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% Dummy setup to test the syntax and the generated proof tasks
\begin{definition}\label{subseteq}
$A\subseteq B$ iff $A = B$.
\end{definition}
\begin{definition}\label{pow}
$\pow{A} = \emptyset$.
\end{definition}
\begin{definition}\label{cons}
$\cons{a}{B} = \emptyset$.
\end{definition}
\begin{definition}\label{times}
$A\times B = \emptyset$.
\end{definition}
\begin{definition}\label{fld}
$\fld{R} = \emptyset$.
\end{definition}
\begin{definition}\label{preimg}
$\preimg{R}{A} = \emptyset$.
\end{definition}
\begin{axiom}\label{lmao}
$x\in\emptyset$.
\end{axiom}
\begin{inductive}\label{fin}
Define $\fin{A}\subseteq\pow{A}$ inductively as follows.
\begin{enumerate}
\item $\emptyset \in\fin{A}$.
\item If $a\in A$ and $B\in\fin{A}$, then $\cons{a}{B}\in\fin{A}$.
\end{enumerate}
\end{inductive}
\begin{lemma}\label{fin_mono}
Let $A, B$ be sets.
Suppose $A\subseteq B$.
Then $\fin{A}\subseteq\fin{B}$.
\end{lemma}
\begin{inductive}\label{tracl}
Define $\tracl{R}\subseteq\fld{R}\times\fld{R}$ inductively as follows.
\begin{enumerate}
\item If $w\in R$, then $w\in\tracl{R}$.
\item If $(a, b)\in\tracl{R}$ and $(b,c)\in R$, then $(a,c)\in\tracl{R}$.
\end{enumerate}
\end{inductive}
\begin{inductive}\label{quasirefltracl}
Define $\qrefltracl{R}\subseteq\fld{R}\times\fld{R}$ inductively as follows.
\begin{enumerate}
\item If $a\in\fld{R}$, then $(a,a)\in\qrefltracl{R}$.
\item If $(a, b)\in\qrefltracl{R}$ and $(b,c)\in R$, then $(a,c)\in\qrefltracl{R}$.
\end{enumerate}
\end{inductive}
\begin{inductive}\label{refltracl}
Define $\refltracl{A}{R}\subseteq A\times A$ inductively as follows.
\begin{enumerate}
\item If $a\in A$, then $(a,a)\in\refltracl{A}{R}$.
\item If $(a, b)\in\refltracl{A}{R}$ and $(b,c)\in R$, then $(a,c)\in\refltracl{A}{R}$.
\end{enumerate}
\end{inductive}
\begin{inductive}\label{acc}
Define $\acc{R}\subseteq \fld{R}$ inductively as follows.
\begin{enumerate}
\item If $a\in\fld{R}$ and $\preimg{R}{\{a\}}\in\pow{\acc{R}}$, then $a\in\acc{R}$.
\end{enumerate}
\end{inductive}
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