Is this modification of connexity necessary, or redundant in the definition of partial ordering? My question pertains to Fundamentals of Mathematics, Volume 1
Foundations of Mathematics: The Real Number System and Algebra, Edited by H. Behnke, F. Bachmann, K. Fladt, W. Süss and H. Kunle Vol-1 A8.3 Special Properties of Relations
The terms in use are defined as:
$$\begin{aligned}
\text{transitive:}&\ \forall_{x}\forall_{y}\forall_{z}\left(\left(x\mathrm{r}y\land y\mathrm{r}z\right)\implies x\mathrm{r}z\right)\\
\text{identitive:}&\ \forall_{x}\forall_{y}\left(\left(x\mathrm{r}y\land y\mathrm{r}x\right)\implies x=z\right)\\
\text{connex:}&\ \forall_{x}\forall_{y}\left(x\mathrm{r}y\lor y\mathrm{r}x\right)
\end{aligned}$$
The following statement seems to have a redundancy:

Relations which are transitive, identitive and connex are called orderings in the sense of $\le$ (example: $x\le y$). For orderings in the sense of $<$ the requirements of indentitivity and connexity are replaced by $\forall_{x}\lnot x\mathrm{r}x$ and $\forall_{x}\forall_{y}\left(x\ne y\implies\left(x\mathrm{r}y\lor y\mathrm{r}x\right)\right).$

In particular, the condition $x\ne y$ seems to have already been given.  For if $x=y$ then $x\mathrm{r}y\iff x\mathrm{r}x\iff y\mathrm{r}x,$ which is precluded by $\forall_{x}\lnot x\mathrm{r}x$. Am I correct in this understanding?
 A: It is necessary that in the definition of ordering in the sense of $<$, the connex property is replaced by the condition $\forall x \forall y (x \neq y \implies (x\,R\,y \lor y\, R\, x))$. Otherwise, there would not exist any ordering in the sense of $<$, because the connex property and the irreflexive property (i.e. $\forall x \, \lnot(x\,R\,x)$) are incompatible: if one of the two holds, the other is false.
Indeed, suppose that an ordering in the sense of $<$ is defined as a relation $R$ that is

*

*transitive: $\forall x \forall y \forall z ((x\,R\,y \land y\,R\,z) \implies x\,R\,z)$

*irreflexive: $\forall x \, \lnot (x\,R\,x)$

*connex: $\forall x \forall y (x\,R\,y \lor y\, R\, x)$.

Proposition. In every non-empty set, there is no ordering in the sense of $<$.
Proof. By contradiction, suppose that $R$ is an ordering in the sense of $<$ on a non-empty set $S$. Let $z \in S$ (it exists because $S$ is not empty). By the connex property taking $x = y =z$, we have $z\,R\,z \lor z\,R\,z$, that is, $z\,R\,z$. But this is impossible because of the irreflexive property.$\qquad\square$

What is wrong in your reasoning? You claim that, because of irreflexivity, we already know that $x \neq y$ and so this hypothesis $x \neq y$ in the condition $\forall x \forall y (x \neq y \implies (x\,R\,y \lor y\, R\, x))$ is superfluous and can be dropped, since it always holds.
Wait! The convex property says that for every $x$ and $y$ in the set $S$ where the relation $R$ is defined, we have $x\,R\,y$ or $y\, R\, x$. By irreflexivity, this implies that $x \neq y$. But convexity is talking about every $x$ and $y$ in the set $S$, also when $x = y$. Summing up, irreflexivity and convexity together are saying that whenever you take whatsoever $x$ and $y$ in $S$, necessarily we have $x \neq y$; but this is absurd, because you can always choose $x = y$ in $S$.
To avoid this contradiction, in the definition of ordering in the sense of $<$, the hypothesis $x \neq y$ added to the convex condition is not redundant, but necessary.
Said differently, you are right when you say that if you assume irreflexivity and convexity, then
\begin{align}\tag{*}
\text{for every $x$ and $y$ we have $x \neq y$}
\end{align}
Your error is that you did not notice that $(*)$ is claiming something impossible, and hence irreflexivity and convexity cannot coexist: there is no relation that is both irreflexive and connex.

A terminological remark. The terminology used in your reference is quite unusual. Often, the "ordering in the sense of $\leq$" is called total or linear (nonstrict) order, the "ordering in the sense of $<$" is called total (or linear) strict order, the "connex" property is called total or linear or strong connnected, the "identitive" property is called antisymmetric.
