reserve N,M,K for ExtNat;

theorem
  N <= M implies ex K st M = N + K
proof
  assume A1: N <= M;
  per cases by Th3;
  suppose M is Nat;
    then reconsider m = M as Nat;
    m in NAT by ORDINAL1:def 12;
    then N in NAT by A1, Th5;
    then reconsider n = N as Nat;
    consider k being Nat such that
      A2: m = n + k by A1, NAT_1:10;
    reconsider K = k as ExtNat;
    take K;
    m = (n qua ExtNat) + k by A2;
    hence thesis;
  end;
  suppose A3: M = +infty;
    take M;
    thus thesis by A3, XXREAL_3:def 2;
  end;
end;
