reserve X for non empty set,
  F for with_the_same_dom Functional_Sequence of X, ExtREAL,
  seq,seq1,seq2 for ExtREAL_sequence,
  x for Element of X,
  a,r for R_eal,
  n,m,k for Nat;

theorem Th2:
  (for n be Nat holds seq1.n <= seq2.n) implies (
  inferior_realsequence seq1).k <= (inferior_realsequence seq2).k & (
  superior_realsequence seq1).k <= (superior_realsequence seq2).k
proof
  reconsider k1 = k as Element of NAT by ORDINAL1:def 12;
  assume
A1: for n be Nat holds seq1.n <= seq2.n;
A2: now
    let n be Nat;
A3: (seq2^\k1).n = seq2.(n+k) by NAT_1:def 3;
    (seq1^\k1).n = seq1.(n+k) by NAT_1:def 3;
    hence (seq1^\k1).n <= (seq2^\k1).n by A1,A3;
  end;
  then sup(seq1^\k1) <= sup(seq2^\k1) by MESFUNC5:55;
  then
A4: (superior_realsequence seq1).k <= sup(seq2^\k1) by RINFSUP2:27;
  inf(seq1^\k1) <= inf(seq2^\k1) by A2,Th1;
  then (inferior_realsequence seq1).k <= inf(seq2^\k1) by RINFSUP2:27;
  hence thesis by A4,RINFSUP2:27;
end;
