A little more work is needed to prove that op l is closed on Σ.
with respect to clause 2 of the above definition, it is necessary to prove that op l (a 1 , a 2 ) = ⊥ or for all labels l , |hole l (op l (a 1 , a 2 ))| ≤ 1, but it is straightforward to see this is the case.