% 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}
