In case you did not get it.

来源: 于 2011-08-06 12:57:56 [旧帖] [给我悄悄话] 本文已被阅读：次

f(1+0)<=0*f(1)+f(f(1))

-f(1+0)>=-[0*f(1)+f(f(1))] (not <=)