
theorem
  for a being Ordinal, n, m being Nat
  holds n*^exp(omega,a) (+) m*^exp(omega,a) = (n+m)*^exp(omega,a)
proof
  let a be Ordinal, n, m be Nat;
  per cases;
  suppose A1: n = 0;
    hence n*^exp(omega,a) (+) m*^exp(omega,a)
       = 0 (+) m*^exp(omega,a) by ORDINAL2:35
      .= (n+m)*^exp(omega,a) by A1, Th82;
  end;
  suppose A2: n <> 0;
    then A3: 0 in n & n in omega by XBOOLE_1:61, ORDINAL1:11, ORDINAL1:def 12;
    omega -exponent last CantorNF (n*^exp(omega,a))
       = omega -exponent last({}^<% n*^exp(omega,a) %>) by A2, Th69
      .= omega -exponent(n*^exp(omega,a)) by AFINSQ_1:92
      .= a by A3, ORDINAL5:58;
    hence n*^exp(omega,a) (+) m*^exp(omega,a)
       = n*^exp(omega,a) +^ m*^exp(omega,a) by Th86
      .= (n+^m)*^exp(omega,a) by ORDINAL3:46
      .= (n+m)*^exp(omega,a) by CARD_2:36;
  end;
end;
