[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?