theorem Th3:
  (for n be Nat holds seq1.n <= seq2.n) implies lim_inf seq1 <=
  lim_inf seq2 & lim_sup seq1 <= lim_sup seq2
proof
  assume for n be Nat holds seq1.n <= seq2.n;
  then for n be Nat holds (inferior_realsequence seq1).n <= (
  inferior_realsequence seq2).n & (superior_realsequence seq1).n <= (
  superior_realsequence seq2).n by Th2;
  hence thesis by Th1,MESFUNC5:55;
end;
