[Anindya wrote](https://forum.azimuthproject.org/discussion/comment/18393/#Comment_18393):

> Surely \\(\mathcal{X}(x, y)\\) should be \\(\tt{true}\\) if \\(x \mid y\\) and \\(\tt{false}\\) otherwise.

Aha, I misread Property 1! It says that \\(I \leq \mathcal{X}(x,x)\\), but I swapped one \\(x\\) out for a \\(y\\), which would mean that everything has to be less than or equal to \\(\mathrm{true}\\) if we take that as the identity. But as you've made me aware, that's not the case -- only \\(\mathcal{X}(x, x)\\) is so constrained, and this corresponds to how categories require the presence of all identity arrows.

Let me see if I can recover this once more!

> Surely \\(\mathcal{X}(x, y)\\) should be \\(\tt{true}\\) if \\(x \mid y\\) and \\(\tt{false}\\) otherwise.

Aha, I misread Property 1! It says that \\(I \leq \mathcal{X}(x,x)\\), but I swapped one \\(x\\) out for a \\(y\\), which would mean that everything has to be less than or equal to \\(\mathrm{true}\\) if we take that as the identity. But as you've made me aware, that's not the case -- only \\(\mathcal{X}(x, x)\\) is so constrained, and this corresponds to how categories require the presence of all identity arrows.

Let me see if I can recover this once more!