Name for relation property: If $xRy$ and $xRz$ and $x \not =y$, then $yRz$.

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.

• Looks the same as "Euclidean" to me. Commented Mar 4, 2018 at 10:57
• It is not, cf. the now-included edit. Commented Mar 4, 2018 at 11:31
• @Rasmus - Why do you say that "The mentioned property does not entail $yRx$ under these circumstances"? Look at my answer. Commented Mar 4, 2018 at 11:47
• Your proposed counterexample is incorrect and maybe only appears correct at first because of the unfortunate choice of variables $x$ and $y$. Let $R=\{(a,a),(a,b)\}$. Then set $x=z=a$ and $y=b$. We then have the hypotheses of prop. $\ne$: $aRb$ and $aRa$ and $a\ne b$, but not the conclusion: $b\not Ra$. Thus, $R$ has neither property so isn't a counterexample. Incidentally, you won't be able to find a counterexample because @Taroccoesbrocco's proof is correct. Commented Mar 4, 2018 at 13:22

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$_1$.

• For specific $x,y,z\in X$, assuming $xRy$ and $xRz$ allows us to conclude both $yRz$ and $zRy$ by if Euclidean$_1$. You show that $yRz$ under the assumption that $R$ is Euclidean$_2$ -- I agree to that. I counter the claim that this is sufficient for Euclidean$_1$: It is not given that $zRy$, cf. the example I included. Commented Mar 4, 2018 at 12:04
• The "counterexample" in your edit actually is not a counterexample: you can always conclude that $yRx$. Indeed, suppose $R$ is Euclidean$_2$ and $xRx$ and $xRy$. There are two cases: either $x \neq y$, and then $yRx$ because $R$ is Euclidean$_2$; or $x = y$, and then $yRx$ because $xRx$. Commented Mar 4, 2018 at 12:24
• I think we've been talking past each other due to a poor choice of variables on my side. I've updated my answer now -- still the same example, but hopefully clarified. Commented Mar 4, 2018 at 12:41
• @Rasmus - The relation $R$ you defined in your last edit does not have the property $\neq$ (i.e. it is not Euclidean$_2$): you have $xRy$ and $xRx$ with $x \neq y$, but $y \not R x$. Commented Mar 4, 2018 at 13:21