reserve i, j, k, l, m, n, t for Nat;

theorem
  k <> 0 implies (n+k) div k = (n div k) + 1
proof
  assume k <> 0;
  then consider t be Nat such that
A1: n = k * (n div k) + t and
A2: t < k by NAT_D:def 1;
  n + k = k * ((n div k) + 1) + t by A1;
  hence thesis by A2,NAT_D:def 1;
end;
