If R is any relation on a set A then $\mathrm{RoR}^{-1}$ is

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Professor PC Thomas Classes · Answer

If $(x, y) \in \operatorname{ROR}^{-1}$ then $\exists z \in A \quad f(x, z), \in R^{-1}$ and $(\mathrm{z}, \mathrm{y}) \in \mathrm{R}$. $ \begin{aligned} & 	herefore(\mathrm{z}, \mathrm{x}) \in \mathrm{R} 	ext { and }(\mathrm{y}, \mathrm{z}) \in \mathrm{R}^{-1} \ & \Rightarrow(\mathrm{y}, \mathrm{x}) \in \mathrm{RoR}^{-1} \ & \Rightarrow \mathrm{RoR}^{-1} 	ext { is symmetric } \end{aligned} $ \ $\operatorname{RoR}^{-1}=\{(1,1),(1,2),(2,1),(2,2),(2,3)$, $(3,2),(3,3)$ } \ is not reflexive and not transitive