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\import{relation.tex}
\import{relation/properties.tex}
\subsection{Quasiorders}
% also called preorder
\begin{abbreviation}\label{quasiorder}
$R$ is a quasiorder iff
$R$ is quasireflexive and transitive.
\end{abbreviation}
% also called preorder
\begin{abbreviation}\label{quasiorder_on}
$R$ is a quasiorder on $A$ iff
$R$ is a binary relation on $A$ and
$R$ is reflexive on $A$ and transitive.
\end{abbreviation}
\begin{struct}\label{quasiordered_set}
A quasiordered set $X$ is a onesorted structure
equipped with
\begin{enumerate}
\item $\lt$
\end{enumerate}
such that
\begin{enumerate}
\item\label{quasiorder_type} $\lt[X]$ is a binary relation on $\carrier[X]$.
\item\label{quasiorder_refl} $\lt[X]$ is reflexive on $\carrier[X]$.
\item\label{quasiorder_tran} $\lt[X]$ is transitive.
\end{enumerate}
\end{struct}
\begin{lemma}\label{quasiorder_transitive_double}
Let $X$ be a quasiordered set.
Let $a, b, c, d \in X$.
Suppose $a\mathrel{\lt[X]} b\mathrel{\lt[X]} c\mathrel{\lt[X]} d$.
Then $a\mathrel{\lt[X]} d$.
\end{lemma}
\begin{proof}
$\lt[X]$ is transitive by \cref{quasiorder_tran}.
Thus $a\mathrel{\lt[X]} c\mathrel{\lt[X]} d$ by \cref{transitive}.
Hence $a\mathrel{\lt[X]} d$ by \cref{transitive}.
Follows by assumption.
\end{proof}
\begin{proposition}\label{subseteqrel_is_quasiorder}
$\subseteqrel{A}$ is a quasiorder on $A$.
\end{proposition}
\begin{proof}
Show for all $w\in\subseteqrel{A}$ we have $w\in A\times A$.
\begin{subproof}
Fix $w$.
Assume $w\in\subseteqrel{A}$.
Take $a,b$ such that $w=(a,b)$ and $a,b\in A$
by \cref{subseteqrel,pair_eq_iff}.
Follows by \cref{times_tuple_intro}.
\end{subproof}
Thus $\subseteqrel{A}$ is a binary relation on $A$
by \cref{subseteq}.
$\subseteqrel{A}$ is reflexive on $A$
by \cref{reflexive_on,subseteqrel,subseteq}.
Show for all $a,b,c$ such that
$a\mathrel{\subseteqrel{A}}b$ and
$b\mathrel{\subseteqrel{A}}c$
we have $a\mathrel{\subseteqrel{A}}c$.
\begin{subproof}
Fix $a,b,c$.
Assume $a\mathrel{\subseteqrel{A}}b$ and
$b\mathrel{\subseteqrel{A}}c$.
Take $x,y$ such that $(a,b)=(x,y)$ and
$x,y\in A$ and $x\subseteq y$ by \cref{subseteqrel}.
Then $a=x$ and $b=y$ by \cref{pair_eq_iff}.
Thus $a\subseteq b$ and $a\in A$ by assumption.
Take $u,v$ such that $(b,c)=(u,v)$ and
$u,v\in A$ and $u\subseteq v$ by \cref{subseteqrel}.
Then $b=u$ and $c=v$ by \cref{pair_eq_iff}.
Thus $b\subseteq c$ and $c\in A$ by assumption.
Thus $a\subseteq c$ by \cref{subseteq_transitive}.
Follows by \cref{subseteqrel}.
\end{subproof}
Thus $\subseteqrel{A}$ is transitive
by \cref{transitive}.
Follows by assumption.
\end{proof}
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