theorem Th6:
  ASeq is non-descending implies P * ASeq is non-decreasing
proof
A1: dom (P * ASeq) = NAT by SEQ_1:1;
  assume
A2: ASeq is non-descending;
  now
    let n,m be Nat;
    assume n <= m;
    then
A3: ASeq.n c= ASeq.m by A2,PROB_1:def 5;
     reconsider nn=n, mm=m as Element of NAT by ORDINAL1:def 12;
    (P * ASeq).nn = P.(ASeq.nn) & (P * ASeq).mm = P.(ASeq.mm)
      by A1,FUNCT_1:12;
    hence (P * ASeq).n <= (P * ASeq).m by A3,PROB_1:34;
  end;
  hence thesis by SEQM_3:6;
end;
