
theorem Th121:
  for a,b being Nat st a < b & b <> 0
  holds (2 * a) div b < 2
proof
  let a,b be Nat;
  assume A1: a < b;
  assume A2: b <> 0;
  per cases;
  suppose 2 * a < b;
    then (2 * a) div b = 0 by NAT_D:27;
    hence thesis;
  end;
  suppose 2 * a >= b;
    then 2 * a - b >= b - b by XREAL_1:13;
    then 2 * a - b >= 0;
    then 2 * a - b in NAT by INT_1:3;
    then reconsider ab=2 * a - b as Nat;
    A4: b div b = (b * 1) div b
               .= 1 by A2, NAT_D:18;
    A5: b mod b = (b * 1) mod b
               .= 0 by NAT_D:13;
    2 * a < 2 * b by A1, XREAL_1:68;
    then (2 * a) - b < (2 * b) - b by XREAL_1:9;
    then A6: ab div b = 0 by NAT_D:27;
    (2 * a) div b = (ab + b) div b
    .= (ab div b) + (b div b) by A5, NAT_D:19
    .= 1 by A4,A6;
    hence thesis;
  end;
end;
