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:

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

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})\$$).

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

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