[John Baez #4](https://forum.azimuthproject.org/discussion/comment/16344/#Comment_16344):
> Some excellent responses! Just one small issue, coming from some mistakes in Seven Sketches. Everything Matthew and Owen just said is true for posets, but not for preorders.
> Remember that a **preorder** is a set with a binary relation ≤ that's reflexive and transitive. A **poset** is a preorder where \\(x \leq y\\) and \\(y \leq x\\) imply \\(x = y\\).

Okay... but I don't see how my alternative definition uses anti-symmetry (i.e. the rule \\(x \leq y\\) and \\(y \leq x\\) imply \\(x = y\\)).

Here's my attempted proof:

**Lemma**: Assume that \\(f\\) and \\(g\\) are monotone and for all \\(a\\) and \\(b\\) we have \\(f(g(b))\leq b\\) and \\(a \leq g(f(a))\\)

We want to show \\(f \dashv g\\), which is to say for all \\(a\\) and \\(b\\):

f(a)\leq b\text{ if and only if } a \leq g(b)
**Proof.** I hope it's okay if I only show \\(f(a)\leq b \Longrightarrow a \leq g(b)\\), since the other direction is quite similar.

Assume \\(f(a)\leq b\\). Then by monotony of \\(g\\) we have \\(g(f(a)) \leq g(b)\\). However, since \\(a \leq g(f(a))\\) by assumption, then we have \\(a \leq g(b)\\) by transitivity.


Since anti-symmetry wasn't used I don't see why this proof doesn't apply to preorders...? I greatly appreciate you taking the time to help me out.