
theorem Th6:
  for n being Nat st n > 1 holds (2 * n -' 2)! * n * (n
  + 1) < (2 * n)!
proof
  let n be Nat;
  assume
A1: n > 1;
  then
A2: 2 * 1 < 2 * n by XREAL_1:68;
  then
A3: 2 * n -' 1 + 1 = 2 * n by XREAL_1:235,XXREAL_0:2;
  2 - 1 < 2 * n - 1 by A2,XREAL_1:9;
  then
A4: 1 < 2 * n -' 1 by A2,XREAL_1:233,XXREAL_0:2;
  2 * n -' 2 + 1 = 2 * n -' 1 -' 1 + 1 by NAT_D:45
    .= 2 * n -' 1 by A4,XREAL_1:235;
  then
A5: (2 * n -' 2)! * (2 * n -' 1) * (2 * n) = (2 * n -' 1)! * (2 * n) by
NEWTON:15
    .= (2 * n)! by A3,NEWTON:15;
  (2 * n -' 2)! > 0 by NEWTON:17;
  then
A6: (2 * n -' 2)! * n > 0 * n & (2 * n -' 2)! * n < (2 * n -' 2)! * (2 * n
  -' 1) by A1,Th3,XREAL_1:68;
  n + 1 < n + n by A1,XREAL_1:6;
  hence thesis by A5,A6,XREAL_1:98;
end;
