> 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\$$