I was curious how the choice of symmetric monoidal poset affects the resulting \\(\mathbf{Bool}\\)-enriched category.

From any poset over \\(\mathbf{Bool}\\), we can make it symmetric monoidal by picking an \\(I\\) and a \\(\otimes\\). For \\(\leq\\), I will just pick the usual \\(\tt{false} \le \tt{true}\\). Although as Matthew notes, there is also the trivial poset where \\( \forall x y, x \le y \\) also and the discrete poset where \\( \forall x, x \le x \\).

The choice for \\(I\\) over \\(\mathbf{Bool}\\) must be either \\(\tt{true}\\) or \\(\tt{false}\\). I'll interpret the effect of each using property a) of \\(\mathcal{V}\\)-enriched categories:

\[ \text{for all objects } x, I\leq\mathcal{X}(x,x) .\]

If \\(I = \tt{false}\\), we've added no restrictions, since \\(\tt{false} \leq \tt{anything}\\). However, if \\(I = \tt{true}\\), then \\(\mathcal{X}(x,x)\\) must be \\(\tt{true}\\) also. So setting \\(I = \tt{true}\\) effectively forces every identity relation to be \\(\tt{true}\\). In other words:

I & \text{effect of a) on $\mathcal{X}(x,y)$} \\\\
\tt{false} & \text{No restrictions} \\\\
\tt{true} & \mathcal{X}(x,x) = \tt{true}

There are 16 possible binary relations \\(\mathbf{Bool} \times \mathbf{Bool} \to \mathbf{Bool}\\), but only 8 symmetric binary relations. We can actually name these 8 relations:

\[ \\{ C_F (\tt{constant~false}), \downarrow (\tt{nor}), \oplus (\tt{xor}), \uparrow (\tt{nand}), \wedge (\tt{and}), \leftrightarrow (\tt{xnor}), \vee (\tt{or}), C_T (\tt{constant~true}) \\} \]

\\( \\{ C_F, \downarrow, \uparrow, C_T \\} \\) fail to satisfy the monoidal unit laws, so we can't use them for our monoidal product \\(\otimes\\). So we are left with \\( \{ \oplus, \wedge, \leftrightarrow, \vee \} \\). There is one valid choice of identity for each of them, giving us four symmetric monoidal structures over our \\(\mathbf{Bool}\\) poset:

\[ \{ (\oplus, \tt{false}), (\wedge, \tt{true}), (\leftrightarrow, \tt{true}), (\vee, \tt{false}) \} \]

With these monoids in hand, let's explore property b) of \\(\mathcal{V}\\)-enriched categories.

\[ \text{for all objects } x,y,z, \mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z). \]

This sets restrictions on \\(\mathcal{X}(x,y)\\) in our category, based on how \\(\otimes\\) acts.

For example, if \\(\otimes = \vee\\), then with \\(\mathcal{X}(x,y) = \tt{true}\\), we can derive \\(\forall z, \mathcal{X}(y,z) = \tt{true}\\) and \\(\forall w, \mathcal{X}(w,x) = \tt{true}\\). So if any \\(\mathcal{X}(x,y) = \tt{true}\\), then all \\(\mathcal{X}(x,y) = \tt{true}\\)!

If \\(\otimes = \wedge\\), then \\(\mathcal{X}(x,z) = \tt{true}\\) whenever \\(\mathcal{X}(x,y) = \tt{true}\\) and \\(\mathcal{X}(y,z) = \tt{true}\\). This kind of looks like the choice of \\(\otimes\\) gives \\(\mathcal{X}\\) closure over composition.

I've discussed the easy two, but I haven't found a good interpretation of the properties of the other two (\\((\oplus, \tt{false})\\) and \\((\leftrightarrow, \tt{true})\\)).

\text{monoid} & \text{effect of b) on $\mathcal{X}(x,y)$} \\\\
(\oplus, \tt{false}) & \text{If $\mathcal{X}(x,y)$, then for all objects $z$, either $\mathcal{X}(x,z)$ or $\mathcal{X}(z,x)$ or both.} \\\\
(\wedge, \tt{true}) & \text{If $\mathcal{X}(x,y)$ and $\mathcal{X}(y,z)$, then $\mathcal{X}(x,z)$.} \\\\
(\leftrightarrow, \tt{true}) & \text{If $\mathcal{X}(x,y) = \mathcal{X}(y,z)$, then $\mathcal{X}(x,z)$.} \\\\
(\vee, \tt{false}) & \text{Either $\mathcal{X}(x,y) = \tt{false}$, or $\mathcal{X}(x,y) = \tt{true}$, for all objects $x,y$.} \\\\

Maybe we can get even more interesting \\(\mathbf{Bool}\\)-categories by considering the other two posets.