182 lines
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8.4 KiB
TeX
182 lines
No EOL
8.4 KiB
TeX
% chktex-file 46
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\section{Introduction}
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This is a summary of the course ``Algebra des Programmierens'' taught by Prof.\ Dr.\ Stefan Milius in the winter term 2023/2024 at the FAU~\footnote{Friedrich-Alexander-Universität Erlangen-Nürnberg}.
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The course is based on~\cite{poll1999algebra} with~\cite{adamek1990abstract} as a reference for category theory.
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Goal of the course is to develop a mathematical theory for semantics of data types and their accompanying proof principles. The chosen environment is the field of category theory.
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\subsection{Functions}
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A function $f : X \rightarrow Y$ is a mapping from the set $X$ (the domain of $f$) to the set $Y$ (the codomain of $f$).
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More concretely $f$ is a relation $f \subseteq X \times Y$ which is
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\begin{itemize}
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\item \emph{left-total}, i.e.\ for all $x \in X$ exists some $y \in Y$ such that $(x,y) \in f$;
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\item \emph{right-unique}, i.e.\ any $(x,y),(x,y') \in f$ imply $y = y'$.
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\end{itemize}
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Often, one is also interested in the symmetrical properties, a function is called
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\begin{itemize}
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\item \emph{injective} or \emph{left-unique} if for every $x,x' \in X$ the implication $f(x) = f(x') \rightarrow x = x'$ holds;
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\item \emph{surjective} or \emph{right-total} if for every $y \in Y$ there exists an $x \in X$ such that $f(x) = y$;
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\item \emph{bijective} if it is injective and surjective.
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\end{itemize}
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\begin{example}
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\begin{enumerate}
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\item The identity function $id_A : A \rightarrow A$, $id_A(x) = x$
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\item The constant function $b! : A \rightarrow B$ for $b \in B$ defined by $b!(x) = b$
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\item The inclusion function $i_A : A \rightarrow B$ for $A \subseteq B$ defined by $i_A(x) = x$
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\item Constants $b : 1 \rightarrow B$, where $1 := {*}$. The function $b$ is in bijection with the set $B$.
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\item Composition of function $f : A \rightarrow B, g : B \rightarrow C$ called $g \circ f : A \rightarrow C$ defined by $(g \circ f)(x) = g(f(x))$.
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\item The empty function $¡ : \emptyset \rightarrow B$
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\item The singleton function $! : A \rightarrow 1$
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\end{enumerate}
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\end{example}
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\subsection{Data Types}
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Programs work with data that should ideally be organized in a useful manner.
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A useful representation for data in functional programming is by means of \emph{algebraic data types}.
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Some basic data types (written in Haskell notation) are
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\begin{minted}{haskell}
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data Bool = True | False
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data Nat = Zero | Succ Nat
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\end{minted}
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These data types are declared by means of constructors, yielding concrete descriptions how inhabitants of these types are created.
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\emph{Parametric data types} are additionally parametrized by another data type, e.g.\
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\begin{minted}{haskell}
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data Maybe a = Nothing | Just a
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data Either a b = Left a | Right b
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data List a = Nil | Cons a (List a)
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\end{minted}
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Such data types (parametric or non-parametric) usually come with a principle for defining functions called recursion and in richer type systems (e.g.\ in a dependently typed setting) with a principle for proving facts about recursive functions called induction.
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Equivalently, every function defined by recursion can be defined via a \emph{fold}-function which satisfies an identity and fusion law, which replace the induction principle. Let us now consider two examples of data types and illustrate this.
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\subsubsection{Natural Numbers}
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The type of natural numbers comes with a fold function $foldn : C \rightarrow (Nat \rightarrow C) \rightarrow Nat \rightarrow C$ for every $C$, defined by
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\begin{alignat*}{2}
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& foldn\;c\;h\;zero & & = c \\
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& foldn\;c\;h\;(suc\;n) & & = h\;(foldn\;c\;h\;n)
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\end{alignat*}
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\begin{example} Let us now consider some functions defined in terms of $foldn$.
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\begin{itemize}
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\item $iszero : Nat \rightarrow Bool$ is defined by
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\[iszero = foldn\;true\;false!\]
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\item $plus : Nat \rightarrow Nat \rightarrow Nat$ is defined by
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\[plus = foldn\;id (\lambda f\;n. succ (f\;n)) \]
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\end{itemize}
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\end{example}
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\begin{proposition}
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$foldn$ satisfies the following two rules
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\begin{enumerate}
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\item \customlabel{law:natident}{\textbf{Identity}}: $foldn\;zero\;succ = id_{Nat}$
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\item \customlabel{law:natfusion}{\textbf{Fusion}}: for all $c : C$, $h, h' : Nat
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\rightarrow C$ and $k : C \rightarrow C'$ with $kc = c'$ and $kh = h'k$ follows $k \circ foldn\;c\;h = foldn\;c'\;h'$, or diagrammatically:
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% https://q.uiver.app/#q=WzAsNSxbMiwwLCJDIl0sWzQsMCwiQyJdLFsyLDIsIkMnIl0sWzQsMiwiQyciXSxbMCwwLCIxIl0sWzQsMCwiYyJdLFswLDIsImsiXSxbNCwyLCJjJyIsMl0sWzAsMSwiaCIsMl0sWzIsMywiaCciLDJdLFsxLDMsImsiLDFdXQ==
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\[
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\begin{tikzcd}[ampersand replacement=\&]
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1 \&\& C \&\& C \\
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\\
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\&\& {C'} \&\& {C'}
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\arrow["c", from=1-1, to=1-3]
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\arrow["k", from=1-3, to=3-3]
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\arrow["{c'}"', from=1-1, to=3-3]
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\arrow["h"', from=1-3, to=1-5]
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\arrow["{h'}"', from=3-3, to=3-5]
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\arrow["k"{description}, from=1-5, to=3-5]
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\end{tikzcd}
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\]
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implies
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% https://q.uiver.app/#q=WzAsMyxbMCwwLCJOYXQiXSxbMiwwLCJDIl0sWzIsMiwiQyciXSxbMSwyLCJrIl0sWzAsMSwiZm9sZG5cXDtjXFw7aCJdLFswLDIsImZvbGRuXFw7YydcXDtoJyIsMl1d
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\[
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\begin{tikzcd}[ampersand replacement=\&]
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Nat \&\& C \\
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\\
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\&\& {C'}
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\arrow["k", from=1-3, to=3-3]
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\arrow["{foldn\;c\;h}", from=1-1, to=1-3]
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\arrow["{foldn\;c'\;h'}"', from=1-1, to=3-3]
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\end{tikzcd}
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\]
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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Both follow by induction over an argument $n : Nat$:
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\begin{enumerate}
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\item~\ref{law:natident}:
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\begin{mycase}
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\case{} $n = zero$
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\[foldn\;zero\;succ\;zero = zero = id\;zero\]
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\case{} $n = succ\;m$
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\begin{alignat*}{1}
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foldn\;zero\;succ\;(succ\;m) & = succ (foldn\;zero\;succ\;m)
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\\&= succ\;m \tag{IH}
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\\&= id (succ\;m)
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\end{alignat*}
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\end{mycase}
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\item~\ref{law:natfusion}:
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\begin{mycase}
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\case{} $n = zero$
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\[k(foldn\;c\;h\;zero) = k\;c = c' = foldn\;c'\;h'\;zero\]
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\case{} $n = succ\;m$
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\begin{alignat*}{1}
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k(foldn\;c\;h\;(succ\;m)) & = k(h(foldn\;c\;h\;m))
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\\&= h'(k(foldn\;c\;h\;m))
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\\&= h'(foldn\;c'\;h'\;m) \tag{IH}
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\\&= foldn\;c'\;h'\;(succ\;m)
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\end{alignat*}
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\end{mycase}
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\end{enumerate}
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\end{proof}
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\begin{example}
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The identity and fusion laws can in turn be used to prove the following induction principle:
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For any predicate $p : Nat \rightarrow Bool$,
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\begin{enumerate}
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\item $p\;zero = true$ and
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\item $p \circ succ = p$
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\end{enumerate}
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implies $p = true!$. This follows by % chktex 40
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\begin{alignat*}{1}
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& p
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\\=\;&p \circ (foldn\;zero\;succ) \tag{\ref{law:natident}}
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\\=\;&foldn\;true\;id \tag{\ref{law:natfusion}}
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\\=\;&true! \circ (foldn\;zero\;succ) \tag{\ref{law:natfusion}}
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\\=\;&true!. \tag{\ref{law:natident}}
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\end{alignat*}
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Where the first application of~\ref{law:natfusion} is justified, since the diagram
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% https://q.uiver.app/#q=WzAsNSxbMiwwLCJOYXQiXSxbNCwwLCJOYXQiXSxbMiwyLCJCb29sIl0sWzQsMiwiQm9vbCJdLFswLDAsIjEiXSxbNCwwLCJ6ZXJvIl0sWzAsMSwic3VjYyJdLFsxLDMsInAiXSxbMCwyLCJwIl0sWzIsMywiaWQiLDFdLFs0LDIsInRydWUiLDJdXQ==
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\[\begin{tikzcd}
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1 && Nat && Nat \\
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\\
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&& Bool && Bool
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\arrow["zero", from=1-1, to=1-3]
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\arrow["succ", from=1-3, to=1-5]
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\arrow["p", from=1-5, to=3-5]
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\arrow["p", from=1-3, to=3-3]
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\arrow["id"{description}, from=3-3, to=3-5]
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\arrow["true"', from=1-1, to=3-3]
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\end{tikzcd}\]
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commutes by the requisite properties of $p$, and the second application of~\ref{law:natfusion} is justified, since the diagram
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% https://q.uiver.app/#q=WzAsNSxbMiwwLCJOYXQiXSxbNCwwLCJOYXQiXSxbMiwyLCJCb29sIl0sWzQsMiwiQm9vbCJdLFswLDAsIjEiXSxbNCwwLCJ6ZXJvIl0sWzAsMSwic3VjYyJdLFsxLDMsInRydWUhIl0sWzAsMiwidHJ1ZSEiXSxbMiwzLCJpZCIsMV0sWzQsMiwidHJ1ZSIsMl1d
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\[\begin{tikzcd}
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1 && Nat && Nat \\
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\\
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&& Bool && Bool
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\arrow["zero", from=1-1, to=1-3]
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\arrow["succ", from=1-3, to=1-5]
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\arrow["{true!}", from=1-5, to=3-5]
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\arrow["{true!}", from=1-3, to=3-3]
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\arrow["id"{description}, from=3-3, to=3-5]
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\arrow["true"', from=1-1, to=3-3]
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\end{tikzcd}\]
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trivially commutes.
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\end{example}
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\subsubsection{Lists} |