\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}