It is proved that (1) if any x∈S 1 contains x∈S2, then S 1 ∪ S2 = S2 is the proper subset of R, so there must be c ∈ R. S 1∪S2

(2) If S2 is included in S 1, it can also be proved that c∈R makes C? S 1∪S2

(3) If the above two conditions are not true, there is x∈S 1 making x? S2, and y∈S2 exists to make Y? S 1, then consider X+Y ∈ R, if it belongs to S 1, then y=(x+y)-x belongs to S 1, which is contradictory! So it does not belong to S 1. Similarly, it doesn't belong to S2, so x+y? S 1∪S2, get the license.