> Shall we pose the natural follow-up question, then, about whether any non-empty posets exist that can't be given a monoidal structure?

Okay... I see how to adapt your refutation of my attempted counter example into a proof that you've found one!

**Theorem**

Jonathan Castello's "bowtie" poset *cannot* have a monoidal preorder associated with it.

![](https://i.imgur.com/Km2PosD.png)

**Proof.**

We first observe that if the bowtie can have a monoid preorder associated with it, there must be *some* element that is the identity \\(I\\) for that monoid.

There are two cases to consider: \\(I \in \\{a,b\\}\\) and \\(I \in \\{c,d\\}\\).

In order to prove the theorem, we must prove both are infeasible.

By symmetry we only have to consider where \\(I = a\\) or \\(I = c\\).

*First case*: Suppose \\(I = a\\).

Since \\(c \leq a\\) and \\(d \leq d\\) then \\(c \otimes d \leq a \otimes d\\). But since \\(a \otimes d = d\\), then \\(c \otimes d \leq d\\).
But by construction of the bowtie diagram:

$$\forall x. x \leq d \implies x = d$$

...thus \\(c \otimes d = d\\).

However, since \\(c \leq c\\) and \\(d \leq a \\) we have \\(c \otimes d \leq c \otimes a\\). But we know \\(c \otimes a = c\\) and \\(\forall x. x \leq c \implies x = c\\), hence \\(c \otimes d = c\\).

But then \\(c = d\\) ⚡️

*Second case*: Suppose \\(I = c\\).

Since \\(c \leq a\\) and \\(b \leq b\\) then \\(c \otimes b \leq a \otimes b\\), and hence \\(b \leq a \otimes b\\). But then \\(a \otimes b = b\\) by construction.

Moreover seince \\(a \leq a\\) and \\(c \leq b\\) then \\(a \leq a \otimes b\\), hence \\(a = a \otimes b\\).

And thus \\(a = b\\) ⚡️

\\(\Box\\)