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Someone answer my question!!!
ok, i've also simplified it to an equivalent theorem. If you can prove this, the Shroder-Bernstein is easy to prove.
if A3 is a proper subset of A2, which is a proper subset of A1 , and |A1|=|A3| (or A1~A3), then |A1|=|A2|. ( or A1~A2~A3)
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