How about choosing B=B1−B2 and C=C1−C2, we have BinS(EcapA) by property of S(EcapA), and CinS(E), by property of S(E).
The fact that this works can be seen by observing that both B1cup(C1−A), and B2cup(C2−A) are in fact both disjoint unions. B1,B2 has everything to do with A and C1−A,C2−A has nothing to do with A at all.
ps: I don't know why you use the and and or symbol here instead of intersection and union symbol.
No comments:
Post a Comment