\import{set/cons.tex}
\import{set/regularity.tex}

\subsection{Ordinal successor}

\begin{definition}\label{suc}
    $\suc{x} = \cons{x}{x}$.
\end{definition}

\begin{proposition}\label{suc_intro_self}
    $x\in\suc{x}$.
\end{proposition}

\begin{proposition}\label{suc_intro_in}
    Suppose $x\in y$. Then $x\in\suc{y}$.
\end{proposition}

\begin{proposition}\label{suc_elim}
    Suppose $x\in\suc{y}$.
    Then $x = y$ or $x\in y$.
\end{proposition}

\begin{proposition}\label{suc_iff}
    $x\in\suc{y}$ iff $x = y$ or $x\in y$.
\end{proposition}

\begin{proposition}\label{suc_neq_emptyset}
    $\suc{x}\neq \emptyset$.
\end{proposition}

\begin{proposition}\label{suc_subseteq_implies_in}
    Suppose $\suc{x}\subseteq y$. Then $x\in y$.
\end{proposition}

\begin{proposition}\label{suc_neq_self}
    $\suc{x}\neq x$.
\end{proposition}

\begin{proposition}\label{suc_injective}
    Suppose $\suc{x} = \suc{y}$. Then $x = y$.
\end{proposition}
\begin{proof}
    Suppose not.
    $\suc{x}\subseteq \suc{y}$. Hence $x\in\suc{y}$.
    Then $x\in y$.
    $\suc{y}\subseteq \suc{x}$. Hence $y\in\suc{x}$.
    Then $y\in x$.
    Contradiction.
\end{proof}

\begin{proposition}\label{subseteq_self_suc_intro}
    $x\subseteq\suc{x}$.
\end{proposition}

\begin{proposition}\label{suc_subseteq_intro}
    Suppose $x\in y$ and $x\subseteq y$.
    Then $\suc{x}\subseteq y$.
\end{proposition}

\begin{proposition}\label{suc_subseteq_elim}
    Suppose $\suc{x}\subseteq y$.
    Then $x\in y$ and $x\subseteq y$.
\end{proposition}

\begin{proposition}\label{suc_next_subset}
    There exists no $z$ such that $x\subset z\subset \suc{x}$.
\end{proposition}
\begin{proof}
    Follows by \cref{suc,subset,subseteq,subset_witness,suc_elim}.
\end{proof}
