Clarification needed on state q

Question

so on state Q I find out the relation is (p’ exor q’) & (!q’ | !p’) .. so from the exor we can determine that the transition should be on {!p&q} and {p&!q} state. But what about (!q’ | !p’) ??. It can be simplified into !(q’&p’). And we get the transition to {!p&q} and {p&!q} and {} state.

But while checking the answer there is no transition from state q to empty set.

Answer 1

Sure, just note that !q'|!p' is equivalent to p'&!q' | !p'&q' | !p'&!q', since if q' is false p' may be true or false, and if q' is true, then p' must be false.