diff --git a/draft/draft.tex b/draft/draft.tex
index 73ed9b4..3ea97a5 100644
--- a/draft/draft.tex
+++ b/draft/draft.tex
@@ -2497,9 +2497,9 @@ The following example justifies why the $g$ in~\autoref{def:coliftable-ord} shou
 	\begin{align*}
 		\relar (g^\op\comp r\comp f)&\\
 		=&\appr\comp F(g^\op\comp r\comp f)\\
-		=&\appr\comp(Fg^\op)\comp Fr\comp Ff\\
-		=&(Fg^\op)\comp\appr\comp Fr\comp Ff&\by{\autoref{lem:liftable}}\\
-		=&(Fg^\op)\comp\relar r\comp Ff
+		=&\appr\comp(Fg)^\op\comp Fr\comp Ff\\
+		=&(Fg)^\op\comp\appr\comp Fr\comp Ff&\by{\autoref{lem:liftable}}\\
+		=&(Fg)^\op\comp\relar r\comp Ff
 	\end{align*}\qed
 \end{proof}
 \begin{remark}
@@ -2513,22 +2513,22 @@ The following example justifies why the $g$ in~\autoref{def:coliftable-ord} shou
 	\begin{align*}
 		\hat{\relar}(g^\op\comp r\comp f)&\\
 		=&\relar(g^\op\comp r\comp f)\cap(\relar(f^\op\comp r^\op\comp g))^\op\\
-		=&Fg^\op\comp\relar r\comp Ff\cap(Ff^\op\comp\relar r^\op\comp Fg)^\op\\
-		=&Fg^\op\comp\relar r\comp Ff\cap Fg^\op\comp(\relar r^\op)^\op\comp Ff
+		=&(Fg)^\op\comp\relar r\comp Ff\cap((Ff)^\op\comp\relar r^\op\comp Fg)^\op\\
+		=&(Fg)^\op\comp\relar r\comp Ff\cap (Fg)^\op\comp(\relar r^\op)^\op\comp Ff
 	\end{align*}
 	So we are left to prove
 	\begin{gather*}
-		Fg^\op\comp\relar r\comp Ff\cap Fg^\op\comp(\relar r^\op)^\op\comp Ff=Fg^\op\comp(\relar r\cap(\relar r^\op)^\op)\comp Ff
+		(Fg)^\op\comp\relar r\comp Ff\cap (Fg)^\op\comp(\relar r^\op)^\op\comp Ff=(Fg)^\op\comp(\relar r\cap(\relar r^\op)^\op)\comp Ff
 	\end{gather*}
-	because then we will have $\hat{\relar}(g^\op\comp r\comp f)=Fg^\op\comp\hat{\relar}\comp Ff$. We have
+	because then we will have $\hat{\relar}(g^\op\comp r\comp f)=(Fg)^\op\comp\hat{\relar}\comp Ff$. We have
 	\begin{align*}
-		(t,s)\in &Fg^\op\comp\relar r\comp Ff\cap Fg^\op\comp(\relar r^\op)^\op\comp Ff,\\
-		&\iff(t,s)\in Fg^\op\comp\relar r\comp Ff\quad\&\quad (t,s)\in Fg^\op\comp(\relar r^\op)^\op\comp Ff,\\
+		(t,s)\in &(Fg)^\op\comp\relar r\comp Ff\cap (Fg)^\op\comp(\relar r^\op)^\op\comp Ff,\\
+		&\iff(t,s)\in (Fg)^\op\comp\relar r\comp Ff\quad\&\quad (t,s)\in (Fg)^\op\comp(\relar r^\op)^\op\comp Ff,\\
 		&\iff Ff(t) \mathrel{(\relar r)} Fg(s) \quad\&\quad Ff(t) \mathrel{(\relar r^\op)^\op} Fg(s),\\
 		&\iff Ff(t) \mathrel{(\relar\cap(\relar r^\op)^\op)} Fg(s),\\
-		&\iff t\mathrel{(Fg^\op\comp(\relar\cap(\relar r^\op)^\op)\comp Ff)} s.
+		&\iff t\mathrel{((Fg)^\op\comp(\relar\cap(\relar r^\op)^\op)\comp Ff)} s.
 	\end{align*}
-	It worth noting that when we have $(t,s)\in Fg^\op\comp\relar r\comp Ff$, it mean that there exist $t'$ and $s'$ that $(t,t')\in Ff$, $(t',s')\in \relar r$, and $(s',s)\in Fg^\op$. Since $Ff$ and $Fg$ are functions, then $t'=Ff(t)$ and $s'=Fg(s)$, and these are unique elements. The uniqueness enables us in the above reasoning to go from the second line to the third line.\qed 
+	It worth noting that when we have $(t,s)\in (Fg)^\op\comp\relar r\comp Ff$, it mean that there exist $t'$ and $s'$ that $(t,t')\in Ff$, $(t',s')\in \relar r$, and $(s',s)\in Fg^\op$. Since $Ff$ and $Fg$ are functions, then $t'=Ff(t)$ and $s'=Fg(s)$, and these are unique elements. The uniqueness enables us in the above reasoning to go from the second line to the third line.\qed 
 \end{proof}
 \begin{prop}
 	If $\appr$ is an order structure on $F$ that for every sets $X$ and $Y$, $(\Hom(X,FY),\appr)$ is antisymmetric as well (making the posets), then the symmetrization of the left-lax Barr relator of $F$ and $\appr$ is normal.