Well, we can try to see if the yanking axiom holds.

The equations we want to test are,

\\[

(Id\_X \times \cap\_X) \circ (\cup\_X \times Id\_X)

= Id\_X =

(\cap\_X \times Id\_X) \circ (Id\_X \times \cup\_X).

\\]

The equations we want to test are,

\\[

(Id\_X \times \cap\_X) \circ (\cup\_X \times Id\_X)

= Id\_X =

(\cap\_X \times Id\_X) \circ (Id\_X \times \cup\_X).

\\]