reserve
  a,b,c,d,e for Ordinal,
  m,n for Nat,
  f for Ordinal-Sequence,
  x for object;
reserve S,S1,S2 for Sequence;

theorem Th54:
  for A being finite Ordinal-Sequence holds Sum^ (A^<%a%>) = (Sum^ A) +^ a
  proof
    let A be finite Ordinal-Sequence;
    consider f being Ordinal-Sequence such that
A1: Sum^(A^<%a%>) = last f & dom f = succ dom(A^<%a%>) & f.0 = 0 &
    for n being Nat st n in dom(A^<%a%>)
    holds f.(n+1) = f.n+^(A^<%a%>).n by Def8;
    consider g being Ordinal-Sequence such that
A2: Sum^ A = last g & dom g = succ dom A & g.0 = 0 &
    for n being Nat st n in dom A holds g.(n+1) = g.n +^ A.n by Def8;
    dom <%a%> = 1 by AFINSQ_1:def 4; then
A3: dom (A^<%a%>) = (dom A)+^1 & (dom A)+^1 = succ dom A
    by ORDINAL2:31,ORDINAL4:def 1;
    reconsider dA = dom A as Element of NAT by ORDINAL1:def 12;
A4: dom A in succ dom A by ORDINAL1:6;
    defpred P[Nat] means $1 in succ dom A implies f.$1 = g.$1;
A5: P[0] by A1,A2;
A6: P[n] implies P[n+1] proof assume that
A7:   P[n] and
A8:   n+1 in succ dom A;
      Segm(n+1) = succ Segm n by NAT_1:38; then
A9:   n in dom A by A8,ORDINAL3:3; then
   n in succ dom A by A4,ORDINAL1:10; then
      g.(n+1) = g.n +^ A.n & f.(n+1) = f.n+^(A^<%a%>).n by A1,A2,A3,A9;
      hence thesis by A7,A9,A4,ORDINAL1:10,ORDINAL4:def 1;
    end;
A10: P[n] from NAT_1:sch 2(A5,A6);
    thus Sum^ (A^<%a%>) = f.((dom A)+^1) by A1,A3,ORDINAL2:6
    .= f.(dA+1) by CARD_2:36
    .= f.(dom A)+^(A^<%a%>).len A by A1,A3,ORDINAL1:6
    .= f.(dom A) +^ a by AFINSQ_1:36
    .= g.(dom A) +^ a by A10,ORDINAL1:6
    .= (Sum^ A) +^ a by A2,ORDINAL2:6;
  end;
