Do quasitransitivity and completeness imply transitivity? Let $X$ be a set and let $R$ be a binary relation on $X$, i.e. $R \subseteq X^2$. For $(a,b) \in X^2$, let $a R b$ denote $(a,b) \in R$. Let $P$ be the antisymmetric subset of $R$. Define the properties:

*

*Completeness (C): $a \lnot R b \implies b R a$


*Quasitransitivity (Q): $a P b P c \implies a P c$


*Transitivity (T): $a R b R c \implies a R c$
Here's my confusion: I can't tell if C+Q imply T.
Argument 1. C+Q imply T.
Suppose $a \lnot R b \lnot R c$. Completeness implies $c P b P a$; Quasitransitivity implies $c P a$. So $R$ is transitive since $c R b R a$ and $c R a$.
Argument 2. C+Q do not imply T.
Let $\succ_i$ and $\succ_j$ be complete strict transitive relations on $X$, and suppose they are as follows: $b \succ_i c \succ_i a$ and $c \succ_j a \succ_j b$.
Define $R$ as follows: for all $(a,b) \in X^2, a R b \iff a \succ_k b \text{ for some } k \in \{i,j\}$. Clearly, $R$ is complete. It is also quasitransitive,  since $a \lnot R b \lnot R c$ implies that $a \succ_k c$ for all $k$ due to transitivity of $\succ_k$, and thus $c \lnot R a$.  But it is not transitive: we have $a R b R c$ and yet $a \lnot R c$.
Where is my mistake?
 A: I do not understand your "Argument 1". We assume that $R$ satisfies:

*

*Completeness: rephrase, you have that for all $a,b$, either $aRb$ or $bRa$ (or possibly both).


*Quasitransitivity: if $aRb$ and $bRc$, and in addition, $b\not Ra$ and $c\not Rb$, then $aRc$.
You define $P$ to be $P=\{(a,b)\in R\mid (b,a)\notin R\}$.  So $R$ is quasitransitive if and only if $P$ is transitive.
You want to show transitivity of $R$. So you are trying to show that $cRb$ and $bRa$ imply $cRa$ You show this to be the case when $a\not Rb$ and when $b\not Rc$.  Then it just follows from quasitransitivity; but what if one of them holds?  You should check what happens if $aRb$ and $b\not Rc$; if $a\not Rb$ and $b Rc$; and if $aRb$ and $bRc$. Why do you start by assuming that $a\not Rb$ and that $b\not R c$?
Take $X=\{a,b,c\}$, $R = \{(a,a), (b,b), (c,c), (a,b), (b,a), (b,c), (c,b), (a,c)\}$. This is total, but not transitive (since $(c,b)$ and $(b,a)$ are in $R$, but $(c,a)$ is not).
The relation $P$ you define is just $P=\{(a,c)\}$. This is transitive by vacuity, so $R$ is quasitransitive.
How does your Argument 1 fail here? We take $cRb$ and $bRa$. And then... your argument does not apply, because we do not have $a\not Rb$ and $b\not R c$. In fact, we have both $aRb$ and $bRc$.
Are you perhaps assuming that $P$ will be total when $R$ is total? As you can see from the example, that is not the case. In fact, $P$ is never total, since it is never reflexive. It is also not trichotomic in general.
