Complementary of a preorder relation We have $R = (A, A, G )$ a preorder relation (reflexive and transitive) which is defined on set $A$ and $ ρ_R  = R \cap R^{-1} $ an equivalence relation through $R$. Therefore the relation $\overline{R}$ defined on the quotient set $A/ ρ_R $ through 
$[a]_{ρ_R} \, \overline{R} \, [b]_{ρ_R} \iff a\,R\,b$  
is well defined and represents a partial order relation on $A/ ρ_R $.
I don't understand why is that, since $ \overline{R} $ can't be reflexive or transitive.
 A: First let me make clear that the formula in the question serves as the definition of $\overline R$. In particular, other than you thought, this overline notation does not denote a complement.
As an example, consider the relation "$xRy \iff$ person $x$ is at most as tall as person $y$" with the following people:


*

*$a=\text{Adam}$, $170~\rm cm$

*$b=\text{Barbara}$, $180~\rm cm$

*$c=\text{Charles}$, $180~\rm cm$

*$d=\text{Doris}$, $190~\rm cm$


We obviously have $aRa,aRb,aRc,aRd,bRb,bRc,bRd,cRb,cRc,cRd,dRd$.
Now $\rho_r = R \cap R^{-1}$ is the equivalence relation "$x$ is exactly as tall as $y$", with the three equivalence classes


*

*$[a]_{\rho_R} = \{a\}$

*$[b]_{\rho_R} = [c]_{\rho_R} = \{b,c\}$

*$[d]_{\rho_R} = \{d\}$


Now using the definition of $\overline R$, we get


*

*$[a]_{\rho_R} \overline R [a]_{\rho_R}$ because $aRa$.

*$[a]_{\rho_R} \overline R [b]_{\rho_R}$ because $aRb$.

*$[a]_{\rho_R} \overline R [d]_{\rho_R}$ because $aRd$.

*$[b]_{\rho_R} \overline R [b]_{\rho_R}$ because $bRb$.

*$[b]_{\rho_R} \overline R [d]_{\rho_R}$ because $bRd$.

*$[d]_{\rho_R} \overline R [d]_{\rho_R}$ because $dRd$.


This is obviously a partial order (indeed, in this case even a total order) of the three equivalence classes.
In the general case, you can see that $\overline R$ is indeed a partial order as follows:


*

*$\overline R$ is actually well defined (this has to be checked!): If $[a]_{\rho_R} = [b]_{\rho_R}$ and $[c]_{\rho_R} = [d]_{\rho_R}$ then we have to show that $[a]_{\rho_R} \overline R [c]_{\rho_R}$ iff $[b]_{\rho_R} \overline R [d]_{\rho_R}$.
If $[a]_{\rho_R} = [b]_{\rho_R}$ then we have both $aRb$ and $bRa$, and from $[c]_{\rho_R} = [d]_{\rho_R}$ we get $cRd$ and $dRc$.
Now by definition of $\overline R$, $[a]_{\rho_R} \overline R [c]_{\rho_R}$ iff $aRc$. Now thanks to transitivity of $R$, we get from $bRa$ and $aRc$ that $bRc$, and then with $cRd$ that $bRd$. The direction $bRd \implies aRc$ works analogously. But by definition of $\overline R$, $bRd \iff [b]_{\rho_R} \overline R [d]_{\rho_R}$.

*Reflexivity: Due to reflexivity of $a$, we have $aRa$, and thus by definition of $\overline R$ we get $[a]_{\rho_R}\overline R [a]_{\rho_R}$.

*Transitivity: If $[a]_{\rho_R} \overline R [b]_{\rho_R}$ and $[b]_{\rho_R} \overline R [c]_{\rho_R}$, then we have by the definition of $\overline R$ that $aRb$ and $bRc$, and thus by transitivity of $R$ we have $aRc$ which by definition of $\overline R$ means $[a]_{\rho_R} \overline R [c]_{\rho_R}$.

*Antisymmetry: If $[a]_{\rho_R} \overline R [b]_{\rho_R}$ and $[b]_{\rho_R} \overline R [a]_{\rho_R}$, then by definition of $\overline R$ we have $aRb$ and $bRa$, but that means that $a$ and $b$ are equivalent according to $\rho_R$, thus $[a]_{\rho_R} = [b]_{\rho_R}$.
