Given a0 and a1, since S∈Obs(C), there exists an a2∈A such that for all e∈E, we have a2∈if(S,a0,a1). Then, since T∈Obs(C), there exists an a3∈A such that for all e∈E, we have a3∈if(S,a0,a2). Unpacking and combining these, we get for all e∈E, a3∈if(S∪T,a0,a1). Since we could construct such an a3 from an arbitrary a0,a1∈A, we know that S∪T∈Obs(C). □
I think there’s a typo here. Should be a3∈if(T,a0,a2), not a3∈if(S,a0,a2).
In 4.1:
I think there’s a typo here. Should be a3∈if(T,a0,a2), not a3∈if(S,a0,a2).
(also not sure how to copy latex properly).
Yep. Fixed. Thanks.