my question is about the example the tutors have chosen in their presentation (solutions for sheet 8). If I got the Exist Algorithm right, there are always three possibilities:
- label(e) > label(phi)
- label(e) = label(phi)
- label(e) < label(phi)
according to slide 38 of the BDD-slide-set. Only in 2. the algorithm uses Apply(OR,l,h).
I just tried to remake the exercise sheet and wanted to correct myself with the step-by-step solution by the tutors, but now I am confused. I dont understand the step from Screenshot 1 to Screenshot 2 because in my opinion in the left subtree case 3 (label(e) < label(phi)) takes place and therefore there is no Apply to be applied. But in Screenshot 2 you can see that the tutors also used Apply(OR,...) here. Also I dont understand the rest of the left subtree in Screenshot 2. I completely agree on the right subtree since in Screenshot 1 you can see that label(e) = label(phi) and there Apply has to be applied.
I hope my question is clear enough.