Given a binary relation (dyadic relation) $R$ over two sets $X$ and $Y$,

The complement of the converse relation is the converse of the complement,in other words:

$$\overline{R^{T}}={\overline R^{T}}$$

Where $\bar R$ is the complement of $R$ and $R^T$ is the converse of $R$.


Then:$$\overline{R^{T}}=\left\{\left(y,x\right):x \not \mathrel{R} y\right\} \tag{I}$$

On the other hand:

$${\overline R}=\left\{\left(x,y\right):x \not \mathrel{R} y\right\}$$ Then:

$${\overline R^{T}}=\left\{\left(y,x\right):x \not \mathrel{R} y\right\} \tag{II}$$

Which proves $\left(\text{I}\right)=\left(\text{II}\right)$ are equal,however I'm not sure if the last line is true (it's does not make sense that much to me),can some check the proof and if there is an alternative proof please let me know.

  • $\begingroup$ No, the complement of a relation is not the relation . $\endgroup$ – William Elliot May 8 at 11:57
  • $\begingroup$ @WilliamElliot,what?! $\endgroup$ – user771003 May 8 at 12:08
  • $\begingroup$ Oh yeah, that not so noticeable slash. $\endgroup$ – William Elliot May 8 at 12:22

Using $R'$ for the reverse or converse of R, then
$X×Y - R' = \left\{ (x,y) \in X×Y : \text{not} \;\;yRx \right\}$,
$(X×Y - R)' = \left\{ (y,x) \in X×Y : \text{not}\;\; xRy \right\}$.
Clearly they are equal.

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  • $\begingroup$ Is my proof right?if not then why? $\endgroup$ – user771003 May 8 at 12:47

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