
theorem Th26:
  for a being Ordinal, n being Nat holds Sum^(n --> a) = n *^ a
proof
  let a be Ordinal, n be Nat;
  consider fi being Ordinal-Sequence such that
    A1: Sum^(n --> a) = last fi & dom fi = succ dom(n --> a) & fi.0 = 0 and
    A2: for k being Nat st k in dom(n --> a)
      holds fi.(k+1) = fi.k +^ (n --> a).k by ORDINAL5:def 8;
  A4: now
    let C be Ordinal;
    assume succ C in succ n;
    then A5: C in n by ORDINAL3:3;
    n in omega by ORDINAL1:def 12;
    then C in omega by A5, ORDINAL1:10;
    then reconsider k = C as Nat;
    A6: k in dom(n --> a) by A5;
    thus fi.succ C = fi.succ Segm C
      .= fi.Segm(k+1) by NAT_1:38
      .= fi.k +^ (n --> a).k by A2, A6
      .= fi.C +^ a by A5, FUNCOP_1:7;
  end;
  now
    let C be Ordinal;
    assume A7: C in succ n & C <> 0 & C is limit_ordinal;
    succ n in omega by ORDINAL1:def 12;
    then C in omega by A7, ORDINAL1:10;
    hence fi.C = union sup(fi|C) by A7; :: by contradiction
  end;
  hence thesis by A1, A4, ORDINAL2:def 15;
end;
