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Is the relation $R:=\{(1,2),(1,3)\}$ transitive on $M=\{1,2,3\}$ with $R\subseteq M\times M$?

I think it's transitive, because we don't have elements that satisfy $xRy \land yRz $ and therefore $\forall x,y,z \in M: xRy \land yRz \implies xRz$ is always true. Is this right?

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    $\begingroup$ Yes, that’s correct. In this kind of situation you can say that it’s vacuously true. $\endgroup$ Oct 19, 2020 at 19:12
  • $\begingroup$ Alternatively: There are no such $x,y,z$ of $M$ where $x\operatorname R y\wedge y\operatorname R z\wedge \neg(x\operatorname R z)$. $\endgroup$ Oct 20, 2020 at 0:13

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It is transitive, and this is vacuously true for the reason you have said.

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