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**emakarov** In fact, $\displaystyle \bar{R}$ is not only a subset, but is equal to $\displaystyle (A\times A)-R$ (by definition).

To get a better intuition, consider an example, such as $\displaystyle xRy$ iff $\displaystyle x-y$ is even ($\displaystyle x,y$ are integers), or $\displaystyle A\,R\,B$ iff $\displaystyle A\cap B\ne\emptyset$ ($\displaystyle A,B$ are sets). Take a pair $\displaystyle (x,y)$ from $\displaystyle \bar{R}$ and see if $\displaystyle (y,x)\in\bar{R}$. Do this several times and then try to have a different outcome for whether $\displaystyle (y,x)\in\bar{R}$.