[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.

\$$\Box\$$

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.