Prove or disprove: $\lvert\mathcal{R}\rvert=\lvert\mathcal{R}^{-1}\rvert$

Question

Prove or disprove:

If $\mathcal{R}$ is a relation then $\lvert\mathcal{R}\rvert=\lvert\mathcal{R}^{-1}\rvert$.

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I think it is true but I do not know how to prove it.

Facts:

• $\mathcal{R}^{-1}=\{(y,x)\mid(x,y)\in\mathcal{R}\}$.
• $\mathcal{R}\subseteq A\times B$.
• $\lvert\mathcal{R}\rvert\leq\lvert A\times B\rvert$.
• $(x,y)\in\mathcal{R}\implies(x,y)\in A\times B$.

I do not know how to define $\mathcal{R}$. I would know it if it is an equivalence relation, but it is not necessary like an equivalence relation.

Answer 1

${\cal R}^{-1}$ is obtained from ${\cal R}$ by "flipping" all the ordered pairs in ${\cal R}$. This leaves the number of the ordered pairs unchanged.