
theorem Th27:
  for A being finite Ordinal-Sequence, a being Ordinal holds
    Sum^ (A|a) c= Sum^ A
proof
  let A be finite Ordinal-Sequence, a be Ordinal;
  per cases;
  suppose dom A c= a;
    hence thesis by RELAT_1:68;
  end;
  suppose A1: a c= dom A;
    then reconsider a as finite Ordinal;
    consider f1 being Ordinal-Sequence such that
      A2: Sum^(A|a) = last f1 & dom f1 = succ dom(A|a) & f1.0 = 0 and
      A3: for n being Nat st n in dom(A|a) holds f1.(n+1) = f1.n +^ (A|a).n
      by ORDINAL5:def 8;
    consider f2 being Ordinal-Sequence such that
      A4: Sum^ A = last f2 & dom f2 = succ dom A & f2.0 = 0 and
      A5: for n being Nat st n in dom A holds f2.(n+1) = f2.n +^ A.n
      by ORDINAL5:def 8;
    defpred P[Nat] means $1 in dom f1 implies f1.$1 = f2.$1;
    A6: P[0] by A2, A4;
    A7: for n being Nat st P[n] holds P[n+1]
    proof
      let n be Nat;
      assume A8: P[n];
      assume n+1 in dom f1;
      then A9: succ n in dom f1 by Lm5;
      n in succ n by ORDINAL1:6;
      then A10: f1.n = f2.n by A8, A9, ORDINAL1:10;
      A11: n in dom(A|a) by A2, A9, ORDINAL3:3;
      then A12: n in dom A by RELAT_1:57;
      thus f1.(n+1) = f1.n +^ (A|a).n by A3, A11
        .= f2.n +^ A.n by A10, A11, FUNCT_1:47
        .= f2.(n+1) by A5, A12;
    end;
    A13: for n being Nat holds P[n] from NAT_1:sch 2(A6,A7);
    A14: last f1 = f1.dom(A|a) & last f2 = f2.dom A by A2, A4, ORDINAL2:6;
    then A15: last f1 = f2.dom(A|a) by A2, A13, ORDINAL1:6
      .= f2.a by A1, RELAT_1:62;
    Segm a c= Segm dom A by A1;
    then consider k being Nat such that
      A16: dom A = a + k by NAT_1:10, NAT_1:39;
    defpred Q[Nat] means a + $1 <= dom A implies f2.a c= f2.(a+$1);
    A17: Q[0];
    A18: for n being Nat st Q[n] holds Q[n+1]
    proof
      let n be Nat;
      assume A19: Q[n];
      assume A20: a + (n+1) <= dom A;
      then a + n + 1 < dom A + 1 by NAT_1:13;
      then A21: f2.a c= f2.(a+n) by A19, XREAL_1:6;
      Segm(a+(n+1)) c= Segm dom A by A20, NAT_1:39;
      then a+n+1 c= dom A;
      then succ(a+n) c= dom A by Lm5;
      then f2.(a+n+1) = f2.(a+n) +^ A.(a+n) by A5, ORDINAL1:21;
      then f2.(a+n) c= f2.(a+n+1) by ORDINAL3:24;
      hence thesis by A21, XBOOLE_1:1;
    end;
    for n being Nat holds Q[n] from NAT_1:sch 2(A17,A18);
    hence thesis by A2, A4, A14, A15, A16;
  end;
end;
