Let $X$ be a set an $R \subset X \times X$ a binary relation on $X$. Then $R$ is Euclidean if for all $x,y,z\in X$, if $xRy$ and $xRz$, then $yRz$.
Through modal logic, I've run into a property of binary relations which are close to Euclidean, but with an extra requirement:
Property$\not =$: For all $x,y,z\in X,$ if $xRy$ and $xRz$ and $x\not =y$, then $yRz$
Does this property have a name?
In the paper, it is referred to as "R1", but the paper does not make use of that term else and searches have been futile.
Edit: Property $\not =$ is not the same as being Euclidean. Consider a set $X=\{x,y\}$ and the relation $R=\{(x,x),(x,y)\}$ on $X$. I.e., $xRx, xRy$ and not $yRx$.
Graphically,
$R$ satisfies Property $\not =$ as the antecedent is not satisfied: We have $xRx$ and $xRy$, but the premise $x \not = x$ is not satisfied. Thus $R$ has property $\not =$.
$R$ is not Euclidean: From $xRx$ and $xRy$, Euclideaness would imply $yRx$. But that is not the case.
Hence the two properties are not the same.
The two definitions of Euclidean relation are equivalent. More precisely, given a binary relation $R \subseteq X \times X$, we say that:
$R$ is Euclidean$_1$ if for all $x,y,z\in X$, if $xRy$ and $xRz$, then $yRz$;
$R$ is Euclidean$_2$ if for all $x,y,z\in X$, if $xRy$ and $xRz$ and $x \neq y$, then $yRz$.
Clearly, if $R$ is Euclidean$_1$ then $R$ is Euclidean$_2$ (the hypothesis in the definition of Euclidean$_2$ is just a special case of the hypothesis in the definition of Euclidean$_1$).
Conversely, suppose that $R$ is Euclidean$_2$. We want to show that $R$ is Euclidean$_1$. Let $x, y, z \in X$ be such that $xRy$ and $xRz$. If $x \neq y$ then $y R z$ because $R$ is Euclidean$_2$. If $x = y$ then $yRz$ because $xRz$ (we can replace $y$ with $x$ since they are equal). In any case, we conclude that $yRz$. Therefore, $R$ is Euclidean$_1$.