[Artur Grezsiak](https://forum.azimuthproject.org/discussion/comment/16936/#Comment_16936) nicely solved Puzzle 33.

Yes, an equivalence relation is a preorder! If I wanted to show off - which luckily I never want to do - I could have said:

**Definition.** An **equivalence relation** \$$\sim\$$ on a set \$$X\$$ is a preorder on \$$X\$$ that is **symmetric**, meaning that for all \$$x,y \in X\$$, \$$x \sim y\$$ implies \$$y \sim x\$$.

It's slightly surprising that equivalence relations are preorders, because often we think about **posets**, which are preorders that are **antisymmetric**, meaning that for all \$$x,y \in X\$$, \$$x \le y\$$ and \$$y \le x\$$ imply \$$x = y\$$.

We use the symmetrical symbol \$$\sim\$$ for preorders that are symmetric, and the asymmetric symbol \$$\le\$$ for preorders that are antisymmetric. However, one beauty of preorders is that they put posets and equivalence relations under the same roof! A theorem about preorders is a theorem about both \$$\le\$$ and \$$\sim\$$!

I posed this puzzle elsewhere, but I'll pose it again:

**Puzzle.** Can a preorder be both symmetric and antisymmetric? If so, what's it like?