because $0 - e$ is \textit{not} a subterm of $-e$


-\newpage
 \paragraph{Free Variables}
 For Arithmetic Expressions
 \begin{align*}
@@ -36,7 +35,7 @@ For Arithmetic Expressions
 For Boolean Expressions
 \begin{align*}
    FV(b_1 \; \texttt{op} \; b_2)  & = FV(b_1) \cup FV(b_2) \\
-    FV(\texttt{not} b)             & = FV(b)                \\
+    FV(\texttt{not}\; b)           & = FV(b)                \\
    FV(b_1 \; \texttt{or} \; b_2)  & = FV(b_1) \cup FV(b_2) \\
    FV(b_1 \; \texttt{and} \; b_2) & = FV(b_1) \cup FV(b_2)
 \end{align*}
@@ -56,7 +55,7 @@ And finally for Statements:

 A substitution $f[x \mapsto e]$ replaces each free occurrence of variable $x$ in $f$ by $e$, where $f$ is any expression.

-Detailed rules for arithmetic expressions:
+Detailed rules for arithmetic expressions (for the last, if variable $y$ is $x$, it is replaced, otherwise not):
 \begin{align*}
    (e_1 \; \texttt{op} \; e_2)[x \mapsto e] & \equiv (e_1[x \mapsto e] \; \texttt{op} \; e_2[x \mapsto e]) \\
    n[x \mapsto e]                           & \equiv n                                                     \\
@@ -1,7 +1,7 @@
 \newpage
 \subsection{Operational Semantics}
-Big-step semantics describe how the \bi{overall} results of the execution are obtained and use Natural semantics rules.
-Small-step semantics describe how the individual steps of the computations take place and use Structural Operational Semantics (SOS)
+\textbf{Big-step semantics} describe how the \bi{overall} results of the execution are obtained and use Natural semantics rules.
+\textbf{Small-step semantics} describe how the individual steps of the computations take place and use Structural Operational Semantics (SOS)

 \subsubsection{Big-Step Semantics}
 \paragraph{Transition Systems}
@@ -2,8 +2,8 @@
 Each inference rule is actually a rule scheme, where the meta-variables are placeholders for statements, states, etc.
 Each rule scheme describes infinitely many \bi{rule instances}.

-A rule is \bi{instantiated} when all meta-variables are replaced with syntactic elements
-Assignment rule scheme vs. instance
+A rule is \bi{instantiated} when all meta-variables are replaced with syntactic elements.
+Assignment rule scheme vs. instance:
 \[
    \begin{prooftree}
        \infer0[\textsc{Ass}$_{NS}$]{\langle x := e, \sigma \rangle \rightarrow \sigma[x \mapsto \cA\llbracket e \rrbracket \sigma]}