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

theorem
  n <= k & k < n + n implies k div n = 1
proof
  assume that
A1: n <= k and
A2: k < n + n;
A3: k = n + (k - n) .= n * 1 + (k-'n) by A1,XREAL_1:233;
  k - n < n + n - n by A2,XREAL_1:9;
  hence thesis by A3,NAT_D:def 1;
end;
